Overview
In earlier classes, you have studied measures of central tendency such as mean, mode, median of ungrouped and grouped data. In addition to these measures, we often need to calculate a second type of measure called a measure of dispersion which measures the variation in the observations about the middle value- mean or median etc.
This chapter is concerned with some important measures of dispersion such as mean deviation, variance, standard deviation etc., and finally analysis of frequency distributions.
Measures of dispersion
         Mean deviation about their median M is given by
         \(\text { M.D }( M )=\frac{\left|x_i- M \right|}{n} \dots(2)\)
Coefficient of variation:
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Let \(a , b , c \in N\) and \(a < b < c\). Let the mean, the mean deviation about the mean and the variance of the 5 observations \(9,25, a, b, c\) be 18,4 and \(\frac{136}{5}\), respectively. Then \(2 a+b-c\) is equal to [JEE Main 2024 (Online) 8th April Evening Shift]
\(
\begin{aligned}
&\begin{aligned}
& a, b, c \in N \quad a<b<c \\
& \overline{ x }=\text { mean }=\frac{9+25+ a + b + c }{5}=18 \\
& a+b+c=56
\end{aligned}\\
&\text { Mean deviation }=\frac{\Sigma\left| x _{ i }-\overline{ x }\right|}{ n }=4
\end{aligned}
\)
\(
\begin{aligned}
& =9+7+|18- a |+|18- b |+|18- c |=20 \\
& =|18- a |+|18- b |+|18- c |=4 \\
& \text { Variance }=\frac{\Sigma\left| x _{ i }-\overline{ x }\right|^2}{ n }=\frac{136}{5} \\
& =81+49+|18-a|^2+|18-b|^2+|18-c|^2=136 \\
& =(18-a)^2+(18-b)^2+(18-c)^2=6
\end{aligned}
\)
Possible values \((18- a )^2=1, \quad(18- b )^2=1 \quad(18- c )^2=4; a < b < c\) \(18-a=1 \quad 18-b=-1 \quad 18-c=-2\)
\(
a =17 \quad b=19 \quad c =20
\)
\(
\begin{aligned}
& 2 a+b-c \\
& 34+19-20 \\
& =33
\end{aligned}
\)
Let the mean and the standard deviation of the probability distribution
\(
\begin{array}{|c|c|c|c|c|}
\hline X & \alpha & 1 & 0 & -3 \\
\hline P ( X ) & \frac{1}{3} & K & \frac{1}{6} & \frac{1}{4} \\
\hline
\end{array}
\)
be \(\mu\) and \(\sigma\), respectively. If \(\sigma-\mu=2\), then \(\sigma+\mu\) is equal to ____. [JEE Main 2024 (Online) 5th April Evening Shift]
Mean \((\mu)=\Sigma x_i P\left(x_i\right)\)
Standard deviation \((\sigma)=\sqrt{\left(\Sigma x_i^2 P\left(x_i\right)\right)-\mu^2}\)
\(
\begin{aligned}
& \Rightarrow \quad \mu=\frac{1}{3} \alpha+K-\frac{3}{4} \\
& \sigma=\sqrt{\left(\frac{1}{3} \alpha^2+K+0+\frac{9}{4}\right)-\left(\frac{1}{3} \alpha+K-\frac{3}{4}\right)^2} \\
& \because \Sigma P_i=1 \Rightarrow \frac{1}{3}+K+\frac{1}{6}+\frac{1}{4}=1 \\
& \Rightarrow \quad K=\frac{1}{4} \Rightarrow \mu=\frac{1}{3} \alpha-\frac{1}{2} \\
& \because \sigma-\mu=2 \\
& \sigma^2=(\mu+2)^2 \\
& \frac{1}{3} \alpha^2+\frac{5}{2}-\mu^2=(\mu+2)^2 \\
& \frac{1}{3} \alpha^2+\frac{5}{2}=\left(\frac{1}{3} \alpha-\frac{1}{2}\right)^2+\left(\frac{1}{3} \alpha+\frac{3}{2}\right)^2 \\
& \Rightarrow \alpha=0,6
\end{aligned}
\)
If \(\alpha=0, K=\frac{1}{4} \quad\) If \(\alpha=6, K=\frac{1}{4}\)
\(
\begin{array}{ll}
\mu=-\frac{1}{2}, \sigma=\frac{3}{2} & \mu=\frac{3}{2}, \sigma=\frac{7}{2} \\
\sigma+\mu=1 & \sigma+\mu=5
\end{array}
\)
The variance \(\sigma^2\) of the data
\(
\begin{array}{|l|l|l|l|l|l|l|l|}
\hline x_i & 0 & 1 & 5 & 6 & 10 & 12 & 17 \\
\hline f_i & 3 & 2 & 3 & 2 & 6 & 3 & 3 \\
\hline
\end{array}
\)
is ____. [JEE Main 2024 (Online) 30th January Evening Shift]
\(
\begin{array}{|c|c|c|c|}
\hline x _{ i } & f _{ i } & f _{ i } x _{ i } & f _{ i } x _{ i }{ }^2 \\
\hline 0 & 3 & 0 & 0 \\
\hline 1 & 2 & 2 & 2 \\
\hline 5 & 3 & 15 & 75 \\
\hline 6 & 2 & 12 & 72 \\
\hline 10 & 6 & 60 & 600 \\
\hline 12 & 3 & 36 & 432 \\
\hline 17 & 3 & 51 & 867 \\
\hline & \Sigma f _{ i }=22 & & \sum f _{ i } x _{ i }{ }^2=2048 \\
\hline
\end{array}
\)
\(
\begin{aligned}
& \therefore \quad \Sigma f _{ i } X _{ i }=176 \\
& \text { So } \overline{ x }=\frac{\sum f _{ i } x _{ i }}{\sum f _{ i }}=\frac{176}{22}=8
\end{aligned}
\)
\(
\begin{aligned}
& \text { for } \sigma^2=\frac{1}{N} \sum f _{ i } x _{ i }{ }^2-(\overline{ x })^2 \\
& =\frac{1}{22} \times 2048-(8)^2 \\
& =93.090964 \\
& =29.0909
\end{aligned}
\)
If the mean and variance of the data \(65,68,58,44,48,45,60, \alpha, \beta, 60\) where \(\alpha>\beta\), are 56 and 66.2 respectively, then \(\alpha^2+\beta^2\) is equal to ____. [JEE Main 2024 (Online) 29th January Morning Shift]
\(
\begin{aligned}
& \overline{ x }=56 \\
& \sigma^2=66.2 \\
& \Rightarrow \frac{\alpha^2+\beta^2+25678}{10}-(56)^2=66.2 \\
& \therefore \alpha^2+\beta^2=6344
\end{aligned}
\)
The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking it was found that an observation was read as 10 in place of 12 . If \(\mu\) and \(\sigma^2\) denote the mean and variance of the correct observations respectively, then \(15\left(\mu+\mu^2+\sigma^2\right)\) is equal to _____. [JEE Main 2024 (Online) 27th January Evening Shift]
Let the incorrect mean be \(\mu^{\prime}\) and standard deviation be \(\sigma^{\prime}\)
We have
\(
\mu^{\prime}=\frac{\Sigma x _{ i }}{15}=12 \Rightarrow \Sigma x _{ i }=180
\)
As per given information correct \(\Sigma x _{ i }=180-10+12\)
\(
\Rightarrow \mu(\text { correct mean })=\frac{182}{15}
\)
Also
\(
\begin{aligned}
& \sigma^{\prime}=\sqrt{\frac{\Sigma x _{ i }^2}{15}-144}=3 \Rightarrow \Sigma x _{ i }^2=2295 \\
& \text { Correct } \Sigma x _{ i }^2=2295-100+144=2339 \\
& \sigma^2(\text { correct variance })=\frac{2339}{15}-\frac{182 \times 182}{15 \times 15}
\end{aligned}
\)
Required value
\(
\begin{aligned}
& =15\left(\mu+\mu^2+\sigma^2\right) \\
& =15\left(\frac{182}{15}+\frac{182 \times 182}{15 \times 15}+\frac{2339}{15}-\frac{182 \times 182}{15 \times 15}\right) \\
& =15\left(\frac{182}{15}+\frac{2339}{15}\right) \\
& =2521
\end{aligned}
\)
The mean and standard deviation of the marks of 10 students were found to be 50 and 12 respectively. Later, it was observed that two marks 20 and 25 were wrongly read as 45 and 50 respectively. Then the correct variance is ____. [JEE Main 2023 (Online) 13th April Evening Shift]
The initial mean is given by:
\(
\bar{x}=50
\)
So, the total sum of the marks initially was:
\(
\sum x_i=\bar{x} \times n=50 \times 10=500
\)
We later realize that two marks were incorrectly read as 45 and 50 , when they should have been 20 and 25 . Therefore, the corrected sum of the marks is:
\(
\sum x_{i \text { correct }}=\sum x_i-45-50+20+25=500-45-50+20+25=450
\)
The initial variance is given as:
\(
\sigma^2=144
\)
We know that variance is calculated as the mean of the squares minus the square of the mean. Therefore, rearranging gives:
\(
\frac{\sum x_i^2}{n}=\sigma^2+\bar{x}^2=144+50^2=2644
\)
Then, the sum of the squares of the initial marks is:
\(
\sum x_i^2=n \times \frac{\sum x_i^2}{n}=10 \times 2594=26440
\)
The corrected sum of the squares of the marks is calculated by subtracting the squares of the incorrect marks and adding the squares of the correct marks:
\(
\sum x_{i \text { correct }}^2=\sum x_i^2-45^2-50^2+20^2+25^2=26400-45^2-50^2+20^2+25^2=22940
\)
Now we can calculate the corrected variance. The variance is the mean of the squares minus the square of the mean. Using the corrected values gives:
\(
\sigma_{\text {correct }}^2=\frac{\sum x_{\text {ivorrect }}^2}{n}-\left(\frac{\sum x i_{\text {correct }}}{n}\right)^2=\frac{22940}{10}-\left(\frac{450}{10}\right)^2=2294-45^2=2294-2025=269
\)
Therefore, the correct variance is 269.
Let the mean of the data
\(
\begin{array}{|l|c|c|c|c|c|}
\hline x & 1 & 3 & 5 & 7 & 9 \\
\hline \text { Frequency }(f) & 4 & 24 & 28 & \alpha & 8 \\
\hline
\end{array}
\)
be 5 . If \(m\) and \(\sigma^2\) are respectively the mean deviation about the mean and the variance of the data, then \(\frac{3 \alpha}{m+\sigma^2}\) is equal to _____. [JEE Main 2023 (Online) 13th April Morning Shift]
\(
\begin{aligned}
& 5=\bar{x}=\frac{\sum x_i f_i}{\sum f_i}=\frac{4+72+140+7 \alpha+72}{64+\alpha} \\
& \Rightarrow 320+5 \alpha=288+7 \alpha \Rightarrow 2 \alpha=32 \Rightarrow \alpha=16
\end{aligned}
\)
\(
\begin{aligned}
\sum f_i & =80 \\
\text { M.D } & =\frac{\sum f_i\left|x_i-5\right|}{\sum f_i} \\
& =\frac{4+4+24 \times 2+0+16 \times 2+8 \times 4}{80} \\
& =\frac{8}{5}
\end{aligned}
\)
\(
\begin{aligned}
\sigma^2 & =\frac{\sum f_i\left|x_i-5\right|}{\sum f_i} \\
& =\frac{4+16+24 \times 4+0+16 \times 4+8 \times 16}{80}=\frac{22}{5}
\end{aligned}
\)
So, \(\frac{3 \alpha}{m+\sigma^2}=\frac{3 \times 16}{\frac{8}{5}+\frac{22}{5}}=8\)
Let the positive numbers \(a_1, a_2, a_3, a_4\) and \(a_5\) be in a G.P. Let their mean and variance be \(\frac{31}{10}\) and \(\frac{m}{n}\) respectively, where \(m\) and \(n\) are co-prime. If the mean of their reciprocals is \(\frac{31}{40}\) and \(a_3+a_4+a_5=14\), then \(m+n\) is equal to ______. [JEE Main 2023 (Online) 12th April Morning Shift]
5 positive numbers \(a_1, a_2, \ldots, a_5\) are in geometric progression. So, let the numbers be \(\frac{a}{r^2}, \frac{a}{r}, a, a r, a r^2\) According to question,
\(
\frac{\frac{a}{r^2}+\frac{a}{r}+a+a r+a r^2}{5}=\frac{31}{10} \dots(1)
\)
\(
\begin{aligned}
& \frac{\frac{r^2}{a}+\frac{r}{a}+\frac{1}{a}+\frac{1}{a r}+\frac{1}{a r^2}}{5}=\frac{31}{40} \\
& \Rightarrow \frac{\frac{1}{a}\left(r^2+r+1+\frac{1}{r}+\frac{1}{r^2}\right)}{5}=\frac{31}{40} \quad \ldots(2)
\end{aligned}
\)
By (1) & (2), we get
\(
a^2=4 \Rightarrow a=2
\)
So,
\(
\begin{aligned}
& \left(r^2+r+1+\frac{1}{r}+\frac{1}{r^2}\right)=\frac{31}{4} \\
& \Rightarrow\left(r+\frac{1}{r}\right)^2+r+\frac{1}{r}=\frac{27}{4}+2 \\
& \Rightarrow t^2+t=\frac{35}{4}
\end{aligned}
\)
where, \(t=r+\frac{1}{r}\)
\(
\begin{aligned}
& 4 t^2+4 t-35=0 \\
& \Rightarrow t=\frac{5}{2} \Rightarrow r=2
\end{aligned}
\)
So, numbers are \(\frac{1}{2}, 1,2,4,8\)
Variance is
\(
\begin{aligned}
& =\frac{\frac{1}{4}+1+4+16+64}{5}-\left(\frac{\frac{1}{2}+1+2+4+8}{5}\right)^2 \\
& =\frac{186}{25}=\frac{m}{n} \\
& m+n=211
\end{aligned}
\)
Hence this is the required option.
If the mean of the frequency distribution
\(
\begin{array}{|l|c|c|c|c|c|}
\hline \text { Class: } & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\
\hline \text { Frequency: } & 2 & 3 & x & 5 & 4 \\
\hline
\end{array}
\)
is 28, then its variance is __________. [JEE Main 2023 (Online) 10th April Morning Shift]
Given mean is 28
So, \(\frac{2 \times 5+3 \times 15+x \times 25+5 \times 35+4 \times 45}{14+x}=28\)
\(
\begin{aligned}
& \Rightarrow \frac{10+45+25 x+175+180}{14+x}=28 \\
& \Rightarrow 310+25 x=392+28 x \\
& \Rightarrow 3 x=18 \Rightarrow x=6
\end{aligned}
\)
\(
\begin{aligned}
& \therefore \text { Variance }=\left(\frac{\sum x_i^2 f_i}{\sum f_i}\right)-(\text { mean })^2 \\
& =\left(\frac{2 \times 5^2+3 \times 15^2+6 \times 25^2+5 \times 35^2+4 \times 45^2}{20}\right)-(28)^2 \\
& =\left(\frac{50+675+3750+6125+8100}{20}\right)-(28)^2 \\
& =\left(\frac{18700}{20}\right)-(28)^2 \\
& =935-784=151
\end{aligned}
\)
Let the mean and variance of 8 numbers \(x, y, 10,12,6,12,4,8\) be 9 and 9.25 respectively. If \(x>y\), then \(3 x-2 y\) is equal to ____. [JEE Main 2023 (Online) 8th April Morning Shift]
Concept :
(a) Mean \(=\frac{\Sigma x_i}{n}\)
(b) Variance \(=\frac{\Sigma\left(x_i-\bar{x}\right)^2}{n}\)
\(
\begin{array}{|c|c|c|}
\hline x_i & \left(x_i-\bar{x}\right) & \left(x_i-\bar{x}\right)^2 \\
\hline x & x-9 & (x-9)^2 \\
\hline y & y-9 & (y-9)^2 \\
\hline 10 & 1 & 1 \\
\hline 12 & 3 & 9 \\
\hline 6 & -3 & 9 \\
\hline 12 & 3 & 9 \\
\hline 4 & -5 & 25 \\
\hline 8 & -1 & 1 \\
\hline x+y+92 & & (x-9)^2+(y-9)^2+54 \\
\hline
\end{array}
\)
Now, mean \((\bar{x})=9\)
\(
\begin{aligned}
& \Rightarrow \frac{x+y+52}{8}=9 \\
& \Rightarrow x+y=20
\end{aligned}
\)
Also, variance \(=9.25\)
\(
\begin{aligned}
& \Rightarrow \frac{(x-9)^2+(y-9)^2+54}{8}=9.25 \\
& \Rightarrow x^2+y^2+81+81-2 \times 9(x+y)=20 \\
& \Rightarrow x^2+y^2-18 \times 20=-142 \\
& \Rightarrow x^2+y^2=218 \\
& \Rightarrow x^2+(20-x)^2=218 \\
& \Rightarrow x^2+400+x^2-40 x=218 \\
& \Rightarrow 2 x^2-40 x+182=0
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow x=\frac{40 \pm 12}{4} \\
& \Rightarrow x=13 \text { or } x=7 \Rightarrow y=7 \text { or } y=13
\end{aligned}
\)
But \(x>y\)
\(\therefore x=13\) and \(y=7\)
So, \(3 x-2 y=39-14=25\)
If the mean and variance of the frequency distribution
\(
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline x_i & 2 & 4 & 6 & 8 & 10 & 12 & 14 & 16 \\
\hline f_i & 4 & 4 & \alpha & 15 & 8 & \beta & 4 & 5 \\
\hline
\end{array}
\)
\(
\text { are } 9 \text { and } 15.08 \text { respectively, then the value of } \alpha^2+\beta^2-\alpha \beta \text { is }
\) _____. [JEE Main 2023 (Online) 6th April Evening Shift]
\(
\begin{array}{lllll}
\hline x_i & f_i & x_i^2 & f_i x_i & f_i x_i^2 \\
\hline 2 & 4 & 4 & 8 & 16 \\
\hline 4 & 4 & 16 & 16 & 64 \\
\hline 6 & \alpha & 36 & 6 \alpha & 36 \alpha \\
\hline 8 & 15 & 64 & 120 & 960 \\
\hline 10 & 8 & 100 & 80 & 800 \\
\hline 12 & \beta & 144 & 12 \beta & 144 \beta \\
\hline 14 & 4 & 196 & 56 & 784 \\
\hline 16 & 5 & 256 & 80 & 1280
\end{array}
\)
\(
\begin{aligned}
& \therefore \Sigma f_i=40+\alpha+\beta \\
& \Sigma f_i x_i=360+6 \alpha+12 \beta \\
& \Sigma f_i x_i^2=3904+36 \alpha+144 \beta
\end{aligned}
\)
Given, mean \(=9\)
\(
\begin{aligned}
& \Rightarrow \frac{\Sigma f_i x_i}{\Sigma f_i}=9 \\
& \Rightarrow \frac{360+6 \alpha+12 \beta}{40+\alpha+\beta}=9 \\
& \Rightarrow 360+6 \alpha+12 \beta=9(40+\alpha+\beta) \\
& \Rightarrow 3 \beta=3 \alpha
\end{aligned}
\)
\(
\Rightarrow \alpha=\beta \ldots \ldots \ldots(i)
\)
\(
\begin{aligned}
& \text { Variance }=15.08 \\
& \Rightarrow \frac{\Sigma f_i x_i^2}{\Sigma f_i}-\left(\frac{\Sigma f_i x_i}{\Sigma f_i}\right)^2=15.08 \\
& \Rightarrow \frac{3904+36 \alpha+144 \beta}{40+\alpha+\beta}-(9)^2=15.08
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \frac{3904+180 \alpha}{40+2 \alpha}=81+15.08 \quad[\because \alpha=\beta] \\
& \Rightarrow 3904+180 \alpha=96.08(40+2 \alpha) \\
& \Rightarrow 3904+180 \alpha=3843.2+192.16 \alpha \\
& \Rightarrow 60.8=12.16 \alpha \\
& \Rightarrow \alpha=5=\beta \\
& \therefore \alpha^2+\beta^2-\alpha \beta=25+25-25=25
\end{aligned}
\)
If the variance of the frequency distribution
\(
\begin{array}{|l|l|l|l|l|l|l|l|}
\hline x_i & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline \text { Frequency } f_i & 3 & 6 & 16 & \alpha & 9 & 5 & 6 \\
\hline
\end{array}
\)
is 3 , then \(\alpha\) is equal to ____. [JEE Main 2023 (Online) 31st January Morning Shift]
\(
\begin{array}{|c|c|c|c|c|}
\hline x _{ i } & f _{ i } & \begin{array}{l}
\begin{array}{c}
d_i= \\
x_i-5
\end{array}
\end{array} & f _{ i } d _{ i }^2 & f _{ i } d _{ i } \\
\hline 2 & 3 & -3 & 27 & -9 \\
\hline 3 & 6 & -2 & 24 & -12 \\
\hline 4 & 16 & -1 & 16 & -16 \\
\hline 5 & \alpha & 0 & 0 & 0 \\
\hline 6 & 9 & 1 & 9 & 9 \\
\hline 7 & 5 & 2 & 20 & 10 \\
\hline 8 & 6 & 3 & 54 & 18 \\
\hline
\end{array}
\)
\(
\begin{aligned}
& \sigma_{ x }^2=\sigma_{ d }^2=\frac{\sum f _{ i }^2}{\sum f _{ i }}-\left(\frac{\sum f _{ i } d _{ i }}{\sum f _{ i }}\right)^2 \\
& =\frac{150}{45+\alpha}-0=3 \\
& \Rightarrow 150=135+3 \alpha \\
& \Rightarrow 3 \alpha=15 \Rightarrow \alpha=5
\end{aligned}
\)
The mean and variance of 7 observations are 8 and 16 respectively. If one observation 14 is omitted and a and b are respectively mean and variance of remaining 6 observation, then \(a +3 b-5\) is equal to ____. [JEE Main 2023 (Online) 30th January Morning Shift]
\(
\begin{aligned}
& \sum x_i=7 \times 8=56 \\
& \frac{\sum x_i^2}{n}-\left(\frac{\sum x_i}{n}\right)^2=16 \\
& \frac{\sum x_i^2}{7}-64=16 \\
& \sum x_i^2=560
\end{aligned}
\)
when 14 is omitted
\(
\sum x_i=56-14=42
\)
New mean \(=a=\frac{\sum x_i}{6}=7\)
\(
\sum x_i^2=560-196=364
\)
new variance, \(b=\frac{\sum x_i^2}{6}-\left(\frac{\sum x_i}{6}\right)^2\)
\(
\begin{aligned}
& =\frac{364}{6}-49=\frac{35}{3} \\
& 3 b=35
\end{aligned}
\)
\(
a+3 b-5=7+35-5=37
\)
Let \(X=\{11,12,13, \ldots, 40,41\}\) and \(Y=\{61,62,63, \ldots, 90,91\}\) be the two sets of observations. If \(\bar{x}\) and \(\bar{y}\) are their respective means and \(\sigma^2\) is the variance of all the observations in \(X \cup Y\), then \(\left|\bar{x}+\bar{y}-\sigma^2\right|\) is equal to _____. [JEE Main 2023 (Online) 29th January Evening Shift]
\(
\bar{x}=\frac{\sum_{i=11}^{41} i }{31}=\frac{11+41}{2}=26 \quad(31 \text { elements })
\)
\(
\bar{y}=\frac{\sum_{j=61}^{91} j}{31}=\frac{61+91}{2}=76 \quad(31 \text { elements })
\)
Combined mean, \(\mu=\frac{31 \times 26+31 \times 76}{31+31}\)
\(
\begin{aligned}
& =\frac{26+76}{2}=51 \\
& \sigma^2=\frac{1}{62} \times\left(\sum_{i=1}^{31}\left(x_i-\mu\right)^2+\sum_{i=1}^{31}\left(y_i-\mu\right)^2\right)=705
\end{aligned}
\)
Since, \(x_i \in X\) are in A.P. with 31 elements & common difference 1 , same is \(y_i \in y\), when written in increasing order.
