Statistics

There are 27 numbers in the list: $3x, 6x, 9x, \ldots, 81x$.

Because there are an odd number of numbers, the median is the $14^\text{th}$ number in the list.

Each number increases by $3x$, so the $14^\text{th}$ number is: $3x + (14-1) \times 3x = 3x + 39x = 42x$ So, the median is $42x$.

The mean deviation about the median means we find the average of the distances from each value to $42x$: $\frac{\,|3x-42x| + |6x-42x| + \ldots + |81x-42x|\,}{27} = 91$

So, $|3x-42x| + |6x-42x| + \ldots + |81x-42x| = 27 \times 91$

Every term inside the absolute value is a multiple of $|x|$.
The differences $|3x-42x|, |6x-42x|, \ldots, |81x-42x|$ will be $|39x|, |36x|, \ldots, |0x|, \ldots, |39x|$.

So the total sum is:
$|x| \times (39 + 36 + \ldots + 0 + 3 + 6 + \ldots + 39) = 27 \times 91$

The numbers in the bracket add up to $546$, so: $|x| \times 546 = 27 \times 91$

$|x| = \frac{27 \times 91}{546} = \frac{9}{2}$

Let us take assume mean $(A)=5$ and $h=2$

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$ \begin{aligned} \because \quad \bar{x} & =A+\frac{\Sigma f_i \mu_i}{\Sigma f_i} \times h \\ \bar{x} & =5+0=5 \\ \mathrm{SD} & =\sigma=h \times \sqrt{\frac{\Sigma f_i \mu_i^2}{N}-\left(\frac{\Sigma f_i \mu_i}{N}\right)^2} \\ \sigma & =2 \times \sqrt{\frac{22}{15}} \\ \mathrm{CV} & =\frac{\sigma}{\bar{x}} \times 100=\frac{2 \times \sqrt{\frac{22}{15}} \times 100}{5} \\ & =\frac{2 \times \sqrt{2} \times \sqrt{11}}{\sqrt{3} \times \sqrt{5} \times 5 \times \sqrt{5}} \times 100 \times \sqrt{5} \\ & =\sqrt{\frac{110}{3}} \times 8 \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \bar{X}=\frac{2+3+5+7+11+13+17+19+22}{9} \\ & \quad=11 \\ & \left|x_i-\bar{x}\right|=9,8,6,4,0,2,6,8,11 \end{aligned}\\ &\text { ∴ Mean deviation from mean }\\ &\begin{aligned} & =\frac{\Sigma\left|x_i-\bar{x}\right|}{9} \\ & =\frac{9+8+6+4+0+2+6+8+11}{9} \end{aligned} \end{aligned} $

$ =\frac{54}{9}=6 $

Step 1: List all values and probabilities

We have $X$ taking the values 2, 3, 5, 7, and 12. Their probabilities are $3k$, $k$, $k$, $2k$, and $k$.

Step 2: Make a table with useful information

$ \begin{array}{lllll} \hline x_i & p_i & p_ix_i & x_i^2 & p_ix_i^2 \\ \hline 2 & 3k & 6k & 4 & 12k \\ 3 & k & 3k & 9 & 9k \\ 5 & k & 5k & 25 & 25k \\ 7 & 2k & 14k & 49 & 98k \\ 12 & k & 12k & 144 & 144k \\ \hline & & 40k & & 288k \\ \hline \end{array} $

Step 3: Find value of $k$

Add up all probabilities:

$ 3k + k + k + 2k + k = 8k $

Set equal to 1:

$ 8k = 1 $

So, $ k = \frac{1}{8} $

Step 4: Find the mean and expected value of $x^2$

$ \Sigma p_i x_i = 40k $

$ \Sigma p_i x_i^2 = 288k $

Step 5: Write the formula for standard deviation

$ \mathrm{SD} = \sqrt{ \Sigma p_i x_i^2 - \left( \Sigma p_i x_i \right)^2 } $

Step 6: Substitute the values and solve

$ \mathrm{SD} = \sqrt{ 288k - (40k)^2 } $ $ = \sqrt{ \frac{288}{8} - (5)^2 } $ $ = \sqrt{36 - 25} $ $ = \sqrt{11} $

Data $x_i: 3,4,5,6,7,8,10,13$

$ \begin{aligned} & \begin{array}{l} \Sigma x_i=3+4+5+6+7+8+10+13=56 \\ \Sigma x_i^2=9+16+25+36+49+64 +100+169 \end{array} \\ & \quad=468 \\ & \text { Variance }=\frac{\Sigma x_i^2}{n}-\left(\frac{\Sigma x_i}{n}\right)^2 \\ & \qquad=\frac{468}{8}-\left(\frac{56}{8}\right)^2=58.5-49=9.5 \end{aligned} $

$ \begin{aligned} & \because P(X=K)=\frac{e^{-\lambda} \lambda^K}{K!} \\ & \because \quad P(X=3)=P(X=5) \\ & \Rightarrow \quad \frac{e^{-\lambda} \lambda^3}{3!}=\frac{e^{-\lambda} \lambda^5}{5!} \\ & \Rightarrow \quad \lambda^2=\frac{120}{6}=20 \\ & \Rightarrow \lambda=\sqrt{20}=2 \sqrt{5} \\ & \therefore P(X=4)=\frac{e^{-\lambda} \lambda^4}{4!}=\frac{e^{-2 \sqrt{5}},(\sqrt{20})^4}{24} \\ & \quad=\frac{400}{24 e^{2 \sqrt{5}}}=\frac{50}{3 e^{2 \sqrt{5}}} \end{aligned} $

Given, $a=9, d=6$

$\therefore$ Variance,

$ \begin{aligned} P & =\frac{6^2\left(n^2-1\right)}{12} \\ & =3\left(n^2-1\right) \\ \Rightarrow \quad n^2-1 & =\frac{P}{3} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Now, variance of $2,4,6, \ldots, 2 n$

$ \begin{gathered} a=2, d=2 \\ \text { Variance, } \frac{\left(4^2\left(n^2-1\right)\right.}{12} \\ =\frac{4\left(n^2-1\right)}{12}=\frac{n^2-1}{3} \\ =\frac{\frac{P}{3}}{3}=\frac{P}{9} \end{gathered} $

Given data,

$ \begin{array}{lllllll} \hline \boldsymbol{x}_{\boldsymbol{i}} & 2 & 9 & 8 & 3 & 5 & 7 \\ \hline \boldsymbol{f}_{\boldsymbol{i}} & 5 & 3 & 1 & 6 & 6 & 1 \\ \hline \end{array} $

$ \begin{array}{ccc} \hline \boldsymbol{x}_{\boldsymbol{i}} & \boldsymbol{f}_{\boldsymbol{i}} & \begin{array}{c} \text { Cumulative } \\ \text { frequency } \end{array} \\ \hline 2 & 5 & 5 \\ \hline 9 & 3 & 8 \\ \hline 8 & 1 & 9 \\ \hline 3 & 6 & 15 \\ \hline 5 & 6 & 21 \\ \hline 7 & 1 & 22 \\ \hline \end{array} $

Total frequency $N=22$ (even)

$ \Rightarrow \quad \frac{N}{2}=\frac{22}{2}=11 $

Since, the cumulative frequency just greater than $\frac{N}{2}$ i.e., 11 is 15 and the value of $x$ corresponding to 15 is 3 .

$\therefore$ Median $=3$

Now, mean deviation

$ \left.=\frac{1}{N} \Sigma f_i \right\rvert\, x_i-\text { Median } \mid $

So, $\Sigma f_i\left|x_i-3\right|=44$

So, mean deviation $=\frac{44}{22