\(
\begin{aligned}
& \therefore \sum_{i=1}^{31}\left(x_i-\mu\right)^2=\sum_{i=1}^{31}\left(y_i-\mu\right)^2 \\
& =10^2+11^2+\ldots .+40^2 \\
& =\frac{40 \times 41 \times 81}{6}-\frac{9 \times 10 \times 19}{6}=21855 \\
& \therefore\left|\bar{x}+\bar{y}-\sigma^2\right|=|26+76-705|=603
\end{aligned}
\)
Let the mean and the variance of 20 observations \(x_1, x_2, \ldots, x_{20}\) be 15 and 9 , respectively. For \(\alpha \in R\), if the mean of \(\left(x_1+\alpha\right)^2,\left(x_2+\alpha\right)^2, \ldots,\left(x_{20}+\alpha\right)^2\) is 178 , then the square of the maximum value of \(\alpha\) is equal to ______. [JEE Main 2022 (Online) 29th July Morning Shift]
Given \(\sum_{\frac{i=1}{20}}^{20} x_i=15 \Rightarrow \sum_{i=1}^{20} x_i=300\)
and \(\sum_{\frac{i=1}{20}}^{20} x_i^2-(\bar{x})^2=9 \Rightarrow \sum_{i=1}^{20} x_i^2=4680\)
\(
\begin{aligned}
& \text { Mean }=\frac{\left(x_i+\alpha\right)^2+\left(x_2+\alpha\right)^2+\ldots .+\left(x_{20}+\alpha\right)^2}{20}=178 \\
& \Rightarrow \frac{\sum_{i=1}^{20} x_i^2+2 \alpha \sum_{i=1}^{20} x_i+20 \alpha^2}{20}=178
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow 4680+600 \alpha+20 \alpha^2=3560 \\
& \Rightarrow \alpha^2+30 \alpha+56=0 \\
& \Rightarrow \alpha^2+28 \alpha+2 \alpha+56=0 \\
& \Rightarrow(\alpha+28)(\alpha+2)=0 \\
& \alpha_{\max }=-2 \Rightarrow \alpha_{\max }^2=4 .
\end{aligned}
\)
Let \(x_1, x_2, x_3, \ldots, x_{20}\) be in geometric progression with \(x_1=3\) and the common ratio \(\frac{1}{2}\). A new data is constructed replacing each \(x_i\) by \(\left(x_i-i\right)^2\). If \(\bar{x}\) is the mean of new data, then the greatest integer less than or equal to \(\bar{x}\) is _____. [JEE Main 2022 (Online) 28th July Morning Shift]
\(
\begin{aligned}
& \sum x _0^1=\frac{3\left(1-\left(\frac{1}{2}\right)\right)^{20}}{1 \frac{-1}{2}}=6\left(1-\frac{1}{2^{20}}\right) \\
& =\sum_{ i =1}^{20}\left( x _{ i } – i \right)^2 \\
& =\sum_{ i =1}^{20}\left( x _{ i }\right)^2+( i )^2-2 x _{ i } i \\
& \text { Now }=\sum_{ i =1}^{20}\left( x _{ i }\right)^2=\frac{9\left(1-\left(\frac{1}{4}\right)\right)^{20}}{1-\frac{1}{4}}=12\left(1-\frac{1}{2^{40}}\right)
\end{aligned}
\)
\(
\sum_{ i =1}^{20} i ^2=\frac{1}{6} \times 20 \times 21 \times 41=2870
\)
\(
\begin{aligned}
& \sum_{i=1}^{20} x_i i=s=3+2.3 \frac{1}{2}+3.3 \frac{1}{2^2}+4.3 \frac{1}{2^3}+\ldots \ldots . A G P \\
& =6\left(2-\frac{22}{2^{20}}\right) \\
& \bar{x}=\frac{12-\frac{12}{2^{40}}+2870-12\left(2-\frac{22}{2^{20}}\right)}{20} \\
& \bar{x}=\frac{2858}{20}+\left(\frac{-12}{2^{40}}+\frac{22}{2^{20}}\right) \times \frac{1}{20} \\
& {[\bar{x}]=142}
\end{aligned}
\)
The mean and variance of 10 observations were calculated as 15 and 15 respectively by a student who took by mistake 25 instead of 15 for one observation. Then, the correct standard deviation is _____. [JEE Main 2022 (Online) 27th July Morning Shift]
\(
\begin{aligned}
& \text { Given } \frac{\sum_{i=1}^{10} x_i}{10}=15 \ldots . . \text { (1) } \\
& \Rightarrow \sum_{i=1}^{10} x_i=150 \\
& \text { and } \frac{\sum_{i=1}^{10} x_i^2}{10}-15^2=15 \\
& \Rightarrow \sum_{i=1}^{10} x_i^2=2400
\end{aligned}
\)
Replacing 25 by 15 we get
\(
\begin{aligned}
& \sum_{i=1}^9 x_i+25=150 \\
& \Rightarrow \sum_{i=1}^9 x_i=125
\end{aligned}
\)
\(
\therefore \text { Correct mean }=\frac{\sum_{i=1}^9 x_i+15}{10}=\frac{125+15}{10}=14
\)
Similarly, \(\sum_{i=1}^2 x_i^2=2400-25^2=1775\)
\(
\begin{aligned}
& \therefore \text { Correct variance }=\frac{\sum_{i=1}^9 x_i^2+15^2}{10}-14^2 \\
& =\frac{1775+225}{10}-14^2=4
\end{aligned}
\)
\(\therefore\) Correct \(S . D=\sqrt{4}=2\).
The mean and standard deviation of 40 observations are 30 and 5 respectively. It was noticed that two of these observations 12 and 10 were wrongly recorded. If \(\sigma\) is the standard deviation of the data after omitting the two wrong observations from the data, then \(38 \sigma^2\) is equal to _____. [JEE Main 2022 (Online) 26th July Evening Shift]
\(
\begin{aligned}
& \mu=\frac{\sum x_i}{40}=30 \Rightarrow \sum x_i=1200 \\
& \sigma^2=\frac{\sum x_i^2}{40}-(30)^2=25 \Rightarrow \sum x_i^2=37000
\end{aligned}
\)
After omitting two wrong observations
\(
\begin{aligned}
& \sum y_i=1200-12-10=1178 \\
& \sum y_i^2=37000-144-100=36756 \\
& \text { Now } \sigma^2=\frac{\sum y_i^2}{38}-\left(\frac{\sum y_i}{38}\right)^2 \\
& =\frac{36756}{38}-\left(\frac{1178}{38}\right)^2=-31^2 \\
& =38 \sigma^2=36756-36518=238
\end{aligned}
\)
Suppose a class has 7 students. The average marks of these students in the mathematics examination is 62 , and their variance is 20 . A student fails in the examination if he/she gets less than 50 marks, then in worst case, the number of students can fail is _____. [JEE Main 2022 (Online) 28th June Evening Shift]
According to given data
\(
\text { Variance } = \frac{\sum_{i=1}^7\left(x_i-62\right)^2}{7}=20
\)
\(
\Rightarrow \sum_{i=1}^7\left(x_i-62\right)^2=140
\)
So for any \(x _{ i },\left(x_i-62\right)^2 \leq 140\)
\(
\Rightarrow x_i>50 \forall i=1,2,3, \ldots .7
\)
So no student is going to score less than 50 .
Note:
\(
\Rightarrow x_1-62^2+x_2-62^2+\ldots .+x_7-62^2=140
\)
If \(x _1=49\)
\(
|49-62|^2=169
\)
then, \(x _2-62^2+\ldots .+ x _7-62^2=\) Negative Number which is not possible, therefore, no student can fail.
The mean and standard deviation of 15 observations are found to be 8 and 3 respectively. On rechecking it was found that, in the observations, 20 was misread as 5. Then, the correct variance is equal to _____. [JEE Main 2022 (Online) 28th June Morning Shift]
We have Variance \(=\) \(\frac{\sum_{r=1}^{15} x_r^2}{15}-\left(\frac{\sum_{r=1}^{15} x_r}{15}\right)^2\)
Now, as per information given in equation
\(
\frac{\sum x _{ r }^2}{15}-8^2=3^2 \Rightarrow \sum x _{ r }^2=\log 5
\)
Now, the new \(\sum x _{ r }^2=\log 5-5^2+20^2=1470\)
And, new \(\sum x_r=(15 \times 8)-5+(20)=135\)
\(\therefore\) Variance \(=\frac{1470}{15}-\left(\frac{135}{15}\right)^2=98-81=17\)
Alternate:
\(
\frac{\sum x_i^2}{15}-8^2=9 \Rightarrow \sum x_i^2=15 \times 73=1095
\)
Let \(\bar{x}_c\) be corrected mean \(\bar{x}_c=9\)
\(
\sum x_c^2=1095-25+400=1470
\)
Correct variance \(=\frac{1470}{15}-(9)^2=98-81=17\)
If the mean deviation about the mean of the numbers \(1,2,3\), \(\ldots\) n , where n is odd, is \(\frac{5(n+1)}{n}\), then \(\)n\(\) is equal to ______. [JEE Main 2022 (Online) 25th June Evening Shift]
\(
\begin{aligned}
& \text { Mean }=\frac{n \frac{(n+1)}{2}}{n}=\frac{n+1}{2} \\
& \text { M.D. }=\frac{2\left(\frac{n-1}{2}+\frac{n-3}{2}+\frac{n-5}{2}+\ldots 0\right)}{n}=\frac{5(n+1)}{n} \\
& \Rightarrow((n-1)+(n-3)+(n-5)+\ldots 0)=5(n+1) \\
& \Rightarrow\left(\frac{n+1}{4}\right) \cdot(n-1)=5(n+1)
\end{aligned}
\)
So, \(n=21\)
The mean of 10 numbers \(7 \times 8,10 \times 10,13 \times 12,16 \times 14, \ldots \ldots .\). is _____. [JEE Main 2021 (Online) 31st August Morning Shift]
\(
\begin{aligned}
& 7 \times 8,10 \times 10,13 \times 12,16 \times 14 \ldots \ldots \\
& T_n=(3 n+4)(2 n+6)=2(3 n+4)(n+3) \\
& =2\left(3 n^2+13 n+12\right)=6 n^2+26 n+24
\end{aligned}
\)
\(
\begin{aligned}
& S _{10}=\sum_{n=1}^{10} T_n=6 \sum_{n=1}^{10} n^2+26 \sum_{n=1}^{10} n+24 \sum_{n=1}^{10} 1 \\
& =\frac{6(10 \times 11 \times 21)}{6}+26 \times \frac{10 \times 11}{2}+24 \times 10 \\
& =10 \times 11(21+13)+240 \\
& =3980 \\
& \text { Mean }=\frac{S_{10}}{10}=\frac{3980}{10}=398
\end{aligned}
\)
An online exam is attempted by 50 candidates out of which 20 are boys. The average marks obtained by boys is 12 with a variance 2 . The variance of marks obtained by 30 girls is also 2 . The average marks of all 50 candidates is 15 . If \(\mu\) is the average marks of girls and \(\sigma^2\) is the variance of marks of 50 candidates, then \(\mu\) \(+\sigma^2\) is equal to _____. [JEE Main 2021 (Online) 27th August Evening Shift]
\(
\begin{aligned}
& \sigma_b^2=2 \text { (variance of boys) } n_1=\text { no. of boys } \\
& \bar{x}_b=12 \quad n _2=\text { no. of girls } \\
& \sigma_{ g }^2=2 \\
& \bar{x}_g=\frac{50 \times 15-12 \times \sigma_b}{30}=\frac{750-12 \times 20}{30}=17=\mu
\end{aligned}
\)
variance of combined series
\(
\begin{aligned}
& \sigma^2=\frac{ n _1 \sigma_{ b }^2+ n _2 \sigma_{ g }^2}{ n _1+ n _2}+\frac{ n _1 \cdot n _2}{\left( n _1+ n _2\right)^2}\left(\overline{ x }_{ b }-\overline{ x }_{ g }\right)^2 \\
& \sigma^2=\frac{20 \times 2+30 \times 2}{20+30}+\frac{20 \times 30}{(20+30)^2}(12-17)^2 \\
& \sigma^2=8 \\
& \Rightarrow \mu+\sigma^2=17+8=25
\end{aligned}
\)
Let \(n\) be an odd natural number such that the variance of \(1,2,3,4, \ldots . ., n\) is 14 . Then \(n\) is equal to _____. [JEE Main 2021 (Online) 27th August Morning Shift]
\(
\begin{aligned}
& \text { Given: variance }=14 \\
& \Rightarrow \frac{\sum_{i=1}^n x_i^2}{n}-(\bar{x})^2=14 \\
& \Rightarrow \frac{1^2+2^2+\ldots+n^2}{n}-\left(\frac{1+2+3+\ldots+n}{n}\right)^2=14 \\
& \Rightarrow \frac{n(n+1)(2 n+1)}{6}-\left(\frac{\left(\frac{n(n+1)}{2}\right)}{n}\right)^2=14 \\
& \Rightarrow \frac{(n+1)(2 n+1)}{6}-\frac{(n+1)^2}{4}=14 \\
& \Rightarrow \frac{(n+1)}{2}\left(\frac{(2 n+1)}{3}-\frac{(n+1)}{2}\right)=14 \\
& \Rightarrow(n+1)(4 n+2-3 n-3)=168 \\
& \Rightarrow n^2-1=168 \\
& \therefore n=13
\end{aligned}
\)
Alternate:
\(
\frac{n^2-1}{12}=14 \Rightarrow n=13
\)
Let the mean and variance of four numbers \(3,7, x\) and \(y(x>y)\) be 5 and 10 respectively. Then the mean of four numbers \(3+2 x, 7+2 y, x+y\) and \(x-y\) is ____. [JEE Main 2021 (Online) 26th August Evening Shift]
\(
\begin{aligned}
& 5=\frac{3+7+x+y}{4} \Rightarrow x+y=10 \\
& \operatorname{Var}( x )=10=\frac{3^2+7^2+x^2+y^2}{4}-25 \\
& 140=49+9+x^2+y^2 \\
& x^2+y^2=82 \\
& x+y=10 \\
& \Rightarrow(x, y)=(9,1)
\end{aligned}
\)
Four numbers are \(21,9,10,8\)
\(
\text { Mean }=\frac{48}{4}=12
\)
Consider the following frequency distribution :
\(
\begin{array}{|l|c|c|c|c|c|}
\hline \text { Class: } & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 \\
\hline \text { Frequency : } & \alpha & 110 & 54 & 30 & \beta \\
\hline
\end{array}
\)
If the sum of all frequencies is 584 and median is 45 , then | \(\alpha-\beta\) | is equal to ____. [JEE Main 2021 (Online) 25th July Morning Shift]
\(\because\) Sum of frequencies \(=584\)
\(
\Rightarrow \alpha+\beta=390
\)
Now, median is at \(\frac{584}{2}=292^{\text {th }}\)
\(\because\) Median \(=45\) (lies in class 40 – 50 )
\(
\begin{aligned}
& \Rightarrow \alpha+110+54+15=292 \\
& \Rightarrow \alpha=113, \beta=277 \\
& \Rightarrow|\alpha-\beta|=164
\end{aligned}
\)
Consider the following frequency distribution :
\(
\begin{array}{|l|c|c|c|c|c|}
\hline \text { Class : } & 0-6 & 6-12 & 12-18 & 18-24 & 24-30 \\
\hline \text { Frequency : } & a & b & 12 & 9 & 5 \\
\hline
\end{array}
\)
If mean \(=\frac{309}{22}\) and median \(=14\), then the value \((a-b)^2\) is equal to _______. [JEE Main 2021 (Online) 22th July Evening Shift]
\(
\begin{array}{|l|l|l|l|}
\hline \text { Class } & \text { Frequency } & x_i & f_i x_i \\
\hline 0-6 & a & 3 & 3 a \\
\hline 6-12 & b & 9 & 9 b \\
\hline 12-18 & 12 & 15 & 180 \\
\hline 18-24 & 9 & 21 & 189 \\
\hline 24-30 & 5 & 27 & 135 \\
\hline & N=(26+a+b) & & (504+3 a+9 b) \\
\hline
\end{array}
\)
\(
\begin{aligned}
& \text { Mean }=\frac{3 a+9 b+180+189+135}{a+b+26}=\frac{309}{22} \\
& \Rightarrow 66 a+198 b+11088=309 a+309 b+8034 \\
& \Rightarrow 243 a+111 b=3054 \\
& \Rightarrow 81 a+37 b=1018 \rightarrow(1)
\end{aligned}
\)
Now, Median \(=12+\frac{\frac{a+b+c}{2}-(a+b)}{12} \times 6=14\)
\(
\begin{aligned}
& \Rightarrow \frac{13}{2}-\left(\frac{a+b}{4}\right)=2 \\
& \Rightarrow \frac{a+b}{4}=\frac{9}{2} \\
& \Rightarrow a+b=18 \rightarrow(2)
\end{aligned}
\)
\(
\text { From equation (1) and (2)a=8, } b=10 \therefore(a-b)^2=(8-10)^2=4
\)
The mean age of 25 teachers in a school is 40 years. A teacher retires at the age of 60 years and a new teacher is appointed in his place. If the mean age of the teachers in this school now is 39 years, then the age (in years) of the newly appointed teacher is ____. [JEE Main 2021 (Online) 18th March Morning Shift]
\(
\operatorname{Mean}(\bar{x})=\frac{x_1+x_2 \ldots .+x_n}{n}=\frac{\sum x}{n}
\)
Here, Mean \(=40\) of 25 teachers
\(
\begin{aligned}
& \therefore 40=\frac{\sum x}{25} \\
& \Rightarrow \sum x=40 \times 25=1000
\end{aligned}
\)
After retireing of a 60 year old teacher, total age of 24 teachers,
\(
x_1+x_2+\ldots \ldots x_{24}=1000-60=940
\)
Now a new teacher of age A year is appointed.
\(\therefore\) Now total age of this 25 teachers
\(
x_1+x_2+x_3+\ldots .+x_{25}=940+A
\)
\(\therefore\) Mean age \(=\frac{940+A}{25}\)
According to question,
\(
\begin{aligned}
& \frac{940+A}{25}=39 \\
& \Rightarrow A =35
\end{aligned}
\)
Consider a set of \(3 n\) numbers having variance 4 . In this set, the mean of first \(2 n\) numbers is 6 and the mean of the remaining \(n\) numbers is 3 . A new set is constructed by adding 1 into each of first 2 n numbers, and subtracting 1 from each of the remaining \(n\) numbers. If the variance of the new set is \(k\), then \(9 k\) is equal to ____. [JEE Main 2021 (Online) 17th March Evening Shift]
Let first 2 n observations are \(x _1, x _2\) \(\ldots\) \(x_{2 n}\)
and last n observations are \(y _1, y _2\) \(\ldots\) \(y _{ n }\)
Now, \(\frac{\sum x_i}{2 n}=6, \frac{\sum y_i}{n}=3\)
\(
\begin{aligned}
& \Rightarrow \sum x_i=12 n, \sum y_i=3 n \\
& \therefore \frac{\sum x_i+\sum y_i}{3 n}=\frac{15 n}{3 n}=5
\end{aligned}
\)
Now, \(\frac{\sum x_i^2+\sum y_i^2}{3 n}-5^2=4\)
\(
\Rightarrow \sum x_i^2+\sum y_i^2=29 \times 3 n=87 n
\)
Now, mean is \(\frac{\sum\left(x_i+1\right)+\sum\left(y_i-1\right)}{3 n}=\frac{15 n+2 n-n}{3 n}=\frac{16}{3}\)
Now, variance is \(\frac{\sum\left(x_i+1\right)^2+\sum\left(y_i-1\right)^2}{3 n}-\left(\frac{16}{3}\right)^2\)
\(
\begin{aligned}
& =\frac{\sum x_i^2+\sum y_i^2+2\left(\sum x_i-\sum y_i\right)+3 n}{3 n}-\left(\frac{16}{3}\right)^2 \\
& =\frac{87 n+2(9 n)+3 n}{3 n}-\left(\frac{16}{3}\right)^2 \\
& =29+6+1-\left(\frac{16}{3}\right)^2 \\
& =\frac{324-256}{9}=\frac{68}{9}=k \\
& \Rightarrow 9 k =68
\end{aligned}
\)
Therefore, the correct answer is 68.
Consider the statistics of two sets of observations as follows:
\(
\begin{array}{lccc}
& \text { Size } & \text { Mean } & \text { Variance } \\
\text { Observation I } & 10 & 2 & 2 \\
\text { Observation II } & n & 3 & 1
\end{array}
\)
If the variance of the combined set of these two observations is \(\frac{17}{9}\), then the value of \(n\) is equal to _____. [JEE Main 2021 (Online) 16th March Evening Shift]
For group – 1 : \(\frac{\sum x_i}{10}=2 \Rightarrow \sum x_i=20\)
\(
\frac{\sum x_i^2}{10}-(2)^2=2 \Rightarrow \sum x_i^2=60
\)
For group – 2 : \(\frac{\sum y_i}{n}=3 \Rightarrow \sum y_i=3 n\)
\(
\frac{\sum y_i^2}{n}-3^2=1 \Rightarrow \sum y_i^2=10 n
\)
Now, combined variance
\(
\begin{aligned}
& \sigma^2=\frac{\sum\left(x_i^2+y_i^2\right)}{10+n}-\left(\frac{\sum\left(x_i+y_i\right)}{10+n}\right)^2 \\
& \Rightarrow \frac{17}{9}=\frac{60+10 n}{10+n}-\frac{(20+3 n)^2}{(10+n)^2} \\
& \Rightarrow 17\left(n^2+20 n+100\right)=9\left(n^2+40 n+200\right) \\
& \Rightarrow 8 n^2-20 n-100=0 \\
& \Rightarrow 2 n^2-5 n-25=0 \Rightarrow n=5
\end{aligned}
\)
Let \(X _1, X _2, \ldots . . ., X _{18}\) be eighteen observations such that \(\sum_{i=1}^{18}\left(X_i-\alpha\right)=36\) and \(\sum_{i=1}^{18}\left(X_i-\beta\right)^2=90\), where \(\alpha\) and \(\beta\) are distinct real numbers. If the standard deviation of these observations is 1 , then the value of \(\mid \alpha\) \(-\beta \mid\) is _____. [JEE Main 2021 (Online) 26th February Evening Shift]
Given, \(\sum_{i=1}^{18}\left(x_1-\alpha\right)=36\)
\(
\begin{aligned}
& \Rightarrow \sum x_i-18 \alpha=36 \\
& \Rightarrow \sum x_i=18(\alpha+2) \ldots(1)
\end{aligned}
\)
Also, \(\sum_{i=1}^{18}\left(x_1-\beta\right)^2=90\)
\(
\Rightarrow \sum x_i^2+18 \beta^2-2 \beta \sum x_i=90
\)
\(
\begin{aligned}
& \Rightarrow \sum x_i^2+18 \beta^2+2 \beta \times 18(\alpha+2)=90 \text { (using equation (1)) } \\
& \Rightarrow \sum x_i^2=90-18 \beta^2+36 \beta(\alpha+2) \\
& \text { Given, } \sigma^2=1 \Rightarrow \frac{1}{18} \sum x_i^2-\left(\frac{\sum x_i}{18}\right)^2=1 \\
& =\frac{1}{18}\left(90-18 \beta^2+36 \alpha \beta+72 \beta\right)-\left(\frac{18(\alpha+2)}{18}\right)^2=1 \\
& \Rightarrow 90-18 \beta^2+36 \alpha \beta+72 \beta-18(\alpha+2)^2=18 \\
& \Rightarrow 5-\beta^2+2 \alpha \beta+4 \beta-(\alpha+2)^2=1 \\
& \Rightarrow 5-\beta^2+2 \alpha \beta+4 \beta-\alpha^2-4-4 \alpha=1 \\
& \Rightarrow \alpha^2-\beta^2+2 \alpha \beta+4 \beta-4 \alpha=0 \\
& \Rightarrow(\alpha-\beta)(\alpha-\beta+4)=0 \\
& \Rightarrow \alpha-\beta=-4 \\
& \therefore|\alpha-\beta|=4(\alpha \neq \beta)
\end{aligned}
\)
If the variance of 10 natural numbers \(1,1,1, \ldots . ., 1, k\) is less than 10 , then the maximum possible value of \(k\) is _____. [JEE Main 2021 (Online) 24th February Evening Shift]
\(
\begin{aligned}
& \sigma^2=\frac{\sum x^2}{n}-\left(\frac{\sum x}{n}\right)^2 \\
& \sigma^2=\frac{\left(9+k^2\right)}{10}-\left(\frac{9+k}{10}\right)^2<10 \\
& \left(90+k^2\right) 10-\left(81+k^2+8 k\right)<1000 \\
& 90+10 k^2-k^2-18 k-81<1000 \\
& 9 k^2-18 k+9<1000 \\
& (k-1)^2<\frac{1000}{9} \Rightarrow k-1<\frac{10 \sqrt{10}}{3} \\
& k<\frac{10 \sqrt{10}}{3}+1 \\
& k \leq 11
\end{aligned}
\)
Maximum integral value of \(k =11\).
Consider the data on \(x\) taking the values \(0,2,4,8, \ldots . ., 2^n\) with frequencies \({ }^n C_0,{ }^n C_1,{ }^n C_2, \ldots,{ }^n C_n\) respectively. If the mean of this data is \(\frac{728}{2^n}\), then \(n\) is equal to _____. [JEE Main 2020 (Online) 6th September Evening Slot]
\(
\begin{aligned}
& \text { Mean }=\frac{\sum x_1 \cdot f_1}{\sum f_1} \\
& =\frac{0 \cdot{ }^n C_0+2 \cdot{ }^n C_1+2^2 \cdot{ }^n C_2+\ldots+2^n \cdot{ }^n C_n}{{ }^n C_0+{ }^n C_1+\ldots+{ }^n C_n}
\end{aligned}
\)
We know,
\(
(1+ x )^{ n }={ }^n C_0+{ }^n C_1 x+{ }^n C_2 x^2+\ldots+{ }^n C_n x^n \dots(1)
\)
Put \(x=2\), at (1) we get
\(
\Rightarrow 3^n-1=2 \cdot{ }^n C_1+2^2 \cdot{ }^n C_2+\ldots+2^n \cdot{ }^n C_n
\)
And Putting \(x=1\) in (1), we get
\(
\begin{aligned}
& 2^n={ }^n C_0+{ }^n C_1+\ldots+{ }^n C_n \\
& \therefore \text { Mean }=\frac{3^n-1}{2^n}
\end{aligned}
\)
According to question,
\(
\begin{aligned}
& \frac{3^n-1}{2^n}=\frac{728}{2^n} \\
& \Rightarrow 3^n=729 \\
& \Rightarrow n=6
\end{aligned}
\)
If the variance of the following frequency distribution :
\(
\begin{array}{|c|l|c|l|}
\hline \text { Class } & 10-20 & 20-30 & 30-40 \\
\hline \text { Frequency } & 2 & x & 2 \\
\hline
\end{array}
\)
is 50 , then \(x\) is equal to _____. [JEE Main 2020 (Online) 4th September Evening Slot]
\(
\begin{array}{|c|c|c|c|c|c|}
\hline x_i & f_i & f_i x_i & x_i-\bar{x} & \left(x_i-\bar{x}\right)^2 & f_i\left(x_i-\bar{x}\right)^2 \\
\hline 15 & 2 & 30 & -10 & 100 & 200 \\
\hline 25 & x & 25 x & 0 & 0 & 0 \\
\hline 35 & 2 & 70 & 10 & 100 & 200 \\
\hline & \sum f_i=4+x & \sum f_i x_i=100+25 x & & & \sum f_i\left(x_i-\bar{x}\right)^2=400 \\
\hline
\end{array}
\)
Mean:
\(
\begin{aligned}
& \bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{100+25 x}{4+x} \\
& \Rightarrow \bar{x}=25
\end{aligned}
\)
Now, variance
\(
\begin{aligned}
& =\frac{\sum f_i\left(x_i-\bar{x}\right)^2}{\sum f_i}=\frac{400}{4+x} \\
& \Rightarrow \frac{400}{4+x}=50 \\
& \Rightarrow 200+50 x=400 \\
& \therefore x=4
\end{aligned}
\)
If the variance of the terms in an increasing A.P., \(b_1, b_2, b_3, \ldots, b_{11}\) is 90 , then the common difference of this A.P. is ____. [JEE Main 2020 (Online) 2nd September Evening Slot]
\(
\begin{aligned}
& \text { Let the common difference }= d \\
& \text { and } b_1=a \\
& b_2=a+d \\
& b_3=a+2 d \\
& \ldots b_{11}=a+10 d \\
& \text { Variance }=\frac{\sum a_i^2}{11}-\left(\frac{\sum a_i}{11}\right)^2=90 \\
& \Rightarrow \frac{a^2+(a+d)^2+\ldots+(a+10 d)^2}{11}-\left(\frac{a+(a+d)+\ldots+(a+10 d)}{11}\right)^2=90 \\
& \Rightarrow 11\left[11 a^2+385 d^2+110 a d\right]-[11 a+55 d]^2=10890 \\
& \Rightarrow 1210 d^2=10890 \\
& \Rightarrow d^2=9 \\
& \Rightarrow d= \pm 3
\end{aligned}
\)
As A.P is increasing so \(d\) should be positive
\(
\therefore d=3
\)
If the mean and variance of eight numbers \(3,7,9,12,13,20, x\) and \(y\) be 10 and 25 respectively, then x.y is equal to ______. [JEE Main 2020 (Online) 7th January Evening Slot]
\(
\begin{aligned}
& \text { Mean }=\frac{3+7+9+12+13+20+x+y}{8}=10 \\
& 16= x + y \ldots(1)
\end{aligned}
\)
\(
\begin{aligned}
& \text { Variance }\left(\sigma^2\right)=25 \\
& \Rightarrow \frac{3^2+7^2+9^2+12^2+13^2+20^2+x^2+y^2}{8}-100=25 \\
& \Rightarrow 125 \times 8=9+49+81+144+169+400+x^2+y^2-800 \\
& \Rightarrow x^2+y^2=148
\end{aligned}
\)
We know, \((x+y)^2=x^2+y^2+2 x y\)
\(
\begin{aligned}
& \Rightarrow 256=148+2 x y \\
& \Rightarrow x . y=54
\end{aligned}
\)
If the variance of the first \(n\) natural numbers is 10 and the variance of the first \(m\) even natural numbers is 16 , then \(m+n\) is equal to _____. [JEE Main 2020 (Online) 7th January Morning Slot]
Variance \(\sigma^2=\frac{\sum x_i^2}{N}-\mu^2\)
variance of \((1,2, \ldots . . n)\)
\(
10=\frac{1^2+2^2+\ldots .+n^2}{n}-\left(\frac{1+2+3+\ldots .+n}{n}\right)^2
\)
on solving we get \(n =11\)
variance of \(2,4,6 \ldots . . . .2 m=16\)
\(
\begin{aligned}
& \Rightarrow \frac{2^2+4^2+\ldots .+(2 m)^2}{m}-(m+1)^2=16 \\
& \Rightarrow m^2=49 \\
& \Rightarrow m=7 \\
& \therefore m+n=18
\end{aligned}
\)
If the variance of the frequency distribution
\(
\begin{array}{|l|l|l|l|l|l|l|}
\hline x & c & 2 c & 3 c & 4 c & 5 c & 6 c \\
\hline f & 2 & 1 & 1 & 1 & 1 & 1 \\
\hline
\end{array}
\)
is 160 , then the value of \(c \in N\) is [JEE Main 2024 (Online) 9th April Evening Shift]
\(
\begin{array}{|c|c|c|c|}
\hline x_i & f\left(x_i\right) & x(f(x)) & x^2 f(x) \\
\hline \text { C } & 2 & 2 C & 2 C ^2 \\
\hline \text { 2C } & 1 & 2 C & 4 C ^2 \\
\hline \text { 3C } & 1 & 3 C & 9 C ^2 \\
\hline \text { 4C } & 1 & 4 C & 16 C ^2 \\
\hline \text { 5C } & 1 & 5 C & 25 C ^2 \\
\hline \text { 6C } & 1 & 6 C & 36 C ^2 \\
\hline
\end{array}
\)
\(
\begin{aligned}
& \sigma^2=E\left(x^2\right)-[E(x)], \sum f\left(x_i\right)=7 \\
& E(x)=\sum x f(x)=22 C \\
& E\left(x^2\right)=\sum x^2 f(x)=92 C^2 \\
& \sigma^2=160=\frac{92 C^2}{7}-\left(\frac{22 C}{7}\right)^2 \\
& \Rightarrow C= \pm 7 \text { but } C \in N \\
& \Rightarrow C=7
\end{aligned}
\)
The frequency distribution of the age of students in a class of 40 students is given below.
\(
\begin{array}{|l|c|c|c|c|c|c|}
\hline \text { Age } & 15 & 16 & 17 & 18 & 19 & 20 \\
\hline \text { No of Students } & 5 & 8 & 5 & 12 & x & y \\
\hline
\end{array}
\)
If the mean deviation about the median is 1.25 , then \(4 x+5 y\) is equal to : [JEE Main 2024 (Online) 9th April Morning Shift]
\(
\begin{array}{|c|c|c|}
\hline \text { Age } & \text { No. of Students } & \text { CF } \\
\hline 15 & 5 & 5 \\
\hline 16 & 8 & 13 \\
\hline 17 & 5 & 18 \\
\hline 18 & 12 & 30 \\
\hline 19 & x & 30+x \\
\hline 20 & y & 30+x+y \\
\hline
\end{array}
\)
\(
\begin{aligned}
& 30+x+y=40 \\
& x+y=10 \\
& \text { Median }=\left(\frac{n+1}{2}\right)^{\text {th }} \text { observation } \\
& =\frac{40+1}{2}=\frac{41}{2} \\
& \text { Median }=18
\end{aligned}
\)
Mean deviation about median
\(
\begin{aligned}
& 5.3+8.2+5.1+12.0+x \cdot 1+y \cdot 2=1.25 \times 40 \\
& 15+16+5+x+2 y=50 \\
& x+2 y=14 \\
& \quad x+y=10 \\
& \Rightarrow \quad x=4 \\
& \Rightarrow \quad y=6 \\
& 4 x+5 y=24+20 \\
& \quad=44
\end{aligned}
\)
The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation by mistake was taken 8 instead of 12 . The correct standard deviation is [JEE Main 2024 (Online) 6th April Morning Shift]
To find the correct standard deviation, we first need to adjust the mean and sum of squares of the observations based on the correction of the erroneous entry. The original erroneous observation was 8 , and the correct observation is 12 . The mean and standard deviation of the 20 observations before correction were 10 and 2, respectively.
The sum of all 20 original observations ( \(S_{\text {original }}\) ) can be calculated from the mean formula:
\(
\begin{aligned}
& \text { Mean }=\frac{\text { Sum of all observations }}{\text { Number of observations }} \\
& 10=\frac{S_{\text {original }}}{20} \\
& S_{\text {original }}=10 \times 20=200
\end{aligned}
\)
The corrected sum of observations ( \(S_{\text {correct }}\) ) will replace the incorrect observation (8) with the correct one (12):
\(
S_{\text {correct }}=S_{\text {original }}-8+12=200-8+12=204
\)
The corrected mean ( \(\mu_{\text {correct }}\) ) is:
\(
\mu_{\text {correct }}=\frac{S_{\text {correct }}}{20}=\frac{204}{20}=10.2
\)
To find the corrected standard deviation, we need the sum of squares of the deviations from the mean for both the original and corrected data. The original sum of squares ( \(S S_{\text {original }}\) ) is calculated from the original standard deviation formula, where \(\sigma=2\) :
\(
\begin{aligned}
& \sigma^2=\frac{S S}{n} \\
& 4=\frac{S S_{\text {original }}}{20} \\
& S S_{\text {original }}=4 \times 20=80
\end{aligned}
\)
To calculate the corrected sum of squares ( \(S S_{\text {correct }}\) ), we need to adjust \(S S_{\text {original }}\) by removing the square of the deviation of the incorrect observation and adding the square of the deviation of the correct observation:
\(
\begin{array}{ll}
S S_{\text {correct }} & =S S_{\text {original }}-(8-10)^2+(12-10.2)^2 \\
S S_{\text {correct }} & =80-(-2)^2+(1.8)^2 \\
S S_{\text {correct }} & =80-4+3.24=79.24
\end{array}
\)
Finally, the corrected standard deviation ( \(\left.\sigma_{\text {correct }}\right)\) is:
\(
\begin{aligned}
\sigma_{\text {correct }} & =\sqrt{\frac{S S_{\text {correct }}}{20}} \\
\sigma_{\text {correct }} & =\sqrt{\frac{79.24}{20}} \\
\sigma_{\text {correct }} & =\sqrt{3.962}
\end{aligned}
\)
The corrected standard deviation is closest to the value given in Option \(b\), which is
\(
\sqrt{3.96}
\)
Let \(\alpha, \beta \in R\). Let the mean and the variance of 6 observations \(-3,4,7,-6, \alpha, \beta\) be 2 and 23 , respectively. The mean deviation about the mean of these 6 observations is : [JEE Main 2024 (Online) 4th April Morning Shift]
\(
\begin{aligned}
& \text { Mean }=\frac{-3+4+7+(-6)+\alpha+\beta}{6}=2 \\
& \Rightarrow \alpha+\beta=10 \\
& \text { Variance }=\frac{\sum x_i^2}{n}-\left(\frac{\bar{x}}{n}\right)^2=23 \\
& \Rightarrow \sum x_i^2=27 \times 6 \\
& \Rightarrow 9+16+49+36+\alpha^2+\beta^2=162 \\
& \Rightarrow \alpha^2+\beta^2=52
\end{aligned}
\)
We get \(\alpha\) and \(\beta\) as 4 and 6
So, mean deviation about mean
\(
\begin{aligned}
& =\frac{|-3-2|+|4-2|+|7-2|+|-6-2|+|4-2|+|6-2|}{6} \\
& =\frac{5+2+5+8+2+4}{6} \\
& =\frac{13}{3}
\end{aligned}
\)
Consider 10 observations \(x_1, x_2, \ldots, x_{10}\) such that \(\sum_{i=1}^{10}\left(x_i-\alpha\right)=2\) and \(\sum_{i=1}^{10}\left(x_i-\beta\right)^2=40\), where \(\alpha, \beta\) are positive integers. Let the mean and the variance of the observations be \(\frac{6}{5}\) and \(\frac{84}{25}\) respectively. Then \(\frac{\beta}{\alpha}\) is equal to : [JEE Main 2024 (Online) 1st February Evening Shift]
We have given \(\bar{x}(\) mean \()=\frac{6}{5}\)
\(
\begin{aligned}
& \text { Variance }=\frac{84}{25} \\
& \sum_{i=1}^{10}\left(x_i-\alpha\right)=2 \\
& \Rightarrow x_1+x_2+\ldots+x_{10}-10 \alpha=2 \\
& \Rightarrow \frac{x_1+x_2+\ldots+x_{10}}{10}-\alpha=\frac{2}{10} \\
& \Rightarrow \frac{6}{5}-\alpha=\frac{2}{10} \\
& \Rightarrow \alpha=1
\end{aligned}
\)
\(
\begin{aligned}
& \text { and } \sum_{i=1}^{10}\left(x_i-\beta\right)^2=40 \\
& \left(x_1-\beta\right)^2+\left(x_2-\beta\right)^2+\ldots+\left(x_{10}-\beta\right)^2=40 \\
& x_1^2+x_2^2+\ldots+x_{10}^2+10 \beta^2-2 \beta\left(x_1+x_2+\ldots+x_{10}\right)=40 \\
& \Rightarrow \frac{x_1^2+x_2^2+\ldots+x_{10}^2}{10}+\beta^2-\frac{2 \beta\left(x_1+x_2+\ldots+x_{10}\right)}{10}=4 \\
& \Rightarrow \frac{x_1^2+x_2^2+\ldots+x_{10}^2}{10}-\frac{36}{25}+\frac{36}{25}+\beta^2-2 \beta \times \frac{6}{5}=4 \\
& {\left[\text { Variance }=\frac{\sum_{i=1}^n x_i^2}{n}-(\bar{x})^2\right]} \\
& \Rightarrow \frac{84}{25}+\frac{36}{25}+\beta^2-\frac{12 \beta}{5}-4=0 \\
& \Rightarrow \frac{120}{25}+\beta^2-\frac{12 \beta}{5}-4=0 \\
& \Rightarrow 25 \beta^2-60 \beta+20=0 \\
& \Rightarrow 5 \beta^2-12 \beta+4=0
\end{aligned}
\)
\(
\Rightarrow \beta=2, \frac{2}{5}
\)
Take \(\beta=2\)
\(
\frac{\beta}{\alpha}=\frac{2}{1}=2
\)
Let the median and the mean deviation about the median of 7 observation \(170,125,230,190,210\) a, b be 170 and \(\frac{205}{7}\) respectively. Then the mean deviation about the mean of these 7 observations is : [JEE Main 2024 (Online) 1st February Morning Shift]
\(
\text { Median }=170 \Rightarrow 125, a , b , 170,190,210,230
\)
Mean deviation about Median \(=\)
\(
\begin{aligned}
& \frac{0+45+60+20+40+170-a+170-b}{7}=\frac{205}{7} \\
& \Rightarrow a + b =300 \\
& \text { Mean }=\frac{170+125+230+190+210+a+b}{7}=175
\end{aligned}
\)
Mean deviation About mean \(=\)
\(
\frac{50+175-a+175-b+5+15+35+55}{7}=30
\)
Let the mean and the variance of 6 observations \(a, b, 68,44,48,60\) be 55 and 194, respectively. If \(a>b\), then \(a+3 b\) is [JEE Main 2024 (Online) 31st January Evening Shift]
\(
\begin{aligned}
& \text { mean }=\frac{a+b+68+44+48+60}{6}=55 \\
& a+b+220=55 \times 6 \Rightarrow 330 \\
& a+b=110 \\
& \text { variance }=\frac{\sum x_i^2}{n}-(\bar{x})^2=194 \\
& \frac{a^2+b^2+(68)^2+(44)^2+(48)^2+(60)^2}{6}-(55)^2=194 \\
& a^2+b^2+4624+1936+2304+3600=6 \times(3205+194) \\
& a^2+b^2=6850 \\
& \text { Now }(a+b)^2=(110)^2 \\
& a^2+b^2+2 a b=12100 \\
& 6850+2 a b=12100 \\
& 2 a b=12100-6850 \\
& 2 a b=5250 \\
& a b=2625
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow(110-b) b=2625 \quad(\because a=110-b) \\
& b^2-110 b+2625=0 \\
& b=35,75 \\
& \text { but } a>b \\
& a+b=110
\end{aligned}
\)
So \(b=35\) and \(a=75\)
Now
\(
\begin{aligned}
& \Rightarrow a+3 b \\
& \Rightarrow 75+3(35) \\
& \Rightarrow 180
\end{aligned}
\)
Let M denote the median of the following frequency distribution
\(
\begin{array}{|l|c|c|c|c|c|}
\hline \text { Class } & 0-4 & 4-8 & 8-12 & 12-16 & 16-20 \\
\hline \text { Frequency } & 3 & 9 & 10 & 8 & 6 \\
\hline
\end{array}
\)
Then 20 M is equal to : [JEE Main 2024 (Online) 30th January Morning Shift]
\(
\begin{array}{|c|c|c|}
\hline \text { Class } & \text { Frequency } & \begin{array}{c}
\text { Cumulative } \\
\text { frequency }
\end{array} \\
\hline 0-4 & 3 & 3 \\
\hline 4-8 & 9 & 12 \\
\hline 8-12 & 10 & 22 \\
\hline 12-16 & 6 & 30 \\
\hline
\end{array}
\)
\(
\begin{aligned}
& M =1+\left(\frac{\frac{ N }{2}- C }{ f }\right) h \\
& M =8+\frac{18-12}{10} \times 4 \\
& M =10.4 \\
& 20 M =208
\end{aligned}
\)
If the mean and variance of five observations are \(\frac{24}{5}\) and \(\frac{194}{25}\) respectively and the mean of the first four observations is \(\frac{7}{2}\), then the variance of the first four observations in equal to [JEE Main 2024 (Online) 29th January Evening Shift]
\(
\bar{X}=\frac{24}{5} ; \sigma^2=\frac{194}{25}
\)
Let first four observation be \(x _1, x _2, x _3, x _4\)
Here, \(\frac{x_1+x_2+x_3+x_4+x_5}{5}=\frac{24}{5} \ldots(1)\)
Also, \(\frac{x_1+x_2+x_3+x_4}{4}=\frac{7}{2}\)
\(
\Rightarrow x _1+ x _2+ x _3+ x _4=14
\)
Now from eqn (1)
\(
x _5=10
\)
Now, \(\sigma^2=\frac{194}{25}\)
\(
\begin{aligned}
& \frac{x_1^2+x_2^2+x_3^2+x_4^2+x_5^2}{5}-\frac{576}{25}=\frac{194}{25} \\
& \Rightarrow x_1^2+x_2^2+x_3^2+x_4^2=54
\end{aligned}
\)
Now, variance of first 4 observations
\(
\begin{aligned}
\operatorname{Var} & =\frac{\sum_{i=1}^4 x_i^2}{4}-\left(\frac{\sum_{i=1}^4 x_i}{4}\right)^2 \\
& =\frac{54}{4}-\frac{49}{4}=\frac{5}{4}
\end{aligned}
\)
Let \(a_1, a_2, \ldots a_{10}\) be “10” observations such that \(\sum_{k=1}^{10} a_k=50\) and \(\sum_{\forall k<j} a_k \cdot a_j=1100\). Then the standard deviation of \(a_1, a_2, \ldots, a_{10}\) is equal to [JEE Main 2024 (Online) 27th January Morning Shift]
\(
\begin{aligned}
& \sum_{k=1}^{10} a_k=50 \\
& a_1+a_2+\ldots+a_{10}=50 \dots(i) \\
& \sum_{\forall k<j} a_k a_j=1100 \dots(ii)
\end{aligned}
\)
\(
\begin{aligned}
& \text { If } a_1+a_2+\ldots+a_{10}=50 \\
& \left(a_1+a_2+\ldots+a_{10}\right)^2=2500 \\
& \Rightarrow \sum_{i=1}^{10} a_i^2+2 \sum_{k<j} a_k a_j=2500 \\
& \Rightarrow \sum_{i=1}^{10} a_i^2=2500-2(1100) \\
& \sum_{i=1}^{10} a_i^2=300 \text {, Standard deviation ‘ } \sigma^{\prime} \\
& =\sqrt{\frac{\sum a_i^2}{10}-\left(\frac{\sum a_i}{10}\right)^2}=\sqrt{\frac{300}{10}-\left(\frac{50}{10}\right)^2} \\
& =\sqrt{30-25}=\sqrt{5}
\end{aligned}
\)
The mean and standard deviation of 10 observations are 20 and 8 respectively. Later on, it was observed that one observation was recorded as 50 instead of 40 . Then the correct variance is : [JEE Main 2023 (Online) 15th April Morning Shift]
Step-1: Calculate the sum of the original observations:
\(
\begin{aligned}
& \frac{x_1+x_2+\ldots+x_9+50}{10}=20 \\
& x_1+x_2+\ldots+x_9=150
\end{aligned}
\)
Step-2: Calculate the sum of the squares of the original observations using the original variance:
\(
\begin{aligned}
& \sigma^2=8^2=64 \\
& 64=\frac{x_1^2+x_2^2+\ldots+x_9^2+2500}{10}-400 \\
& x_1^2+x_2^2+\ldots+x_9^2=2140
\end{aligned}
\)
Step-3: Calculate the new mean after correcting the error:
New mean \(=\frac{150+40}{10}=19\)
Step-4: Calculate the new variance using the corrected sum of observations and the corrected sum of squares of observations:
New \(\sigma^2=\frac{2140+1600}{10}-(19)^2\)
\(
\Rightarrow \sigma^2=13
\)
The correct variance after correcting the error is 13 (Option C).
Let the mean of 6 observations \(1,2,4,5, x\) and y be 5 and their variance be 10 . Then their mean deviation about the mean is equal to : [JEE Main 2023 (Online) 11th April Evening Shift]
Given that the mean of the observations \(x, y, 1,2,4,5\) is 5 , we get the equation:
\(
x+y+1+2+4+5=6 \cdot 5 \Rightarrow x+y=18 \dots(1)
\)
We are also given that the variance of the observations is 10 . Using the formula for variance, we have:
\(V=\frac{\Sigma x_i^2}{n}-\bar{x}^2\), where \(n\) is the number of observations, and \(\bar{x}\) is the mean.
Substituting the given values in the variance formula, we get:
\(
10=\frac{\Sigma x_i^2}{6}-5^2=\frac{\Sigma x_i^2}{6}-25 \Rightarrow \Sigma x_i^2=210 \text {. }
\)
Here, \(\Sigma x_i^2\) is the sum of the squares of all the observations, thus:
\(
x^2+y^2+1+4+16+25=210 \Rightarrow x^2+y^2=164 \dots(2)
\)
By solving the system of equations (1) and (2), we get \(x=8\) and \(y=10\).
Now, the mean deviation about the mean \((\bar{x}=5)\) is calculated by taking the average of the absolute differences of each observation from the mean:
\(
M D(5)=\frac{1}{6}[|1-5|+|2-5|+|4-5|+|5-5|+|8-5|+|10-5|]=\frac{16}{6}=\frac{8}{3}
\)
\(
\text { So, the mean deviation about the mean is } \frac{8}{3} \text {, and the correct answer is Option } b, \frac{8}{3} \text {. }
\)
Let sets \(A\) and \(B\) have 5 elements each. Let the mean of the elements in sets \(A\) and \(B\) be 5 and 8 respectively and the variance of the elements in sets \(A\) and \(B\) be 12 and 20 respectively. A new set C of 10 elements is formed by subtracting 3 from each element of A and adding 2 to each element of B . Then the sum of the mean and variance of the elements of C is ____. [JEE Main 2023 (Online) 11th April Morning Shift]
\(
\begin{aligned}
& A=\left\{a_1, a_2, a_3, a_4, a_5\right\} \\
& B=\left\{b_1, b_2, b_3, b_4, b_5\right\}
\end{aligned}
\)
\(
\text { Given, } \sum_{ i =1}^5 ai =25, \sum_{ i =1}^5 bi =40
\)
\(
\frac{\sum_{i=1}^5 a_i^2}{5}-\left(\frac{\sum_{i=1}^5 a_i}{5}\right)^2=12, \frac{\sum_{i=1}^5 b_i^2}{5}-\left(\frac{\sum_{i=1}^5 b_i}{5}\right)^2=20
\)
\(
\Rightarrow \sum_{ i =1}^5 a _{ i }^2=185, \sum_{ i =1}^5 b_{ i }^2=420
\)
Now, \(C =\left\{ C _1, C _2, \ldots C _{10}\right\}\)
\(\therefore\) Mean of \(C , \overline{ C }=\frac{\left(\sum a _{ i }-15\right)+\left(\sum b _{ i }-10\right)}{10}\)
\(
\begin{aligned}
& \overline{ C }=\frac{10+50}{10}=6 \\
& \therefore \sigma^2=\frac{\sum_{i=1}^{10} C_i^2}{10}=(\overline{ C })^2 \\
& =\frac{\sum\left(a_i-3\right)^2+\sum\left(b_i-2\right)^2+}{10}-(6)^2 \\
& =\frac{\sum a_i^2+\sum b_i^2-6 \sum a_i+4 \sum b_i+65}{10}-36 \\
& =\frac{185+420-150+160+65}{10}-36 \\
& =32
\end{aligned}
\)
\(
\therefore \text { Mean }+ \text { Variance }=\overline{ C }+\sigma^2=6+32=38
\)
Let \(\mu\) be the mean and \(\sigma\) be the standard deviation of the distribution
\(
\begin{array}{|c|c|c|c|c|c|c|}
\hline x_i & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline f_i & k+2 & 2 k & k^2-1 & k^2-1 & k^2+1 & k-3 \\
\hline
\end{array}
\)
where \(\sum f_i=62\). If \([x]\) denotes the greatest integer \(\leq x\), then \(\left[\mu^2+\sigma^2\right]\) is equal to : [JEE Main 2023 (Online) 10th April Evening Shift]
We have, \(\Sigma f_i=62\)
\(
\begin{aligned}
& \left.(K+2)+2 K+\left(K^2-1\right)\right)+\left(K^2-1\right)+\left(K^2+1\right)+(K-3)=62 \\
& \Rightarrow 3 K^2+4 K-64=0 \\
& \Rightarrow(3 K+16)(K-4)=0 \\
& \Rightarrow K=4 \\
& \left(\because k=\frac{-16}{3} \text { is not possible }\right)
\end{aligned}
\)
\(
\begin{array}{|r|c|c|c|}
\hline x_i & f_i & f_i x_i & f_i x_i^2 \\
\hline 0 & 6 & 0 & 0 \\
1 & 8 & 8 & 8 \\
2 & 15 & 30 & 60 \\
3 & 15 & 45 & 135 \\
4 & 17 & 68 & 272 \\
5 & 1 & 5 & 25 \\
\hline \text { Total } & 62 & 156 & 500 \\
\hline
\end{array}
\)
\(
\begin{aligned}
& \mu=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{0+8+30+45+68+5}{62}=\frac{156}{62} \\
& \sigma^2=\frac{\Sigma f_i x_i^2}{\Sigma f_i}-\left(\frac{\Sigma f_i x_i}{\Sigma f_i}\right)^2
\end{aligned}
\)
\(
\begin{aligned}
= & \frac{1 \times 8+4 \times 15+9 \times 15+16 \times 17+25 \times 1}{62}-\left(\frac{156}{62}\right)^2 \\
= & \frac{500}{62}-\left(\frac{156}{62}\right)^2=\frac{500}{62}-\mu^2
\end{aligned}
\)
\(
\therefore \sigma^2+\mu^2=\frac{500}{62}
\)
Hence, \(\left[\sigma^2+\mu^2\right]=8\)
Let the mean and variance of 12 observations be \(\frac{9}{2}\) and 4 respectively. Later on, it was observed that two observations were considered as 9 and 10 instead of 7 and 14 respectively. If the correct variance is \(\frac{m}{n}\), where m and n are coprime, then \(m + n\) is equal to : [JEE Main 2023 (Online) 8th April Evening Shift]
\(
\begin{aligned}
& \text { Since, Mean }=\frac{9}{2} \\
& \Rightarrow \Sigma x=\frac{9}{2} \times 12=54 \\
& \text { Also, variance }=4 \\
& \Rightarrow \frac{\sum x^2}{12}=\left[\frac{\sum x_i}{12}\right]^2=4 \\
& \Rightarrow \frac{\sum x^2}{12}=4+\frac{81}{4}=\frac{97}{4} \\
& \Rightarrow \sum x^2=291 \\
& \sum x^{\prime}=54-(9+10)+7+14 \\
& =54-19+21=56
\end{aligned}
\)
\(
\begin{aligned}
& \text { and } \sum x^2=291-(81+100)+49+196 \\
& =291-181+49+196=355 \\
& \text { So, } \sigma_{\text {new }}^2=\frac{\sum x_{\text {new }}^2}{12}-\left(\frac{\sum x_{\text {new }}}{12}\right)^2 \\
& =\frac{355}{12}-\left(\frac{56}{12}\right)^2 \\
& =\frac{4260-3136}{144}=\frac{1124}{144}=\frac{281}{36} \\
& =\frac{m}{n} \\
& \Rightarrow m=281, n=36 \\
& \Rightarrow m+n=281+36=317
\end{aligned}
\)
The mean and variance of a set of 15 numbers are 12 and 14 respectively. The mean and variance of another set of 15 numbers are 14 and \(\sigma^2\) respectively. If the variance of all the 30 numbers in the two sets is 13 , then \(\sigma^2\) is equal to: [JEE Main 2023 (Online) 6th April Morning Shift]
We know that if \(n_1, n_2\) are the sizes, \(\bar{X}_1, \bar{X}_2\) are the means and \(\sigma_1, \sigma_2\) are the standard deviation of the series, then the combine variance of the series.
\(
\begin{aligned}
& \sigma^2=\frac{n_1 \sigma_1^2+n_2 \sigma_2^2}{n_1+n_2}+\frac{n_1 \cdot n_2}{\left(n_1+n_2\right)^2}\left(\bar{X}_1-\bar{X}_2\right)^2 \\
& \Rightarrow 13=\frac{15 \times 14+15 \times \sigma^2}{15+15}+\frac{15 \times 15}{(15+15)^2}(12-14)^2 \\
& \Rightarrow 13=\frac{14+\sigma^2}{2}+\frac{1}{4} \times 4 \\
& \Rightarrow 14+\sigma^2=2 \times 12 \\
& \Rightarrow \sigma^2=10
\end{aligned}
\)
Let \(9=x_1<x_2<\ldots<x_7\) be in an A.P. with common difference d. If the standard deviation of \(x_1, x_2 \ldots, x_7\) is 4 and the mean is \(\bar{x}\), then \(\bar{x}+x_6\) is equal to: [JEE Main 2023 (Online) 1st February Evening Shift]
\(
\begin{aligned}
& \text { Mean } \Rightarrow \bar{x}=\frac{\sum_{i=1}^7 x_i}{7}=\frac{\frac{7}{2}[2 a+6 d]}{7}=a+3 d=x_4 \\
& \text { Variance }=\frac{\sum_{i=1}^7\left(x_i-\bar{x}\right)^2}{7}=(4)^2 \Rightarrow \frac{\sum_{i=1}^7\left(x_i-x_4\right)^2}{7}=16 \\
& \Rightarrow \frac{(3 d)^2+(2 d)^2+d^2+0+d^2+(2 d)^2+(3 d)^2}{7}=16 \\
& =4 d^2=16 \Rightarrow d=2 \\
& \Rightarrow \bar{x}=9+3(2)=15 \\
& x_6=a+5 d=9+5(2)=19 \Rightarrow \bar{x}+x_6=34
\end{aligned}
\)
The mean and variance of 5 observations are 5 and 8 respectively. If 3 observations are 1, 3, 5, then the sum of cubes of the remaining two observations is : [JEE Main 2023 (Online) 1st February Morning Shift]
Let observation \(1,3,5, a, b\)
Mean \(=\frac{9+a+b}{5}=5\)
Variance \(=\frac{a^2+b^2+35}{5}-25=8\)
\(\Rightarrow a+b=16\) and \(a^2+b^2=130\)
\(\therefore a\) and \(b\) are 7 and 9
\(
\therefore a^3+b^3=7^3+9^3=1072
\)
Let the mean and standard deviation of marks of class \(A\) of 100 students be respectively 40 and \(\alpha(>0)\), and the mean and standard deviation of marks of class \(B\) of \(n\) students be respectively 55 and \(30-\alpha\). If the mean and variance of the marks of the combined class of \(100+ n\) studants are respectively 50 and 350 , then the sum of variances of classes \(A\) and \(B\) is : [JEE Main 2023 (Online) 31st January Evening Shift]
\(
\begin{array}{cll}
\text { A } & \text { B } & A + B \\
\overline{ x }_1=40 & \overline{ x }_2=55 & \overline{ x }=50
\sigma_1=\alpha & \sigma_2=30-\alpha & \sigma^2=350 \\
n _1=100 & n _2= n & 100+ n
\end{array}
\)
\(
\begin{aligned}
& \overline{ x }=\frac{100 \times 40+55 n }{100+ n } \\
& 5000+50 n =4000+55 n \\
& 1000=5 n \\
& n =200 \\
& \sigma_1^2=\frac{\sum x _{ i }^2}{100}-40^2 \\
& \sigma_2^2=\frac{\sum x _{ j }^2}{100}-55^2
\end{aligned}
\)
\(
\begin{aligned}
& 350=\sigma^2=\frac{\sum x _{ i }^2+\sum x _{ j }^2}{300}-(\overline{ x })^2 \\
& 350=\frac{\left(1600+\alpha^2\right) \times 100+\left[(30-\alpha)^2+3025\right] \times 200}{300}-(50)^2 \\
& 2850 \times 3=\alpha^2+2(30-\alpha)^2+1600+6050 \\
& 8550=\alpha^2+2(30-\alpha)^2+7650 \\
& \alpha^2+2(30-\alpha)^2=900 \\
& \alpha^2-40 \alpha+300=0 \\
& \alpha=10,30 \\
& \sigma_1^2+\sigma_2^2=10^2+20^2=500
\end{aligned}
\)
Let \(S\) be the set of all values of \(a_1\) for which the mean deviation about the mean of 100 consecutive positive integers \(a_1, a_2, a_3, \ldots, a_{100}\) is 25 . Then \(S\) is: [JEE Main 2023 (Online) 30th January Evening Shift]
let \(a_1\) be any natural number
\(
\begin{aligned}
& a_1, a_1+1, a_1+2, \ldots ., a_1+99 \text { are values of } a_i{ }^{\prime} S \\
& \bar{x}=\frac{a_1+\left(a_1+1\right)+\left(a_1+2\right)+\ldots . .+a_1+99}{100} \\
& =\frac{100 a_1+(1+2+\ldots . .+99)}{100}=a_1+\frac{99 \times 100}{2 \times 100} \\
& =a_1+\frac{99}{2}
\end{aligned}
\)
Mean deviation about mean
\(
\begin{aligned}
& =\frac{\sum_{1=1}^{100}\left|x_i-\bar{x}\right|}{100} \\
& =\frac{2\left(\frac{99}{2}+\frac{97}{2}+\frac{95}{2}+\ldots .+\frac{1}{2}\right)}{100} \\
& =\frac{1+3+\ldots . .+99}{100} \\
& =\frac{\frac{50}{2}[1+99]}{100} \\
& =25
\end{aligned}
\)
So, it is true for every natural no. ‘ \(a_1\) ‘.
Three rotten apples are mixed accidently with seven good apples and four apples are drawn one by one without replacement. Let the random variable X denote the number of rotten apples. If \(\mu\) and \(\sigma^2\) represent mean and variance of X , respectively, then \(10\left(\mu^2+\sigma^2\right)\) is equal to :[JEE Main 2023 (Online) 29th January Morning Shift]
3 rotten apples are mixed with 7 good apples.
\(\therefore\) Total apples \(=10\)
Among those 10 apples 4 are chosen randomly.
\(
\begin{array}{|c|c|c|c|}
\hline x_i & p_i & p_i x_i & p_i\left(x_i\right)^2 \\
\hline 0 & \frac{{ }^7 C_4}{{ }^{10} C_4}=\frac{35}{210} & 0 & 0 \\
\hline 1 & \frac{{ }^3 C_1 \times{ }^7 C_3}{{ }^{10} C_4}=\frac{105}{210} & \frac{105}{210} & \frac{105}{210} \\
\hline 2 & \frac{{ }^3 C_2 \times{ }^7 C_2}{{ }^{10} C_4}=\frac{63}{210} & \frac{126}{210} & \frac{252}{210} \\
\hline 3 & \frac{{ }^3 C_3 \times{ }^7 C_1}{{ }^{10} C_4}=\frac{7}{210} & \frac{21}{210} & \frac{63}{210} \\
\hline
\end{array}
\)
\(x_i=\) Number of rotten apples drawn.
\(p_i=\) Probability of rotten apple
We know,
\(
\begin{aligned}
& \text { Mean }(\mu)=\sum p_i x_i \\
& =0+\frac{105}{210}+\frac{126}{210}+\frac{21}{210} \\
& =\frac{252}{210}=\frac{6}{5}
\end{aligned}
\)
Also,
\(
\begin{aligned}
& \text { Variance }\left(\sigma^2\right)=\left(\sum p_i\left(x_i\right)^2\right)-\mu^2 \\
& =\frac{105}{210}+\frac{252}{210}+\frac{63}{210}-\frac{36}{25} \\
& =\frac{1}{2}+\frac{12}{10}+\frac{3}{10}-\frac{36}{25}=\frac{14}{25} \\
& \therefore 10\left(\mu^2+\sigma^2\right) \\
& =10\left(\left(\frac{6}{5}\right)^2+\frac{14}{25}\right) \\
& =10\left(\frac{36+14}{25}\right) \\
& =10 \times \frac{50}{25}=20
\end{aligned}
\)
The mean and variance of the marks obtained by the students in a test are 10 and 4 respectively. Later, the marks of one of the students is increased from 8 to 12 . If the new mean of the marks is 10.2 , then their new variance is equal to : [JEE Main 2023 (Online) 25th January Morning Shift]
\(
\begin{aligned}
& \bar{x}=10 \& \sigma^2=4, \text { No. of students }=N \text { (let) } \\
& \therefore \quad \frac{\sum x_i}{N}=10 \& \frac{\sum x_i^2}{N}-(10)^2=4
\end{aligned}
\)
Now if one of \(x_i\) is changed from 8 to 12 we have
New mean \(\frac{\sum x_i+4}{N}=10+\frac{4}{N}=10.2\)
\(
\begin{aligned}
& \Rightarrow N=20 \\
& \text { and } \sigma_{\text {new }}^2=\frac{\sum x_i^2-(8)^2+(12)^2}{20}-(10 \cdot 2)^2 \\
& =\frac{\sum x_i^2}{20}+\frac{144-64}{20}-(10 \cdot 2)^2 \\
& =104+4-(10 \cdot 2)^2 \\
& =108-104.04=3.96
\end{aligned}
\)
Let the six numbers \(a_1, a_2, a_3, a_4, a_5, a_6\), be in A.P. and \(a_1+a_3=10\). If the mean of these six numbers is \(\frac{19}{2}\) and their variance is \(\sigma^2\), then \(8 \sigma^2\) is equal to : [JEE Main 2023 (Online) 24th January Evening Shift]
\(a_1, a_2, a_3, a_4, a_5, a_6\) are in AP.
Let
\(
\begin{aligned}
& a_1=a \\
& a_2=a+d \\
& a_3=a+2 d \\
& a_4=a+3 d \\
& a_5=a+4 d \\
& a_6=a+5 d
\end{aligned}
\)
Now Mean of \(a_1, a_2, a_3, a_4, a_5\) and \(a_6\) is
\(
\begin{aligned}
& =\frac{a_1+a_2+a_3+a_4+a_5+a_6}{6}=\frac{19}{2} \\
& \Rightarrow \frac{6 a+15 d}{6}=\frac{19}{2} \\
& \Rightarrow 2 a+5 d=19 \ldots \ldots \text { (1) }
\end{aligned}
\)
Also, given,
\(
\begin{aligned}
& a_1+a_3=10 \\
& \Rightarrow a+a+2 d=10 \\
& \Rightarrow 2 a+2 d=10 \\
& \Rightarrow a+d=5 \ldots . .(2)
\end{aligned}
\)
From equation (1) and (2), we get \(a=2\) and \(d=3\)
\(\therefore\) AP is \(2,5,8,11,14,17\)
Now, Variance \(\left(\sigma^2\right)=\frac{\sum x_i^2}{6}-(\bar{x})^2\)
\(=\frac{2^2+5^2+8^2+11^2+14^2+17^2}{6}-\left(\frac{19}{2}\right)^2\)
\(=\frac{105}{4}\)
\(\therefore 8 \sigma^2=8 \times \frac{105}{4}=210\)
If the mean deviation about median for the numbers \(3,5,7,2 k, 12,16,21,24\), arranged in the ascending order, is 6 then the median is : [JEE Main 2022 (Online) 25th July Evening Shift]
\(
\begin{aligned}
& \text { Median }=\frac{2 k+12}{2}=k+6 \\
& \text { Mean deviation }=\sum \frac{\left|x_i-M\right|}{n}=6 \\
& \Rightarrow \frac{(k+3)+(k+1)+(k-1)+(6-k)+(6-k)+(10-k)+(15-k)+(18-k)}{8} \\
& \therefore \frac{58-2 k}{8}=6 \\
& k=5 \\
& \text { Median }=\frac{2 \times 5+12}{2}=11
\end{aligned}
\)
The number of values of \(a \in N\) such that the variance of \(3,7,12, a, 43-a\) is a natural number is : [JEE Main 2022 (Online) 29th June Evening Shift]
Mean \(=13\)
Variance
\(
\begin{aligned}
& =\frac{9+49+144+ a ^2+(43- a )^2}{5}-13^2 \in N \\
& \Rightarrow \frac{2 a ^2- a +1}{5} \in N
\end{aligned}
\)
\(\Rightarrow 2 a ^2- a +1-5 n =0\) must have solution as natural numbers
its \(D=40 n-7\) always has 3 at unit place
\(\Rightarrow\) D can’t be perfect square
D cannot be a perfect square as all perfect squares will be of the form of \(4 \lambda\) or \(4 \lambda+1\)
So, a cannot be natural number
So, a can’t be integer.
\(\therefore \quad\) Number of values \(=0\)
Let the mean and the variance of 5 observations \(x _1, x _2, x _3, x _4, x _5\) be \(\frac{24}{5}\) and \(\frac{194}{25}\) respectively. If the mean and variance of the first 4 observation are \(\frac{7}{2}\) and a respectively, then \(\left(4 a+x_5\right)\) is equal to: [JEE Main 2022 (Online) 29th June Morning Shift]
Mean \((\bar{x})=\frac{x_1+x_2+x_3+x_4+x_5}{5}\)
Given, \(\frac{x_1+x_2+x_3+x_4+x_5}{5}=\frac{24}{5}\)
\(
\Rightarrow x_1+x_2+x_3+x_4+x_5=24 \dots(1)
\)
Now, Mean of first 4 observation
\(
=\frac{x_1+x_2+x_3+x_4}{4}
\)
Given, \(=\frac{x_1+x_2+x_3+x_4}{4}=\frac{7}{2}\)
\(
\Rightarrow x_1+x_2+x_3+x_4=14 \dots(2)
\)
From equation (1) and (2), we get
\(
\begin{aligned}
& 14+x_5=24 \\
& \Rightarrow x_5=10
\end{aligned}
\)
Now, variance of first 5 observation
\(
\begin{aligned}
& =\frac{\sum x_i^2}{n}-(\bar{x})^2 \\
& =\frac{x_1^2+x_2^2+x_3^2+x_4^2+x_5^2}{5}-\left(\frac{24}{5}\right)^2
\end{aligned}
\)
Given,
\(
\begin{aligned}
& \frac{x_1^2+x_2^2+x_3^2+x_4^2+x_5^2}{5}-\left(\frac{24}{5}\right)^2=\frac{194}{24} \\
& \Rightarrow x_1^2+x_2^2+x_3^2+x_4^2+x_5^2=5\left(\frac{194}{25}+\frac{576}{25}\right) \\
& \Rightarrow x_1^2+x_2^2+x_3^2+x_4^2+x_5^2=154 \\
& \Rightarrow x_1^2+x_2^2+x_3^2+x_4^2+(10)^2=154 \\
& \Rightarrow x_1^2+x_2^2+x_3^2+x_4^2=54
\end{aligned}
\)
Now, variance of first 4 observation
\(
=\frac{x_1^2+x_2^2+x_3^2+x_4^2}{4}-\left(\frac{7}{2}\right)^2
\)
Given,
\(
\begin{aligned}
& \frac{x_1^2+x_2^2+x_3^2+x_4^2}{4}-\left(\frac{7}{2}\right)^2=a \\
& \Rightarrow \frac{54}{4}-\frac{49}{4}=a \\
& \Rightarrow a=\frac{5}{4} \\
& \therefore 4 a+x_5 \\
& =4 \times \frac{5}{4}+10=15
\end{aligned}
\)
The mean and variance of the data \(4,5,6,6,7,8, x, y\), where \(x<y\), are 6 and \(\frac{9}{4}\) respectively. Then \(x^4+y^2\) is equal to : [JEE Main 2022 (Online) 27th June Evening Shift]
\(
\begin{aligned}
& \text { Mean }=\frac{4+5+6+6+7+8+x+y}{8}=6 \\
& \therefore x+y=12 \ldots \ldots \text { (i) }
\end{aligned}
\)
And variance
\(
\begin{aligned}
& =\frac{2^2+1^2+0^2+0^2+1^2+2^2+(x-6)^2+(y-6)^2}{8} \\
& =\frac{9}{4} \\
& \therefore(x-6)^2+(y-6)^2=8 \ldots . . \text { (ii) }
\end{aligned}
\)
From (i) and (ii)
\(
\begin{aligned}
& x =4 \text { and } y =8 \\
& \therefore x^4+y^2=320
\end{aligned}
\)
The mean and standard deviation of 50 observations are 15 and 2 respectively. It was found that one incorrect observation was taken such that the sum of correct and incorrect observations is 70 . If the correct mean is 16 , then the correct variance is equal to : [JEE Main 2022 (Online) 26th June Evening Shift]
\(
\begin{aligned}
& \text { Given } \bar{x}=15, \sigma=2 \Rightarrow \sigma^2=4 \\
& \therefore x_2+x_2+\ldots \ldots+x_{50}=15 \times 50=750 \\
& 4=\frac{x_1^2+x_2^2+\ldots . .+x_{50}^2}{50}-225 \\
& \therefore x_1^2+x_2^2+\ldots .+x_{50}^2=50 \times 229
\end{aligned}
\)
Let \(a\) be the correct observation and \(b\) is the incorrect observation then \(a+b=70\) and \(16=\frac{750-b+a}{50}\)
\(
\therefore a-b=50 \Rightarrow a=60, b=10
\)
\(\therefore\) Correct variance \(=\frac{50 \times 229+60^2-10^2}{50}-256=43\)
The mean of the numbers \(a, b, 8,5,10\) is 6 and their variance is 6.8 . If \(M\) is the mean deviation of the numbers about the mean, then 25 M is equal to : [JEE Main 2022 (Online) 26th June Morning Shift]
\(
\because \bar{x}=6=\frac{a+b+8+5+10}{5} \Rightarrow a+b=7 \dots(i)
\)
And \(\sigma^2=\frac{a^2+b^2+8^2+5^2+10^2}{5}-6^2=6.8\)
\(
\Rightarrow a^2+b^2=25 \ldots . . \text { (ii) }
\)
From (i) and (ii) \(( a , b )=(3,4)\) or \((4,3)\)
Now mean deviation about mean
\(
\begin{aligned}
& M=\frac{1}{5}(3+2+2+1+4)=\frac{12}{5} \\
& \Rightarrow 25 M=60
\end{aligned}
\)
The mean and variance of 7 observations are 8 and 16 respectively. If two observations are 6 and 8 , then the variance of the remaining 5 observations is : [JEE Main 2021 (Online) 31st August Evening Shift]
Let \(8,16, x_1, x_2, x_3, x_4, x_5\) be the observations.
Now, \(\frac{x_1+x_2+\ldots .+x_5+14}{7}=8\)
\(
\Rightarrow \sum_{i=1}^5 x_i=42 \ldots(1)
\)
Also, \(\frac{x_1^2+x_2^2+\ldots x_5^2+8^2+6^2}{7}-64=16\)
\(
\Rightarrow \sum_{i=1}^5 x_i^2=560-100=460 \ldots . .(2)
\)
So variance of \(x _1, x _2, \ldots \ldots, x _5\)
\(
=\frac{460}{5}-\left(\frac{42}{5}\right)^2=\frac{2300-1764}{25}=\frac{536}{25}
\)
The mean and standard deviation of 20 observations were calculated as 10 and 2.5 respectively. It was found that by mistake one data value was taken as 25 instead of 35 . if \(\alpha\) and \(\sqrt{\beta}\) are the mean and standard deviation respectively for correct data, then \((\alpha, \beta)\) is [JEE Main 2021 (Online) 26th August Morning Shift]
Given :
Mean \((\bar{x})=\frac{\sum x_i}{20}=10\)
or \(\Sigma x _{ i }=200\) (incorrect)
or \(200-25+35=210=\Sigma x _{ i }\) (Correct)
Now correct \(\bar{x}=\frac{210}{20}=10.5\)
again given \(S . D=2.5(\sigma)\)
\(\sigma^2=\frac{\sum x_i{ }^2}{20}-(10)^2=(2.5)^2\)
or \(\sum x_i{ }^2=2125\) (incorrect)
or \(\sum x_i{ }^2=2125-25^2+35^2\)
\(=2725\) (correct)
\(\therefore\) correct \(\sigma^2=\frac{2725}{20}-(10.5)^2\)
\(
\sigma^2=26
\)
or \(\sigma=26\)
\(
\therefore \alpha=10.5, \beta=26
\)
Let the mean and variance of the frequency distribution
\(
\begin{array}{ccccc}
x: & x_1=2 & x_2=6 & x_3=8 & x_4=9 \\
f: & 4 & 4 & \alpha & \beta
\end{array}
\)
be 6 and 6.8 respectively. If \(x_3\) is changed from 8 to 7 , then the mean for the new data will be: [JEE Main 2021 (Online) 27th July Evening Shift]
Given \(32+8 \alpha+9 \beta=(8+\alpha+\beta) \times 6\)
\(
\Rightarrow 2 \alpha+3 \beta=16 \ldots . .(i)
\)
Also, \(4 \times 16+4 \times \alpha+9 \beta=(8+\alpha+\beta) \times 6.8\)
\(
\begin{aligned}
& \Rightarrow 640+40 \alpha+90 \beta=544+68 \alpha+68 \beta \\
& \Rightarrow 28 \alpha-22 \beta=96 \\
& \Rightarrow 14 \alpha-11 \beta=48 \ldots . . \text { (ii) }
\end{aligned}
\)
from (i) & (ii)
\(
\alpha=5 \& \beta=2
\)
So, new mean \(=\frac{32+35+18}{15}=\frac{85}{15}=\frac{17}{3}\)
If the mean and variance of the following data : \(6,10,7,13, a, 12, b, 12\) are 9 and \(\frac{37}{4}\) respectively, then \((a-b)^2\) is equal to : [JEE Main 2021 (Online) 27th July Morning Shift]
\(
\begin{aligned}
& \text { Mean }=\frac{6+10+7+13+a+12+b+12}{8}=9 \\
& 60+a+b=72 \\
& a+b=12 \ldots . .(1)
\end{aligned}
\)
\(
\begin{aligned}
& \text { variance }=\frac{\sum x_i^2}{n}-\left(\frac{\sum x_i}{n}\right)^2=\frac{37}{4} \\
& \sum x_i^2=6^2+10^2+7^2+13^2+a^2+b^2+12^2+12^2=a^2+b^2+642 \\
& \frac{a^2+b^2+642}{8}-(9)^2=\frac{37}{4} \\
& \frac{a^2+b^2}{8}+\frac{321}{4}-81=\frac{37}{4} \\
& \frac{a^2+b^2}{8}=81+\frac{37}{4}-\frac{321}{4} \\
& \frac{a^2+b^2}{8}=81-71 \\
& \therefore a^2+b^2+2 a b=144
\end{aligned}
\)
\(
\begin{aligned}
&80+2 a b=144\\
&\begin{aligned}
& \therefore 2 a b=64 \\
& \therefore(a-b)^2=a^2+b^2-2 a b=80-64=16
\end{aligned}
\end{aligned}
\)
The first of the two samples in a group has 100 items with mean 15 and standard deviation 3 . If the whole group has 250 items with mean 15.6 and standard deviation \(\sqrt{13.44}\), then the standard deviation of the second sample is : [JEE Main 2021 (Online) 25th July Evening Shift]
\(
\begin{aligned}
& n _1=100 \\
& m=250 \\
& \bar{X}_1=15 \\
& \bar{X}=15.6 \\
& V_1( x )=9 \\
& \operatorname{Var}( x )=13.44 \\
& \sigma^2=\frac{n_1 \sigma_1^2+n_2 \sigma_2^2}{n_1+n_2}+\frac{n_1 n_2}{\left(n_1+n_2\right)^2}\left(\bar{x}_1-\bar{x}_2\right)^2 \\
& n _2=150, \bar{x}_2=16, V_2( x )=\sigma_2 \\
& 13.44=\frac{100 \times 9+150 \times \sigma_2^2}{250}+\frac{100 \times 150}{(250)^2} \times 1 \\
& \Rightarrow \sigma_2{ }^2=16 \Rightarrow \sigma_2=4
\end{aligned}
\)
If the mean and variance of six observations \(7,10,11,15, a, b\) are 10 and \(\frac{20}{3}\), respectively, then the value of \(|a-b|\) is equal to : [JEE Main 2021 (Online) 20th July Evening Shift]
\(
\begin{aligned}
& 10=\frac{7+10+11+15+a+b}{6} \\
& \Rightarrow a + b =17 \ldots . \text { (i) } \\
& \frac{20}{3}=\frac{7^2+10^2+11^2+15^2+a^2+b^2}{6}-10^2 \\
& a ^2+ b ^2=145 \ldots \ldots \text { (ii) }
\end{aligned}
\)
Solve (i) and (ii) \(a=9, b=8\) or \(a=8, b=9\)
\(
|a-b|=1
\)
The mean of 6 distinct observations is 6.5 and their variance is 10.25 . If 4 out of 6 observations are \(2,4,5\) and 7 , then the remaining two observations are : [JEE Main 2021 (Online) 20th July Morning Shift]
Let other two numbers be \(a ,(21- a )\)
Now,
\(
10.25=\frac{\left(4+16+25+49+a^2+(21-a)^2\right.}{6}-(6.5)^2
\)
(Using formula for variance)
\(
\begin{aligned}
& \Rightarrow 6(10.25)+6(6.5)^2=94+a^2+(21-a)^2 \\
& \Rightarrow a^2+(21-a)^2=221 \\
& \therefore a=10 \text { and }(21-a)=21-10=11
\end{aligned}
\)
So, remaining two observations are 10,11.
Let in a series of \(2 n\) observations, half of them are equal to \(a\) and remaining half are equal to \(-a\). Also by adding a constant \(b\) in each of these observations, the mean and standard deviation of new set become 5 and 20 , respectively. Then the value of \(a^2+b^2\) is equal to : [JEE Main 2021 (Online) 18th March Evening Shift]
Given series
\((a, a, a\), \(\ldots\) \(n\) times \(),(-a,-a,-a\), \(\ldots\) n times)
Now \(\bar{x}=\frac{\sum x_i}{2 n}=0\)
as, \(x_i \rightarrow x_i+b\)
then \(\bar{x} \rightarrow \bar{x}+ b\)
So, \(\bar{x}+b=5 \Rightarrow b=5\)
No change in S.D. due to change in origin
Standard deviation \((\sigma)=\sqrt{\frac{\sum_{i=1}^{2 n}\left(x_i-\bar{x}\right)^2}{2 n}}\)
\(
\begin{aligned}
& =\sqrt{\frac{\sum_{i=1}^{2 n} x_i^2}{2 n}}=\sqrt{\frac{2 n a^2}{2 n}}=\sqrt{a^2} \\
& \therefore 20=\sqrt{a^2} \Rightarrow a=20 \\
& \therefore a ^2+ b ^2=425
\end{aligned}
\)
Consider three observations \(a , b\), and c such that \(b = a + c\). If the standard deviation of \(a +\) \(2, b+2, c+2\) is \(d\), then which of the following is true? [JEE Main 2021 (Online) 16th March Morning Shift]
For \(a , b , c\)
mean \(=\bar{x}=\frac{a+b+c}{3}\)
\(
\bar{x}=\frac{2 b}{3}
\)
We know, S.D. of \(a+2, b+2, c+2=\) S.D. of \(a, b, c=d\)
\(
\begin{aligned}
& d^2=\frac{a^2+b^2+c^2}{3}-\frac{4 b^2}{9} \\
& b^2=3 a^2+3 c^2-9 d^2
\end{aligned}
\)
If \(\sum_{i=1}^n\left(x_i-a\right)=n\) and \(\sum_{i=1}^n\left(x_i-a\right)^2=n a\)
\((n, a>1\) ) then the standard deviation of \(n\) observations \(x _1, x _2, \ldots, x _{ n }\) is : [JEE Main 2020 (Online) 6th September Morning Slot]
\(
\begin{aligned}
& S . D =\sqrt{\frac{\sum_{i=1}^n\left(x_i-a\right)}{n}-\left(\frac{\sum_{i=1}^n\left(x_i-a\right)}{n}\right)^2} \\
& =\sqrt{\frac{n a}{n}-\left(\frac{n}{n}\right)^2} \\
& =\sqrt{a-1}
\end{aligned}
\)
If the mean and the standard deviation of the data \(3,5,7, a, b\) are 5 and 2 respectively, then \(a\) and \(b\) are the roots of the equation : [JEE Main 2020 (Online) 5th September Evening Slot]
\(
\begin{aligned}
& \text { Mean }=\frac{3+5+7+a+b}{5}=5 \\
& \Rightarrow a+ b =10
\end{aligned}
\)
\(
\begin{aligned}
& \text { Variance }=\frac{3^2+5^2+7^2+a^2+b^2}{5}-(5)^2=4 \\
& \Rightarrow a^2+b^2=62 \\
& \Rightarrow(a+b)^2-2 a b=62 \\
& \Rightarrow a b=19
\end{aligned}
\)
So \(a\) and b are the roots of the equation
\(
x^2-10 x+19=0
\)
The mean and variance of 7 observations are 8 and 16 , respectively. If five observations are \(2,4,10,12,14\), then the absolute difference of the remaining two observations is : [JEE Main 2020 (Online) 5th September Morning Slot]
\(
\begin{aligned}
& \bar{x}=\frac{2+4++10+12+14+x+y}{7}=8 \\
& x+y=14 \ldots . \text { (i) } \\
& (\sigma)^2=\frac{\sum\left(x_i\right)^2}{n}-\left(\frac{\sum x_i}{n}\right)^2 \\
& \Rightarrow 16=\frac{4+16+100+144+196+x^2+y^2}{2}-8^2 \\
& \Rightarrow 16+64=\frac{460+x^2+y^2}{7} \\
& \Rightarrow 560=460+x^2+y^2 \\
& \Rightarrow x^2+y^2=100 \ldots \ldots . \text { (ii) }
\end{aligned}
\)
Clearly by (i) and (ii),
\(
\begin{aligned}
& (x+y)^2-2 x y=100 \\
& \Rightarrow(14)^2-2 x y=100 \\
& \Rightarrow 2 x y=96 \\
& \Rightarrow x y=48
\end{aligned}
\)
Now, \(| x – y |=\sqrt{(x+y)^2-4 x y}\)
\(
\begin{aligned}
& =\sqrt{196-192} \\
& =2
\end{aligned}
\)
The mean and variance of 8 observations are 10 and 13.5, respectively. If 6 of these observations are \(5,7,10,12,14,15\), then the absolute difference of the remaining two observations is : [JEE Main 2020 (Online) 4th September Morning Slot]
Let the two remaining observations be x and y .
\(
\begin{aligned}
& \because \bar{x}=10=\frac{5+7+10+12+14+15+x+y}{8} \\
& \Rightarrow x+y=17 \ldots .(1) \\
& \because \operatorname{var}(x)=13.5=\frac{25+49+100+144+196+225+x^2+y^2}{8}-(10)^2 \\
& \Rightarrow x^2+y^2=169 \ldots .(2)
\end{aligned}
\)
From (1) and (2)
\(
(x, y)=(12,5) \text { or }(5,12)
\)
So \(|x-y|=7\)
Let \(x_i(1 \leq i \leq 10)\) be ten observations of a random variable \(X\). If \(\sum_{i=1}^{10}\left(x_i-p\right)=3\) and \(\sum_{i=1}^{10}\left(x_i-p\right)^2=9\)
where \(0 \neq p \in R\), then the standard deviation of these observations is : [JEE Main 2020 (Online) 3rd September Evening Slot]
\(
\begin{aligned}
& \text { Standard deviation }=\sqrt{\text { Variance }} \\
& =\sqrt{\frac{\sum x_1^2}{n}-(\bar{x})^2} \\
& =\sqrt{\frac{\sum_{i=1}^{10}\left(x_i-p\right)^2}{10}-\left(\frac{\sum_{i=1}^{10}\left(x_i-p\right)}{10}\right)^2}
\end{aligned}
\)
[ Standard deviation is free from shifting of origin.]
\(
\begin{aligned}
& =\sqrt{\frac{9}{10}-\left(\frac{3}{10}\right)^2} \\
& =\sqrt{\frac{9}{10}-\frac{9}{100}} \\
& =\sqrt{\frac{90-9}{100}} \\
& =\sqrt{\frac{81}{100}} \\
& =\frac{9}{10}
\end{aligned}
\)
For the frequency distribution:
Variate \(( x ): \begin{array}{llllll} x _1 & x _2 & x _3 & \ldots & x _{15}\end{array}\)
Frequency (f): \(f_1 \quad f_2 \quad f_3 \ldots . . f_{15}\)
where \(0<x_1<x_2<x_3<\ldots<x_{15}=10\) and \(\sum_{i=1}^{15} f_i>0\), the standard deviation cannot be : [JEE Main 2020 (Online) 3rd September Morning Slot]
If variate varries from \(m\) to \(M\) then variance
\(
\sigma^2 \leq \frac{1}{4}(M-m)^2
\)
( \(M =\) upper bound of value of any random variable,
\(m =\) Lower bound of value of any random variable)
Here \(M=10\) and \(m=0\)
\(
\begin{aligned}
& \therefore \sigma^2 \leq \frac{1}{4}(10-0)^2 \\
& \Rightarrow \sigma^2 \leq 25 \\
& \Rightarrow-5 \leq \sigma \leq 5 \\
& \therefore \sigma \neq 6
\end{aligned}
\)
Let \(X=\{x \in N: 1 \leq x \leq 17\}\) and \(Y=\{a x+b: x \in X\) and \(a, b \in R, a>0\}\). If mean and variance of elements of \(Y\) are 17 and 216 respectively then \(a+b\) is equal to : [JEE Main 2020 (Online) 2nd September Morning Slot]
Mean of \(X=\frac{\sum_{x=1}^{17} x}{17}=\frac{17 \times 18}{17 \times 2}=9\)
Mean of \(Y =\frac{\sum_{x=1}^{17}(a x+b)}{17}=17\)
\(
\begin{aligned}
& \Rightarrow a \frac{\sum_{x=1}^{17} x}{17}+b=17 \\
& \Rightarrow 9 a+b=17 \ldots . .(1) \\
&
\end{aligned}
\)
Given \(\operatorname{Var}( Y )=216\)
\(
\begin{aligned}
& \Rightarrow \frac{\sum_{x=1}^{17}(a x+b)^2}{17}-(17)^2=216 \\
& \Rightarrow \sum_{x=1}^{17}(a x+b)^2=8585 \\
& \Rightarrow(a+b)^2+(2 a+b)^2+\ldots .+(17 a+b)^2=8585 \\
& \Rightarrow 105 a^2+b^2+18 a b=505 \ldots .(2)
\end{aligned}
\)
From equation (1) & (2)
\(
\begin{aligned}
& a=3 \& b=-10 \\
& \therefore a+b=-7
\end{aligned}
\)
Let the observations \(x_i(1 \leq i \leq 10)\) satisfy the equations, \(\sum_{i=1}^{10}\left(x_1-5\right)=10\) and \(\sum_{i=1}^{10}\left(x_1-5\right)^2=40\). If \(\mu\) and \(\lambda\) are the mean and the variance of the observations, \(x _1-3, x _2-3, \ldots ., x _{10}-3\), then the ordered pair \((\mu, \lambda)\) is equal to : [JEE Main 2020 (Online) 9th January Morning Slot]
\(
\begin{aligned}
&\begin{aligned}
& \sum_{i=1}^{10}\left(x_1-5\right)=10 \\
& \Rightarrow x _1+ x _2+\ldots .+ x _{10}=60 \ldots(1)
\end{aligned}\\
&\sum_{i=1}^{10}\left(x_1-5\right)^2=40
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow\left(x_1^2+x_2^2+\ldots+x_{10}^2\right)+25 \times 10- \\
& 10\left( x _1+ x _2+\ldots .+ x _{10}\right)=40 \\
& \Rightarrow x_1^2+x_2^2+\ldots+x_{10}^2=390 \ldots . .(2)
\end{aligned}
\)
From question,
\(
\begin{aligned}
& \mu=\frac{\left(x_1-3\right)+\left(x_2-3\right)+\ldots+\left(x_{10}-3\right)}{10} \\
& =\frac{60-3 \times 10}{10}=3
\end{aligned}
\)
And \(\lambda=\) variance \(=\frac{\sum_{i=1}^{10}\left(x_i-3\right)^2}{10}-\mu^2\)
\(
\begin{aligned}
& =\frac{\left(x_1^2+x_2^2+\ldots+x_{10}^2\right)+90-6\left(\sum x_i\right)}{10}-9 \\
& =\frac{390+90-360}{10}-9 \\
& =12-9=3
\end{aligned}
\)
The mean and variance of 20 observations are found to be 10 and 4, respectively. On rechecking, it was found that an observation 9 was incorrect and the correct observation was 11. Then the correct variance is [JEE Main 2020 (Online) 8th January Evening Slot]
Let 20 observation be \(x _1, x _2, \ldots . ., x _{20}\)
\(
\begin{aligned}
& \text { Mean }=\frac{x_1+x_2+, \ldots .+x_{20}}{20}=10 \\
& \Rightarrow x_1+x_2+, \ldots .+x_{20}=200
\end{aligned}
\)
\(
\begin{aligned}
& \text { Variance }=\frac{\sum_{i=1}^{i=n} x_i^2}{n}-(\bar{x})^2 \\
& \Rightarrow 4=\frac{x_1^2+x_2^2+\ldots+x_{20}^2}{20}-10^2 \\
& \Rightarrow x_1^2+x_2^2+\ldots+x_{20}^2=2080
\end{aligned}
\)
Also \(x_1+x_2+\ldots . .+x_{20}-9+11=202\)
new variance will be
\(
\begin{aligned}
& =\frac{x_1^2+x_2^2+\ldots+x_{20}^2-81+121}{20}-\left(\frac{202}{20}\right)^2 \\
& =3.99
\end{aligned}
\)
The mean and the standard deviation (s.d.) of 10 observations are 20 and 2 resepectively. Each of these 10 observations is multiplied by p and then reduced by q , where \(p \neq 0\) and q \(\neq 0\). If the new mean and new s.d. become half of their original values, then \(q\) is equal to [JEE Main 2020 (Online) 8th January Morning Slot]
Let observations are \(x _1, x _2, \ldots, x _{10}\)
Here mean \(=20\) and standard deviation(S.D) \(=2\)
When each of these 10 observations is multiplied by p then new observations are \(px _1, px _2\), ….., \(px _{10}\)
and new mean \(=20\) p and new standard deviation(S.D) \(=2|p|\)
Now when Reduced by q then new observations are
\(
px _1- q , px _2- q , \ldots . ., px _{10}- q
\)
and new mean \(=20 p – q\) and new standard deviation(S.D) \(=2| p |\)
Given \(20 p – q =\frac{20}{2}=10\)
and \(2|p|=\frac{2}{2}=1\)
\(
\Rightarrow p = \pm \frac{1}{2}
\)
If \(p =\frac{1}{2}\) then \(q =0\) (not possible as given \(q \neq 0\) )
If \(p=-\frac{1}{2}\) then \(q=-20\)
If the data \(x_1, x_2, \ldots \ldots, x_{10}\) is such that the mean of first four of these is 11 , the mean of the remaining six is 16 and the sum of squares of all of these is 2,000 ; then the standard deviation of this data is : [JEE Main 2019 (Online) 12th April Morning Slot]
\(
\sigma^2=\frac{\sum x_i^2}{10}-\left(\frac{\sum x_i}{10}\right)^2 \rightarrow(i)
\)
Now \(x_1+x_2+x_3+x_4=44 \& x_5+x_6+\) \(\ldots\) \(+ x _{10}=96\)
Hence \(\sigma^2=\frac{2000}{10}-\left(\frac{140}{10}\right)^2=200-196=4\)
Hence \(\sigma=2\)
If both the mean and the standard deviation of 50 observations \(x _1, x _2, \ldots, x _{50}\) are equal to 16 , then the mean of \(\left(x_1-4\right)^2,\left(x_2-4\right)^2, \ldots . .,\left(x_{50}-4\right)^2\) is: [JEE Main 2019 (Online) 10th April Evening Slot]
\(
\begin{aligned}
& \operatorname{Mean}(\mu)=\frac{\sum x_i}{50}=16 \\
& \therefore \sum x_i=16 \times 50
\end{aligned}
\)
S. D. \((\sigma)=\sqrt{\frac{\sum x_i^2}{50}-(\mu)^2}=16\)
\(
\Rightarrow \frac{\sum x_i{ }^2}{50}=256 \times 2
\)
Required mean \(=\frac{\sum\left(x_i-4\right)^2}{50}\)
\(
\begin{aligned}
& \Rightarrow \frac{\sum x_i{ }^2+16 \times 50-8 \sum x_i}{50} \\
& \Rightarrow 256 \times 2+16-8 \times 16 \\
& \Rightarrow 400
\end{aligned}
\)
If for some \(x \in R\), the frequency distribution of the marks obtained by 20 students in a test is :
\(
\begin{array}{|c|c|c|c|c|}
\hline \text { Marks } & 2 & 3 & 5 & 7 \\
\hline \text { Frequency } & (x+1)^2 & 2 x-5 & x^2-3 x & x \\
\hline
\end{array}
\)
then the mean of the marks is [JEE Main 2019 (Online) 10th April Morning Slot]
Number of students
\(
\begin{aligned}
& \Rightarrow(x+1)^2+(2 x-5)+\left(x^2-3 x\right)+x=20 \\
& \Rightarrow 2 x^2+2 x-4=20 \\
& \Rightarrow x^2+x-12=0 \\
& \Rightarrow(x+4)(x-3)=0 \\
& x=3
\end{aligned}
\)
\(
\begin{array}{|l|c|c|c|c|}
\hline \text { Marks } & 2 & 3 & 5 & 7 \\
\hline \text { No. of students } & 16 & 1 & 0 & 3 \\
\hline
\end{array}
\)
\(
\text { Average marks }=\frac{32+3+21}{20}=\frac{56}{20}=2.8
\)
The mean and the median of the following ten numbers in increasing order \(10,22,26,29\), \(34, x , 42,67,70, y\) are 42 and 35 respectively, then \(\frac{y}{x}\) is equal to [JEE Main 2019 (Online) 9th April Evening Slot]
Given ten numbers are \(10,22,26,29,34, x, 42,67,70, y\).
As the numbers are in increasing order so
\(
\begin{aligned}
& \text { Mediun }=\frac{34+x}{2}=35 \\
& \Rightarrow x =36
\end{aligned}
\)
Also given mean \(=42\)
\(
\begin{aligned}
& \Rightarrow \frac{10+22+26+29+34+x+42+67+70+y}{10}=42 \\
& \Rightarrow \frac{300+x+y}{10}=42 \\
& \Rightarrow 300+ x + y =420 \\
& \Rightarrow 336+ y =420 \\
& \Rightarrow y =84 \\
& \therefore \frac{y}{x}=\frac{84}{36}=\frac{7}{3}
\end{aligned}
\)
If the standard deviation of the numbers \(-1,0,1, k\) is \(\sqrt{5}\) where \(k>0\), then \(k\) is equal to [JEE Main 2019 (Online) 9th April Morning Slot]
\(
\begin{aligned}
& \text { standard deviation }=\sqrt{5} \\
& \therefore \text { Variance }=(\sqrt{5})^2=5 \\
& \text { Also variance }=\frac{\sum x_i^2}{N}-\mu^2 \\
& \text { Where } \mu=\text { Mean }=\frac{-1+0+1+k}{4}=\frac{k}{4} \\
& \therefore \text { Variance }=\frac{(-1)^2+0+1^2+k^2}{4}-\frac{k^2}{16} \\
& \Rightarrow 5=\frac{2+k^2}{4}-\frac{k^2}{16} \\
& \Rightarrow 8+3 k ^2=80 \\
& \Rightarrow k ^2=24=2 \sqrt{6}
\end{aligned}
\)
A student scores the following marks in five tests :
\(
45,54,41,57,43 \text {. }
\)
His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six tests is [JEE Main 2019 (Online) 8th April Evening Slot]
Let the score in the sixth test \(= x\)
Given, Mean \((x)=48\)
\(
\begin{aligned}
& \Rightarrow \frac{45+54+41+57+43+x}{6}=48 \\
& \Rightarrow x=48
\end{aligned}
\)
Standard deviation (SD)
\(
\begin{aligned}
& =\sqrt{\frac{\sum_{i=1}^N\left(x_i-\bar{x}\right)^2}{N}} \\
& =\sqrt{\begin{array}{l}
(45-48)^2+(54-48)^2 \\
+(41-48)^2+(57-48)^2 \\
\frac{+(43-48)^2+(48-48)^2}{6}
\end{array}} \\
&
\end{aligned}
\)
\(
\begin{aligned}
& =\sqrt{\frac{9+36+49+81+25}{6}} \\
& =\sqrt{\frac{200}{6}} \\
& =\sqrt{\frac{100}{3}} \\
& =\frac{10}{\sqrt{3}}
\end{aligned}
\)
The mean and variance of seven observations are 8 and 16 , respectively. If 5 of the observations are \(2,4,10,12,14\), then the product of the remaining two observations is : [JEE Main 2019 (Online) 8th April Morning Slot]
Given mean \((\mu)=8\)
variance \(\left(\sigma^2\right)=16\)
No of observations \(( N )=7\)
Let the two unknown observation \(= x\) and y
We know,
\(
\begin{aligned}
& \sigma^2=\frac{\sum x_i^2}{N}-\mu^2=16 \\
& \Rightarrow \frac{2^2+4^2+10^2+12^2+14^2+x^2+y^2}{7}-(8)^2=16 \\
& \Rightarrow x^2+y^2=100 \text {……(1) }
\end{aligned}
\)
We know,
\(
\begin{aligned}
& \mu=\frac{\sum x_i}{N}=8 \\
& \Rightarrow \frac{2+4+10+12+14+x+y}{7}=8
\end{aligned}
\)
\(
\Rightarrow x+y=14 \dots(2)
\)
As \((x+y)^2=x^2+y^2+2 x y\)
\(
\begin{aligned}
& \Rightarrow(14)^2=100+2 x y \\
& \Rightarrow 196=100+2 x y \\
& \Rightarrow x y=48
\end{aligned}
\)
The mean and the variance of five observations are 4 and 5.20 , respectively. If three of the observations are 3,4 and 4 ; then the absolute value of the difference of the other two observations, is : [JEE Main 2019 (Online) 12th January Evening Slot]
mean \(\bar{x}=4, \sigma^2=5.2, n=5, \cdot x_1=3, x _2=4= x _3\)
\(
\begin{aligned}
& \sum x_i=20 \\
& x _4+ x _5=9 \ldots \ldots \text { (i) } \\
& \frac{\sum x_i^2}{x}-(\bar{x})^2=\sigma \Rightarrow \sum x_i^2=106 \\
& x_4^2+x_5^2=65 \ldots \ldots \text { (ii) }
\end{aligned}
\)
Using (i) and (ii) \(\left( x _4- x _5\right)^2=49\)
\(
\left|x_4-x_5\right|=7
\)
If the sum of the deviations of 50 observations from 30 is 50 , then the mean of these observations is : [JEE Main 2019 (Online) 12th January Morning Slot]
\(
\begin{aligned}
&\begin{aligned}
& \sum_{i=1}^{50}\left(x_i-30\right)=50 \\
& \sum x_i=50 \times 30=50 \\
& \sum x_i=50+50+30
\end{aligned}\\
&\begin{aligned}
& \text { Mean }=\bar{x}=\frac{\sum x_i}{n}=\frac{50 \times 30+50}{50} \\
& =30+1=31
\end{aligned}
\end{aligned}
\)
The outcome of each of 30 items was observed; 10 items gave an outcome \(\frac{1}{2}- d\) each, 10 items gave outcome \(\frac{1}{2}\) each and the remaining 10 items gave outcome \(\frac{1}{2}+ d\) each. If the variance of this outcome data is \(\frac{4}{3}\) then \(| d |\) equals : [JEE Main 2019 (Online) 11th January Morning Slot]
Variance is independent of region. So we shift the given data by \(\frac{1}{2}\).
\(
\begin{aligned}
& \text { so, } \frac{10 d^2+10 \times 0^2+10 d^2}{30}-(0)^2=\frac{4}{3} \\
& \Rightarrow d ^2=2 \Rightarrow|d|=\sqrt{2}
\end{aligned}
\)
If mean and standard deviation of 5 observations \(x _1, x _2, x_3, x_4, x_5\) are 10 and 3 , respectively, then the variance of 6 observations \(x_1, x_2, \ldots ., x_5\) and -50 is equal to [JEE Main 2019 (Online) 10th January Evening Slot]
\(
\begin{aligned}
& \bar{x}=10 \Rightarrow \sum_{i=1}^5 x_i=50 \\
& \text { S.D. }=\sqrt{\frac{\sum_{i=1}^5 x_i^2}{5}-(\bar{x})^2=8} \\
& \Rightarrow \sum_{i=1}^5\left(x_i\right)^2=109 \\
& \text { variance }=\frac{\sum_{i=1}^5\left(x_i\right)^2+(-50)^2}{6}-\left(\sum_{i=1}^5 \frac{x_i-50}{6}\right) \\
& =507.5
\end{aligned}
\)
The mean of five observations is 5 and their variance is 9.20. If three of the given five observations are 1, 3 and 8, then a ratio of other two observations is [JEE Main 2019 (Online) 10th January Morning Slot]
Let two observations are \(x_1 \& x_2\)
\(
\begin{aligned}
& \text { mean }=\frac{\sum x_i}{5}=5 \\
& \Rightarrow 1+3+8+x_1+x_2=25 \\
& \Rightarrow x_1+x_2=13 \quad \ldots .(1)
\end{aligned}
\)
variance \(\left(\sigma^2\right)=\frac{\sum x_i^2}{5}-25=9.20\)
\(
\begin{aligned}
& \Rightarrow \sum x_i^2=171 \\
& \Rightarrow x_1^2+x_2^2=97 \dots(2)
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow\left(x_1+x_2\right)^2-2 x_1 x_2=97 \\
& \Rightarrow 169-2 x_1 x_2=97 \\
& \text { or } x_1 x_2=36 \\
& \therefore x_1: x_2=4: 9
\end{aligned}
\)
A data consists of \(n\) observations : \(x_1, x_2, \ldots \ldots, x_n\).
If \(\sum_{i=1}^n\left(x_i+1\right)^2=9 n\) and
\(
\sum_{i=1}^n\left(x_i-1\right)^2=5 n
\)
then the standard deviation of this data is : [JEE Main 2019 (Online) 9th January Evening Slot]
\(
\begin{aligned}
& \sum_{i=1}^n\left(x_i+1\right)^2=9 n \\
& \Rightarrow \sum_{i=1}^n x_i^2+2 \sum_{i=1}^n x_i+n=9 n \dots(1)\\
& \sum_{i=1}^n\left(x_i-1\right)^2=5 n \\
& \Rightarrow \sum_{i=1}^n x_i^2-2 \sum_{i=1}^n x_i+n=5 n \dots(2)
\end{aligned}
\)
Performing (1) + (2), we get
\(
\begin{aligned}
& 2 \sum_{i=1}^n x_i^2+2 n=14 n \\
& \sum_{i=1}^n x_i^2=6 n
\end{aligned}
\)
Performing (1) \(-(2)\), we get
\(
\begin{aligned}
& \Rightarrow 4 \sum_{i=1}^n x_i=4 n \\
& \Rightarrow \Rightarrow \sum_{i=1}^n x_i=n \\
& \text { S.D }(\sigma)=\sqrt{\frac{\sum x_i^2}{n}-(\bar{x})^2} \\
& \sigma=\sqrt{\frac{6 n}{n}-(1)} \\
& \sigma=\sqrt{5}
\end{aligned}
\)
5 students of a class have an average height 150 cm and variance \(18 cm^2\). A new student, whose height is 156 cm , joined them. The variance (in \(cm ^2\) ) of the height of these six students is : [JEE Main 2019 (Online) 9th January Morning Slot]
Average height of 5 students,
\(
\begin{aligned}
& \bar{x}=\frac{x_1+x_2+x_3+x_4+x_5}{5}=150 \\
& \Rightarrow \sum_{i=1}^5 x_i=750
\end{aligned}
\)
We know,
Variance \((\sigma)=\frac{\sum x_i^2}{5}-(\bar{x})^2\)
given that,
\(
\begin{aligned}
& \frac{\sum x_i^2}{5}-(150)^2=18 \\
& \Rightarrow \sum x_i^2=112590
\end{aligned}
\)
Height of new student, \(x_6=156 cm\)
New average height \(\left(\bar{x}_{\text {new }}\right)=\frac{750+156}{6}=151\)
\(
\begin{aligned}
& \text { New variance }=\frac{\sum_{i=1}^6 x_i^2}{6}-\left(\bar{x}_{\text {new }}\right)^2 \\
& =\frac{112590+(156)^2}{6}-(151)^2 \\
& =22821-22801 \\
& =20
\end{aligned}
\)
The mean and the standard deviation(s.d.) of five observations are9 and 0, respectively. If one of the observations is changed such that the mean of the new set of five observations becomes 10, then their s.d. is : [JEE Main 2018 (Online) 16th April Morning Slot]
Here mean \(=\bar{x}=9\)
\(
\begin{aligned}
& \Rightarrow \bar{x}=\frac{\sum x_i}{n}=9 \\
& \Rightarrow \sum x_i=9 \times 5=45
\end{aligned}
\)
Now, standard deviation \(=0\)
\(\therefore\) all the five terms are same i.e.; 9
Now for changed observation
\(
\begin{aligned}
& \bar{x}_{n e w}=\frac{36+x_5}{5}=10 \\
& \Rightarrow x _5=14
\end{aligned}
\)
\(
\begin{aligned}
& \therefore \sigma_{\text {new }}=\sqrt{\frac{\sum\left(x_i-\bar{x}_{\text {new }}\right)^2}{n}} \\
& =\sqrt{\frac{4(9-10)^2+(14-10)^2}{5}}=2
\end{aligned}
\)
If \(\sum_{i=1}^9\left(x_i-5\right)=9\) and
\(\sum_{i=1}^9\left(x_i-5\right)^2=45\), then the standard deviation of the 9 items \(x_1, x_2, \ldots \ldots ., x_9\) is [JEE Main 2018 (Offline)]
When every number is added or subtracted by a fixed number then the standard Deviation remain unchanged.
so let \(x_i-5=y_i\)
So, new equation is \(\sum_{i=1}^9 y_i=9\)
and \(\sum_{i=1}^9 y_i^2=45\)
As, we know. Standard Deviation (S.D)
\(
\begin{aligned}
& =\sqrt{\frac{\sum_{i=1}^9 y_i^2}{9}-\left(\frac{\sum_{i=1}^9 y i}{9}\right)^2} \\
& =\sqrt{\frac{45}{9}-\left(\frac{9}{9}\right)^2} \\
& =\sqrt{5-1} \\
& =\sqrt{4} \\
& =2
\end{aligned}
\)
If the mean of the data: \(7,8,9,7,8,7, \lambda, 8\) is 8 , then the variance of this data is : [JEE Main 2018 (Online) 15th April Evening Slot]
\(
\begin{aligned}
& \bar{x}=\frac{7+8+9+7+8+7+\lambda+8}{8}=8 \\
& \Rightarrow \frac{54+\lambda}{8}=8 \Rightarrow \lambda=10
\end{aligned}
\)
Now variance \(=\sigma^2\)
\(
\begin{aligned}
& =\frac{(7-8)^2+(8-8)^2+(9-8)^2+(7-8)^2+(8-8)^2+(7-8)^2+(10-8)^2+(8-8)^2}{8} \\
& \Rightarrow \sigma^2=\frac{1+0+1+1+0+1+4+0}{8}=\frac{8}{8}=1
\end{aligned}
\)
Hence, the variance is 1.
The mean of set of 30 observations is 75 . If each observation is multiplied by a non-zero number \(\lambda\) and then each of them is decreased by 25 , their mean remains the same. Then \(\lambda\) is equal to : [JEE Main 2018 (Online) 15th April Morning Slot]
As mean is a linear operation, so if each observation is multiplied by \(\lambda\) and decreased by 25 then the mean becomes \(75 \lambda-25\).
According to the question,
\(
75 \lambda-25=75 \Rightarrow \lambda=\frac{4}{3} \text {. }
\)
The mean age of 25 teachers in a school is 40 years. A teacher retires at the age of 60 years and a new teacher is appointed in his place. If now the mean age of the teachers in this school is 39 years, then the age (in years) of the newly appointed teacher is : [JEE Main 2017 (Online) 8th April Morning Slot]
Assume that the ages of 25 teachers are as follows: \(x_1, x_2, x_3, \ldots \ldots x_{25}\). It is given that the mean age of 25 teachers is 40 years.
Therefore, \(\frac{x_1+x_2+x_3+\ldots . . .+x_{25}}{25}=40 \quad \ldots (1)\)
Since a teacher retires at the age of 60 years and a teacher joined and the new mean age is 39 .
Assume that, the age of the newly joined teacher is \(a\).
Therefore,
\(
\begin{gathered}
\frac{x_1+x_2+x_3+\ldots \ldots . .+x_{25}-60+a}{25}=39 \\
\Rightarrow \frac{x_1+x_2+x_3+\ldots \ldots .+x_{25}}{25}+\frac{-60+a}{25}=39 \quad \ldots 2
\end{gathered}
\)
From equation 1 and equation 2 .
\(
\begin{aligned}
40+ & \frac{-60+a}{25}=39 \\
\Rightarrow & \frac{-60+a}{25}=-1 \\
\Rightarrow & -60+a=-25 \\
\Rightarrow & a=35
\end{aligned}
\)
Therefore, the age of the newly appointed teacher is 35 years.
The sum of 100 observations and the sum of their squares are 400 and 2475 , respectively. Later on, three observations, 3, 4 and 5 , were found to be incorrect. If the incorrect observations are omitted, then the variance of the remaining observations is [JEE Main 2017 (Online) 9th April Morning Slot]
We have
\(
\begin{aligned}
& \sum_{i=1}^{100} x_i=400 \\
& \sum_{i=1}^{100} x_i^2=2425
\end{aligned}
\)
The variance of the remaining observations is
\(
\begin{aligned}
& \sigma^2=\frac{\sum x_i^2}{N}-\left(\frac{\sum x_i}{N}\right)^2 \\
& \Rightarrow \frac{2425}{97}-\left(\frac{388}{97}\right)^2 \\
& \Rightarrow \frac{2425}{97}-16 \\
& \Rightarrow \frac{2425-1552}{97}=\frac{873}{97}=9
\end{aligned}
\)
The mean of 5 observations is 5 and their variance is 124 . If three of the observations are 1,2 and 6 ; then the mean deviation from the mean of the data is : [JEE Main 2016 (Online) 10th April Morning Slot]
Let 5 observations are \(x_1, x_2, x_3, x_4, x_5\)
given, \(x_1=1, x_2=2, x_3=6\)
\(
\begin{aligned}
& \text { Mean }=5 \\
& \therefore \text { Mean }(\bar{x})=\frac{x_1+x_2+x_3+x_4+x_5}{5}=5 \\
& \Rightarrow 1+2+6+ x _4+ x _5=25 \\
& \therefore x _4+ x _5=16 \\
& \Rightarrow\left( x _4-5\right)+\left( x _5-5\right)+10=16 \\
& \Rightarrow\left( x _4-5\right)+\left( x _5-5\right)=6
\end{aligned}
\)
\(\therefore\) Mean deviation about mean,
\(
\begin{aligned}
& =\frac{\sum\left|x_i-\bar{x}\right|}{n} \\
& =\frac{|1-5|+|2-5|+|6-5|+\left|x_4-5\right|+\left|x_5-5\right|}{5} \\
& =\frac{4+3+1+6}{5} \\
& =\frac{14}{5} \\
& =2.8
\end{aligned}
\)
If the mean deviation of the numbers \(1,1+d, \ldots, 1+100 d\) from their mean is 255 , then a value of \(d\) is : [JEE Main 2016 (Online) 9th April Morning Slot]
Given numbers are,
\(
1,1+d, 1+2 d \ldots \ldots 1+100 d
\)
\(\therefore\) Total 101 number are present.
\(
\begin{aligned}
& \therefore n=101 \\
& \therefore \text { mean }(\bar{x})=\frac{1+(1+d)+\ldots . .(1+100 d)}{101} \\
& =\frac{1}{101} \times \frac{101}{2}[1+(1+100 d)] \\
& =1+50 d
\end{aligned}
\)
mean deviation from mean
\(
\begin{aligned}
& =\frac{1}{101}[|1-(1+50 d)|+|(1+d)-(1+50 d)| \ldots . . \mid [1+100 d]-(1+50 d)]\\
& =\frac{2|d|}{101}(1+2+3 \ldots \ldots+50) \\
& =\frac{2|d|}{101} \times \frac{50 \times 51}{2}=\frac{2550}{101}|d| \\
& =\frac{2550}{101}|d|=225 \Rightarrow|d|=10.1
\end{aligned}
\)
If the standard deviation of the numbers \(2,3, a\) and 11 is 3.5 , then which of the following is true? [JEE Main 2016 (Offline)]
The formula for standard deviation (S.D)
\(
=\sqrt{\frac{\sum x_i^2}{n}-\left(\frac{\sum x_i}{n}\right)^2}
\)
Where \(\sum x_i^2=\) Sum of square of the numbers
\(
\begin{aligned}
& =2^2+3^2+a^2+11^2 \\
& =4+9+a^2+121 \\
& =134+a^2 \\
& \sum x_i=\text { Sum of numbers } \\
& =2+3+a+11 \\
& =16+a \\
& \therefore S D=\sqrt{\frac{134+a^2}{4}-\left(\frac{16+a}{4}\right)^2} \\
& \Rightarrow \sqrt{\frac{134+a^2}{4}-\left(\frac{16+a}{4}\right)^2}=3.5=\frac{7}{2}(\text { given }) \\
& \Rightarrow \frac{134+a^2}{4}-\left(\frac{16+a}{4}\right)^2=\frac{49}{4} \\
& \Rightarrow 4\left(134+a^2\right)-\left(256+32 a+a^2\right)=4 \times 49 \\
& \Rightarrow 3 a^2-32 a+84=0
\end{aligned}
\)
The mean of the data set comprising of 16 observations is 16 . If one of the observation valued 16 is deleted and three new observations valued 3,4 and 5 are added to the data, then the mean of the resultant data, is : [JEE Main 2015 (Offline)]
Initially we have 16 observations and among them one is 16 .
So, we have 15 unknowns. Let those are \(a_1, a_2, a_3 \ldots \ldots a_{15}\)
\(\therefore\) Mean of 16 datal set
\(
=\frac{a_1+a_2+\ldots . . a_{15}+16}{16}
\)
According to the question,
\(
\begin{aligned}
& \frac{a_1+a_2+\ldots . a_{15}+16}{16}=16 \\
& \Rightarrow a_1+a_2+\ldots \ldots+a_{15}=256-16=240
\end{aligned}
\)
Now we deleted 16 and replaced by there new numbers 3,4 , and 5 .
So, new mean
\(
\begin{aligned}
& =\frac{a_1+a_2+\ldots \ldots a_{15}+(3+4+5)}{18} \\
& =\frac{240+(3+4+5)}{18} \\
& =\frac{240+12}{18} \\
& =14
\end{aligned}
\)
Alternate:
\(
\begin{aligned}
& \text { Sum of } 16 \text { observations }=16 \times 16=256 \\
& \text { Sum of resultant } 18 \text { observations }=256-16+(3+4+5) \\
& =252 \\
& \text { Mean of observations }=\frac{252}{18}=14 \\
&
\end{aligned}
\)
Let the sum of the first three terms of an A. P, be 39 and the sum of its last four terms be 178. If the first term of this A.P. is 10 , then the median of the A.P. is : [Online April 10, 2015]
\(
\begin{aligned}
& 10+(10+d)+(10+2 d)=39 \\
& \Rightarrow d=3 \\
& \Rightarrow t_n=10+(n-1) d \\
& \Rightarrow t_n=3 n+7
\end{aligned}
\)
Now,
\(
\begin{aligned}
& t_n+t_{n-1}+t_{n-2}+t_{n-3}=178 \\
& \Rightarrow 7 \times 4+3[n+n-1+n-2+n-3]=178 \\
& \Rightarrow 4 n-6=50 \\
& \Rightarrow n=14 \\
& \therefore \text { Median } \\
& =\frac{t_7+t_8}{2} \\
& =\frac{14+3 \times 15}{2}=29.5
\end{aligned}
\)
Alternate:
\(
\begin{aligned}
& a_1+a_2+a_3=39 \\
& \Rightarrow a_1+\left(a_1+d\right)+\left(a_1+2 d\right)=39 \\
& \Rightarrow 3 a_1+3 d=39\left[\because a_1=10\right] \\
& \Rightarrow d=3 \\
& \text { Sum of last four term }=178
\end{aligned}
\)
Sum of last four term \(=178\)
\(
\begin{aligned}
& \text { Their mean }=\frac{178}{4}=44.5 \\
& a_n=44.5+1.5+3=49
\end{aligned}
\)
\(
\text { Median }=\frac{10+49}{2}=\frac{59}{2}=29.5
\)
A factory is operating in two shifts, day and night, with 70 and 30 workers respectively. If per day mean wage of the day shift workers is \(₹ 54\) and per day mean wage of all the workers is \(₹ 60\), then per day mean wage of the night shift workers (in ₹) is : [Online April 10, 2015]
Let average wage of Night shift worker is \(x\)
\(
\begin{aligned}
& 70 \times 54+30 \times x=60 \times 100 \\
& x=74
\end{aligned}
\)
In a set of 2 n distinct observations, each of the observations below the median of all the observations is increased by 5 and each of the remaining observations is decreased by 3 . Then the mean of the new set of observations: [Online April 9, 2014]
\(A s\), there are \(2 n\) numbers. So, let these numbers be,
\(
\Rightarrow x_1, x_2, x_3, \ldots \ldots x_{n-1}, x_n, x_{n+1}, \ldots \ldots . x_{2 n-1}, x_{2 n}
\)
And as we know that mean of numbers is,
\(
\Rightarrow \text { Mean }=\frac{\text { Sum of the numbers }}{\text { Total number of numbers }}
\)
So, mean of \(2 n\) numbers given above will be,
Let mean of \(2 n\) numbers \(=\bar{X}[latex]
[latex]
\Rightarrow \text { Then, } \bar{X}=\frac{x_1+x_2+\ldots \ldots+x_n+x_{n+1}+\ldots \ldots+x_{2 n}}{2 n} \dots(1)
\)
As, given in the question that total terms are \(2 n\) which is even.
\(\Rightarrow\) So, median of 2 n terms will be \(\frac{x_n+x_{n+1}}{2}\).
So, median will lie between terms \(x _n\) and \(x _{n+1}\)
As, we are given that each observation below median is increased by 5
\(\Rightarrow\) So, for each i from \(i =1\) to \(i =n\).
\(\Rightarrow x _i\) becomes \(x _i+5\)
As, we are given that each observation above median is decreased by 3 .
\(\Rightarrow\) So, for each i from \(i =n+1\) to \(i =2 n\).
\(\Rightarrow x _i\) becomes \(x _i-3\)
So, now mean of \(2 n\) observations can be written as,
So, now mean of \(2 n\) observations can be written as,
\(\Rightarrow\) New mean
\(
=\frac{\left(x_1+5\right)+\left(x_2+5\right)+\ldots \ldots \ldots+\left(x_n+5\right)+\left(x_{n+1}-3\right)+\ldots \ldots\left(x_{2 n}-3\right)}{2 n}
\)
Solving above equation. It becomes,
\(
\Rightarrow \text { So, new mean }=\frac{x_1+x_2+\ldots \ldots+x_n+5 n+x_{n+1}+\ldots \ldots+x_{2 n}-3 n}{2 n}
\)
Solving the above equation using equation 1. It becomes,
\(
\Rightarrow \text { New mean }=\frac{x_1+x_2+\ldots \ldots+x_n+x_{n+1}+\ldots \ldots+x_{2 n}}{2 n}+\frac{2 n}{2 n}=\bar{X}+1
\)
So, the new mean of \(2 n\) numbers is 1 more than the previous mean.
Alternate:
There are \(2 n\) observations \(x_1, x_2, \ldots, x_{2 n}\)
So, mean \(=\sum_{i=1}^{2 n} \frac{x_i}{2 n}\)
Let these observations be divided into two parts \(x_1\), \(x_2, \ldots, x_n\) and \(x_{n+1}, \ldots, x_{2 n}\)
Each in \(1^{\text {st }}\) part 5 is added, so total of first part is
\(
\sum_{i=1}^n x_i+5 n
\)
In second part 3 is subtracted from each
So, total of second part is \(\sum_{i=n+1}^{2 n} x_i-3 n\)
Total of \(2 n\) terms are
\(
\sum_{i=1}^n x_i+5 n+\sum_{i=n+1}^{2 n} x_i-3 n=\sum_{i=1}^{2 n} x_i+2 n
\)
\(
\text { Mean }=\sum_{i=1}^{2 n} \frac{x_i+2 n}{2 n}=\sum_{i=1}^{2 n} \frac{x_i}{2 n}+1
\)
So, it increase by 1 .
If the median and the range of four numbers \(\{x, y, 2 x+y, x-y\}\), where \(0<y<x<2 y\), are 10 and 28 respectively, then the mean of the numbers is: [Online April 23, 2013]
Since \(0<y<x<2 y\)
\(
\begin{aligned}
& \therefore y>\frac{x}{2} \Rightarrow x-y<\frac{x}{2} \\
& \therefore x-y<y<x<2 x+y
\end{aligned}
\)
Hence median \(=\frac{y+x}{2}=10\)
\(
\Rightarrow x+y=20 \dots(i)
\)
\(
\begin{aligned}
& \text { And range }=(2 x+y)-(x-y)=x+2 y \\
& \text { But range }=28 \\
& \therefore x+2 y=28 \dots(ii)
\end{aligned}
\)
From equations (i) and (ii),
\(
\begin{aligned}
x=12, y & =8 \\
\therefore \text { Mean } & =\frac{(x-y)+y+x+(2 x+y)}{4}=\frac{4 x+y}{4} \\
& =x+\frac{y}{4}=12+\frac{8}{4}=14
\end{aligned}
\)
The mean of a data set consisting of 20 observations is 40 . If one observation 53 was wrongly recorded as 33 , then the correct mean will be : [Online April 9, 2013]
\(
\text { Mean }=\frac{S}{n}
\)
Here \(S=\) sum of all observations, \(n=\) number of observations
\(
\begin{aligned}
& \therefore 40=\frac{S}{20} \\
& \Rightarrow S=800
\end{aligned}
\)
\(\because 53\) is recorded as 33 so we need to add \((53-33=20)\) to get the correct mean which is
\(
\text { Mean }_{\text {correct }}=\frac{800+20}{20}=41
\)
The median of 100 observations grouped in classes of equal width is 25 . If the median class interval is \(20-30\) and the number of observations less than 20 is 45 , then the frequency of median class is [Online May 19, 2012]
Median is given as
\(
M=l+\frac{\frac{N}{2}-F}{f} \times C
\)
where
\(l=\quad\) lower limit of the median – class
\(f=\) frequency of the median class
\(N=\) total frequency
\(F=\quad\) cumulative frequency of the class just before the median class
\(C=\) length of median class
Now, given, \(M=25, N=100, F=45\),
\(
C=20-30=10, l=20 \text {. }
\)
\(\therefore\) By using formula, we have
\(
\begin{aligned}
& 25=20+\frac{50-45}{f} \times 10 \\
& 25-20=\frac{50}{f} \Rightarrow 5=\frac{50}{f} \Rightarrow f=10
\end{aligned}
\)
The frequency distribution of daily working expenditure of families in a locality is as follows:
\(
\begin{array}{|c|c|c|c|c|c|}
\hline \begin{array}{c}
\text { Expenditure } \\
\text { in ₹. }(x):
\end{array} & 0-50 & 50-100 & 100-150 & 150-200 & 200-250 \\
\hline \begin{array}{c}
\text { No. of } \\
\text { families }(f):
\end{array} & 24 & 33 & 37 & b & 25 \\
\hline
\end{array}
\)
If the mode of the distribution is \(₹ 140\), then the value of \(b\) is [Online May 7, 2012]
Frequency distribution is given as
\(
\begin{array}{|c|c|}
\hline \text { Expenditure } & \text { No. of families }(f) \\
\hline 0-50 & 24 \\
\hline 50-100 & 33 \\
\hline 100-150 & 37 \\
\hline 150-200 & b \\
\hline 200-250 & 25 \\
\hline
\end{array}
\)
Clearly, modal class is \(100-150\), as the maximum frequency occurs in this class.
Given, Mode \(=140\)
We have
\(
\text { Mode }=\ell+\frac{f_0-f_{-1}}{2 f_0-f_{-1}-f_1} \times i
\)
where
\(
\begin{aligned}
& \ell=100, f_0=37, f_{-1}=33, f_1= b \\
& i=50
\end{aligned}
\)
Thus, we get
\(
\begin{aligned}
140= & 100+\left[\frac{37-33}{2(37)-33-b}\right] \times 50 \\
& =100+\left[\frac{4}{74-33-b}\right] \times 50 \\
& =100+\frac{200}{41-b} \\
\Rightarrow \quad & 5740=4300+40 b \Rightarrow b=36
\end{aligned}
\)
The average marks of boys in class is 52 and that of girls is 42. The average marks of boys and girls combined is 50 . The percentage of boys in the class is [JEE Main 2007]
Let the number of boys be \(x\) and that of girls be \(y\).
\(
\begin{aligned}
& \Rightarrow 52 x+42 y=50(x+y) \\
& \Rightarrow 52 x-50 x=50 y-42 y \\
& \Rightarrow \quad 2 x=8 y \Rightarrow \frac{x}{y}=\frac{4}{1} \text { and } \frac{x}{x+y}=\frac{4}{5} \\
& \text { Required \% of boys }=\frac{x}{x+y} \times 100=\frac{4}{5} \times 100=80 \%
\end{aligned}
\)
Let \(x_1, x_2, \ldots \ldots \ldots \ldots x_n\) be n observations such that \(\sum x_i^2=400\) and \(\sum x_i=80\). Then the possible value of n among the following is [JEE Main 2005]
We know that for positive real numbers \(x_1, x_2, \ldots, x_n\), A.M. of \(k^{t h}\) powers of \(x_i^{\prime} s \geq k^{t h}\) the power of
A.M. of \(x_i^{\prime} s\)
\(
\Rightarrow \frac{\sum x_1^2}{n} \geq\left(\frac{\sum x_1}{n}\right)^2 \Rightarrow \frac{400}{n} \geq\left(\frac{80}{n}\right)^2
\)
\(\Rightarrow n \geq 16\). So only possible value for \(n=18\)
If in a frequency distribution, the mean and median are 21 and 22 respectively, then its mode is approximately [JEE 2005]
Find the mode:
Given,
Mean \(=21\) and Median \(=22\)
We know,
\(
\begin{aligned}
\text { Mode } & =3 \text { Median-2Mean } \\
& =3(22)-2(21) \\
& =66-42 \\
& =24
\end{aligned}
\)
The median of a set of 9 distinct observations is 20.5 . If each of the largest 4 observations of the set is increased by 2 , then the median of the new set [JEE 2003]
\(n=9\) then median term \(=\left(\frac{9+1}{2}\right)^{\text {th }}=5^{\text {th }}\) term. Last four observations are increased by 2 . The median is 5 th observation which is remaining unchanged.
\(\therefore \quad\) there will be no change in median.
In a class of 100 students there are 70 boys whose average marks in a subject are 75. If the average marks of the complete class is 72 , then what is the average of the girls? [JEE 2002]
\(
\begin{aligned}
& \text { Total student }=100 ; \\
& \text { for } 70 \text { students total marks }=75 \times 70=5250 \\
& \begin{aligned}
\Rightarrow \quad \text { Total marks of girls }=7200-5250 \\
=1950
\end{aligned} \\
& \text { Average of girls }=\frac{1950}{30}=65
\end{aligned}
\)
The variance of first 50 even natural numbers is [JEE 2014]
First 50 even natural numbers are \(2,4,6 \ldots . ., 100\)
\(
\begin{aligned}
& \text { Variance }=\frac{\sum x_i^2}{N}-(\bar{x})^2 \\
& \Rightarrow \sigma^2=\frac{2^2+4^2+\ldots+100^2}{50}-\left(\frac{2+4+\ldots+100}{50}\right)^2 \\
& =\frac{4\left(1^2+2^2+3^2+\ldots+50^2\right)}{50}-(51)^2 \\
& =4\left(\frac{50 \times 51 \times 101}{50 \times 6}\right)-(51)^2=3434-2601 \Rightarrow \sigma^2=833
\end{aligned}
\)
Let \(\overline{ x }, M\) and \(\sigma^2\) be respectively the mean, mode and variance of \(n\) observations \(x_1, x_2, \ldots, x_n\) and \(d_i=-x_i-a[latex], [latex]i =1,2, \ldots, n\), where a is any number.
Statement I: Variance of \(d_1, d_2, \ldots d_n\) is \(\sigma^2\).
Statement II: Mean and mode of \(d _1, d_2, \ldots . d _{ n }\) are \(-\overline{ x }- a\) and \(- M – a\), respectively. [Online April 19, 2014]
\(
\begin{aligned}
\bar{x} & =\frac{x_1+x_2+x_3+\ldots+x_n}{n} \\
\sigma^2 & =\frac{1}{n} \sum_{i=1}^n\left(x_i-\bar{x}\right)^2
\end{aligned}
\)
Mean of \(d_1, d_2, d_3, \ldots, d_n\)
\(
\begin{aligned}
& =\frac{d_1+d_2+d_3+\ldots .+d_n}{n} \\
& =\frac{\left(-x_1-a\right)+\left(-x_2-a\right)+\left(-x_3-a\right)+\ldots .+\left(-x_n-a\right)}{n}
\end{aligned}
\)
\(
\begin{aligned}
& =-\left[\frac{x_1+x_2+x_3+\ldots .+x_n}{n}\right]-\frac{n a}{n} \\
& =-\bar{x}-a
\end{aligned}
\)
Since, \(d_i=-x_i-a\) and we multiply or subtract each observation by any number the mode remains the same. Hence mode of \(-x_i-a\) i.e. \(d_i\) and \(x_i\) are same. Now variance of \(d_1, d_2, \ldots, d_n\)
\(
\begin{aligned}
& =\frac{1}{n} \sum_{i=1}^n\left[d_i-(-\bar{x}-a)\right]^2 \\
& =\frac{1}{n} \sum_{i=1}^n\left[-x_i-a+\bar{x}+a\right]^2 \\
& =\frac{1}{n} \sum_{i=1}^n\left(-x_i+\bar{x}\right)^2 \\
& =\frac{1}{n} \sum_{i=1}^n\left(\bar{x}-x_i\right)^2=\sigma^2
\end{aligned}
\)
Let \(\overline{ X }\) and M.D. be the mean and the mean deviation about \(\bar{X}\) of \(n\) observations \(x_i, i=1,2\), \(\ldots\) n. If each of the observations is increased by 5 , then the new mean and the mean deviation about the new mean, respectively, are : [Online April 12, 2014]
Let \(x_i\) be \(n\) observations, \(i=1,2, \ldots n\)
Let \(\bar{X}\) be the mean and M.D be the mean deviation about \(\bar{X}\).
If each observation is increased by 5 then new mean will be \(\bar{X}+5\) and new M.D. about new mean will be \(M D\).
\(
\left(\because \text { Mean }=\sum_{i=1}^n \frac{x_i}{n}\right)
\)
All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks of 10 to each of the students. Which of the following statistical measures will not change even after the grace marks were given? [JEE 2013]
If initially all marks were \(x_i\) then
\(
\sigma_1^2=\frac{\sum_i\left(x_i-\bar{x}\right)^2}{N}
\)
Now each is increased by 10
\(
\begin{aligned}
& \sigma_1^2= \frac{\sum_i\left[\left(x_i+10\right)-(x+10)\right]^2}{N}=\frac{\sum_i\left(x_i-\bar{x}\right)^2}{N} \\
&=\sigma_1^2
\end{aligned}
\)
Hence, variance will not change even after the grace marks were given.
In a set of 2 n observations, half of them are equal to ‘ a ‘ and the remaining half are equal to ‘ – a ‘. If the standard deviation of all the observations is 2 ; then the value of \(|a|\) is: [Online April 25, 2013]
Clearly mean \(A =0\)
Now, standard deviation \(\sigma=\sqrt{\frac{\sum(x- A )^2}{2 n}}\)
\(
\begin{aligned}
2 & =\sqrt{\frac{(a-0)^2+(a-0)^2+\ldots \ldots+(0-a)^2+\ldots \ldots}{2 n}} \\
& =\sqrt{\frac{a^2 \cdot 2 n}{2 n}}=|a|
\end{aligned}
\)
Hence, \(|a|=2\)
Mean of 5 observations is 7 . If four of these observations are \(6,7,8,10\) and one is missing then the variance of all the five observations is : [Online April 22, 2013]
Let 5th observation be \(x\).
Given mean \(=7\)
\(
\begin{aligned}
& \therefore 7=\frac{6+7+8+10+x}{5} \\
& \Rightarrow x=4
\end{aligned}
\)
Now, Variance
\(
\begin{gathered}
=\sqrt{\frac{(6-7)^2+(7-7)^2+(8-7)^2+(10-7)^2+(4-7)^2}{5}} \\
=\sqrt{\frac{1^2+0^2+1^2+3^2+3^2}{5}}=\sqrt{\frac{20}{5}}=\sqrt{4}=2
\end{gathered}
\)
Let \(x_1, x_2, \ldots, x_n\) be n observations, and let \(\bar{x}\) be their arithmetic mean and \(\sigma^2\) be the variance.
Statement-1: Variance of \(2 x_1, 2 x_2, \ldots, 2 x_n\) is \(4 \sigma^2\).
Statement-2: Arithmetic mean \(2 x_1, 2 x_2, \ldots, 2 x_n\) is \(4 \bar{x}\). [JEE Main 2012]
A.M. of \(2 x_1, 2 x_2, \ldots, 2 x_{ n }\) is
\(
\begin{aligned}
& \frac{2 x_1+2 x_2+\ldots+2 x_n}{n} \\
& =2\left(\frac{x_1+x_2+\ldots .+x_n}{n}\right)=2 \bar{x} \\
& \quad\left(\because \text { Mean }=\frac{\text { sum of observations }}{\text { Number of observations }}\right)
\end{aligned}
\)
So statement-2 is false.
variance \(\left(2 x_i\right)=2^2\) variance \(\left(x_i\right)=4 \sigma^2\) where \(i=1\), \(2, \ldots \ldots\) n
So statement- 1 is true.
Statement 1: The variance of first \(n\) odd natural numbers is
\(
\frac{n^2-1}{3}
\)
Statement 2: The sum of first n odd natural number is \(n^2\) and the sum of square of first \(n\) odd natural numbers is
\(
\frac{n\left(4 n^2+1\right)}{3} \text {. }
\)
[Online May 26, 2012]
1. Sum of the first \(n\) odd natural numbers:
The \(n\)th odd number is \(2 n-1\). The sum of the first \(n\) odd numbers is:
\(
1+3+5+\cdots+(2 n-1)
\)
This sum is known to be:
\(
S_n=n^2
\)
So, the statement “Sum of the first \(n\) odd natural numbers is equal to \(n^{2 “}\) is true.
2. Sum of the squares of the first \(n\) odd natural numbers:
Given that:
Sum of squares of the first \(n\) odd natural numbers \(=\frac{n\left(4 n^2-1\right)}{3}\)
3. Variance of the first \(n\) odd natural numbers:
Variance \(\sigma^2\) is calculated using the formula:
\(
\sigma^2=\frac{1}{n}\left(\sum_{i=1}^n x_i^2\right)-\left(\frac{1}{n} \sum_{i=1}^n x_i\right)^2
\)
Here, \(x_i\) represents the \(i\) th odd number.
Given:
\(
\begin{aligned}
\sum_{i=1}^n x_i & =n^2 \\
\sum_{i=1}^n x_i^2 & =\frac{n\left(4 n^2-1\right)}{3}
\end{aligned}
\)
Now, let’s compute the variance:
First, find the mean \((\mu)\) :
\(
\mu=\frac{1}{n} \sum_{i=1}^n x_i=\frac{n^2}{n}=n
\)
Next, compute the variance:
\(
\begin{aligned}
\sigma^2 & =\frac{1}{n}\left(\sum_{i=1}^n x_i^2\right)-\mu^2 \\
\sigma^2 & =\frac{1}{n}\left(\frac{n\left(4 n^2-1\right)}{3}\right)-n^2 \\
\sigma^2 & =\frac{4 n^2-1}{3}-n^2 \\
\sigma^2 & =\frac{4 n^2-1}{3}-\frac{3 n^2}{3} \\
\sigma^2 & =\frac{4 n^2-1-3 n^2}{3} \\
\sigma^2 & =\frac{n^2-1}{3}
\end{aligned}
\)
This matches the given variance formula \(\frac{n^2-1}{3}\).
4. Verification of the statements:
– Statement 1: The sum of the first \(n\) odd natural numbers is \(n^2\). (This is true)
– Statement 2: The sum of the first \(n\) odd natural numbers is not equal to \(n^2\). (This is false) Therefore, the correct option is that Statement 2 is false, and we have verified the variance formula as well.
If the mean of \(4,7,2,8,6\) and a is 7 , then the mean deviation from the median of these observations is [Online May 12, 2012]
Given observations are \(4,7,2,8,6, a\) and mean is 7 . We know
\(
\begin{aligned}
& \text { Mean }=\frac{4+7+2+8+6+a}{6} \\
& \Rightarrow \quad 7=\frac{4+7+2+8+6+a}{6} \Rightarrow a=15
\end{aligned}
\)
Now, given observations can be written in ascending order which is \(2,4,6,7,8,15\)
Since, No. of observation is even
\(\therefore\) Median
\(=\frac{\left(\frac{6}{2}\right) \text { th observation }+\left(\frac{6}{2}+1\right) \text { th observation }}{2}\)
\(
\begin{aligned}
& =\frac{\text { 3rd observation }+4 \text { th observation }}{2} \\
& =\frac{6+7}{2}=\frac{13}{2}
\end{aligned}
\)
Now, Mean deviation \(=\frac{\sum_{i=1}^6\left|x_i-\frac{13}{2}\right|}{6}\)
\(
\begin{aligned}
& =\frac{\left|4-\frac{13}{2}\right|+\left|7-\frac{13}{2}\right|+\left|2-\frac{13}{2}\right|+\left|8-\frac{13}{2}\right|+\left|6-\frac{13}{2}\right|+\left|15-\frac{13}{2}\right|}{6} \\
& =\frac{\frac{5}{2}+\frac{1}{2}+\frac{9}{2}+\frac{3}{2}+\frac{1}{2}+\frac{17}{2}}{6}=\frac{18}{6}=3
\end{aligned}
\)
A scientist is weighing each of 30 fishes. Their mean weight worked out is 30 gm and a standarion deviation of 2 gm . Later, it was found that the measuring scale was misaligned and always under reported every fish weight by 2 gm . The correct mean and standard deviation (in gm) of fishes are respectively : [2011RS]
Mean weight of fishes \(=30 gm\)
Standard deviation \(=2 g m[latex]
According to question,
Every fish weight is increased by 2 gm
Therefore, over all mean weight is increased by [latex]30+2=32 gm\)
Standard deviation is not effected by the addition of a constant.
Thus, standard deviation will remain same.
If the mean deviation about the median of the numbers \(a\), \(2 a, \ldots \ldots, 50 a\) is 50 , then \(|a|\) equals [JEE 2011]
Median is the mean of 25 th and 26 th observation
\(
\begin{aligned}
& \therefore M=\frac{25 a+26 a}{2}=25.5 a \\
& M . D(M)=\frac{\sum\left|x_i-M\right|}{N} \\
& \Rightarrow 50=\frac{1}{50}[2 \times|a| \times(0.5+1.5+2.5+\ldots .24 .5)] \\
& \Rightarrow 2500=2|a| \times \frac{25}{2}(25) \\
& \Rightarrow|a|=4
\end{aligned}
\)
For two data sets, each of size 5 , the variances are given to be 4 and 5 and the corresponding means are given to be 2 and 4 , respectively. The variance of the combined data set is [JEE 2010]
\(
\begin{aligned}
& \sigma_x^2=4, \sigma_y^2=5, x=2, y=4 \\
& \frac{1}{5} \sum x_i^2-(2)^2=4 ; \frac{1}{5} \sum y_i^2-(4)^2=5 \\
& \sum x_i^2=40 ; \sum y_i^2=105 \\
& \sum\left(x_i^2+y_i^2\right)=145 \\
& \Rightarrow \sum\left(x_i+y_i\right)=5(2)+5(4)=30
\end{aligned}
\)
Variance of combined data
\(
\begin{aligned}
& =\frac{1}{10} \sum\left(x_i^2+y_i^2\right)-\left(\frac{1}{10} \sum\left(x_i+y_i\right)\right)^2 \\
& =\frac{145}{10}-9=\frac{11}{2}
\end{aligned}
\)
If the mean deviation of the numbers \(1,1+ d\), \(1+2 d, \ldots .1+100 d\) from their mean is 255 , then \(d\) is equal to : [JEE Main 2009]
\(
\begin{aligned}
\text { Mean } & =\frac{101+ d (1+2+3+\ldots \ldots+100)}{101} \\
& =1+\frac{ d \times 100 \times 101}{101 \times 2}=1+50 d
\end{aligned}
\)
Mean deviation from the mean \(=255\)
\(
\begin{aligned}
& \Rightarrow \quad \frac{1}{101}[|1-(1+50 d)|+|(1+d)-(1+50 d)| \\
& +|(1+2 d)-(1+50 d)|+\ldots .+|(1+100 d)-(1+50 d)|]=255 \\
& \Rightarrow \quad 2 d[1+2+3+\ldots+50]=101 \times 255 \\
& \Rightarrow \quad 2 d \times \frac{50 \times 51}{2}=101 \times 255 \\
& \Rightarrow \quad d=\frac{101 \times 255}{50 \times 51}=10.1
\end{aligned}
\)
Statement-1: The variance of first \(n\) even natural numbers is \(\frac{n^2-1}{4}\).
Statement-2 : The sum of first \(n\) natural numbers is \(\frac{n(n+1)}{2}\) and the sum of squares of first \(n\) natural numbers is \(\frac{n(n+1)(2 n+1)}{6}\). [JEE Main 2009]
Step-1: Verify statement 1
We know that the sum of the first \(n\) natural number is \(=\frac{n(n+1)}{2}\) Sum of first \(n\) even natural numbers is \(=2+4+6+\ldots+2 n\)
\(
\begin{aligned}
= & 2(1+2+3+\ldots+n) \\
= & 2 \frac{n(n+1)}{2} \\
= & n(n+1)
\end{aligned}
\)
The mean \((\bar{x})\) of the first \(n\) even natural numbers is \(=\frac{n(n+1)}{n}\) \(=n+1\)
We know that the sum of the square first \(n\) natural number is \(=\frac{n(n+1)(2 n+1)}{6}\)
Sum of the square of first \(n\) even natural numbers
\(
\begin{aligned}
= & 2^2+4^2+6^2+\ldots+(2 n)^2 \\
= & 4\left(1^2+2^2+\ldots n^2\right) \\
& =4 \frac{n(n+1)(2 n+1)}{6} \\
& =\frac{2 n(n+1)(2 n+1)}{3}
\end{aligned}
\)
\(
\begin{aligned}
& \text { Variance }=\frac{1}{n} \sum_{i=1}^n\left(x_i^2\right)-(\bar{x})^2 \\
& =\frac{1}{n} \frac{2 n(n+1)(2 n+1)}{3}-(n+1)^2 \\
& =\frac{2(n+1)(2 n+1)}{3}-(n+1)^2 \\
& =(n+1)\left[\frac{2(2 n+1)-3(n+1)}{3}\right] \\
& =(n+1)\left[\frac{n-1}{3}\right] \\
& =\frac{n^2-1}{3}
\end{aligned}
\)
So statement 1 is incorrect
Step-2: Verify statement 2
We know that the sum of the first \(n\) natural number \(=\frac{n(n+1)}{2}\)
We know that the sum of the squares first \(n\) natural numbers
\(
=\frac{n(n+1)(2 n+1)}{6}
\)
So statement 2 is correct
Hence, option(c) i.e. Statement 1 is False, Statement 2 is True is correct
The mean of the numbers a, b, \(8,5,10\) is 6 and the variance is 6.80 . Then which one of the following gives possible values of \(a\) and \(b\) ? [JEE Main 2008]
\(
\begin{aligned}
& \text { Mean of } a, b, 8,5,10 \text { is } 6 \\
& \Rightarrow \quad \frac{a+b+8+5+10}{5}=10 \\
& \Rightarrow \quad a+b=7 \dots(i)
\end{aligned}
\)
Variance of \(a, b, 8,5,10\) is 6.80
\(
\begin{aligned}
& \Rightarrow \quad \frac{(a-6)^2+(b-6)^2+(8-6)^2+(5-6)^2+(10-6)^2}{5}=6.80 \\
& \Rightarrow \quad a^2-12 a+36+(1-a)^2+21=34 \text { [using eq. (i)] }
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \quad 2 a^2-14 a+24=0 \Rightarrow a^2-7 a+12=0 \\
& \Rightarrow \quad a=3 \text { or } 4 \quad \Rightarrow b=4 \text { or } 3
\end{aligned}
\)
\(\therefore \quad\) The possible values of \(a\) and \(b\) are \(a=3\) and \(b=4\)
or, \(a=4\) and \(b=3\)
Suppose a population \(A\) has 100 observations 101,102 , ……….., 200 and another population B has 100 observations 151,152 , \(\ldots\) .250 . If \(V_{ A }\) and \(V_{ B }\) represent the variances of the two populations, respectively then \(\frac{V_A}{V_B}\) is [JEE 2006]
\(\sigma_x^2=\frac{\sum d_i^2}{n}\) (Here deviations are taken from the mean).
Since \(A\) and \(B\) both have 100 consecutive integers, therefore both have same standard deviation and hence the variance. \(\therefore \frac{V_A}{V_B}=1\)
(As \(\sum d_i^2\) is same in both the cases)
Alternate:
\(V\) ariance \(=\sigma^2=\frac{1}{n} \sum_{i=1}^n\left(x_i-\mu\right)^2\), where \(\mu\) is the mean of the observations. The difference of an observation and mean of observations of \(A\) is same as that of the difference of corresponding observation and mean of observations of B.
\(
\begin{aligned}
& \text { So, } V_A=V_B \\
& \therefore \frac{V_A}{V_B}=1
\end{aligned}
\)
In a series of \(2 n\) observations, half of them equal \(a\) and remaining half equal \(-a\). If the standard deviation of the observations is 2 , then \(|a|\) equals. [JEE Main 2004]
Clearlymean \(A=0\)
Standard deviation \(\sigma=\sqrt{\frac{\sum(x-A)^2}{2 n}}\)
\(
\begin{aligned}
2 & =\sqrt{\frac{(a-0)^2+(a-0)^2+\ldots(0-a)^2+\ldots}{2 n}} \\
& =\sqrt{\frac{a^2 \cdot 2 n}{2 n}}=|a|
\end{aligned}
\)
Hence \(|a|=2\)
Consider the following statements :
(A) Mode can be computed from histogram
(B) Median is not independent of change of scale
(C) Variance is independent of change of origin and scale.
Which of these is / are correct? [JEE 2004]
A histogram is an approximate representation of the distribution of numerical. It was first introduced by Karl Pearson. To construct a histogram, the first step is to “bin” the range of values-that is, divide the entire range of values into a series of intervals-and then count how many values fall into each interval.
Change of origin means some value has been added or subtracted in the observation.
Change of scale means some value is multiplied or divided by observations.
The mode of a set of data values is the value that appears most often. As in the histogram there is the representation of all the data values along with its frequency. So, mode can be easily computed from histogram. We can see the highest frequency so made.
So, option A is correct.
Median is the middle number in a sorted list of numbers. If there is an odd amount of numbers, the median value is the number that is in the middle, with the same amount of numbers below and above.
So, if we change the scale definitely the middle number will change and hence the median will also change.
So, option B is also correct.
Variance is the expectation of the squared deviation of a random variable from its mean. Informally, it measures how far a set of numbers are spread out from their average value.
So, variance remains the same even if we change the origin and scale.
So, option C is incorrect.
Hence, mode can be computed from histogram and median is not independent of change of scale.
So option A and option B are correct options.
In an experiment with 15 observations on \(x\), the following results were available:
\(
\Sigma x^2=2830, \Sigma x=170
\)
One observation that was 20 was found to be wrong and was replaced by the correct value 30 . The corrected variance is [JEE Main 2003]
\(
\begin{aligned}
& \Sigma x=170, \Sigma x^2=2830 \text { increase in } \Sigma x=10, \text { then } \\
& \Sigma x^{\prime}=170+10=180
\end{aligned}
\)
Increase in \(\Sigma x^2=900-400=500\) then
\(
\begin{aligned}
& \Sigma x^{\prime 2}=2830+500=3330 \\
& \text { Variance }=\frac{1}{n} \Sigma x^{\prime 2}-\left(\frac{1}{n} \Sigma x^{\prime}\right)^2 \\
& =\frac{1}{15} \times 3330-\left(\frac{1}{15} \times 180\right)^2=222-144=78 .
\end{aligned}
\)
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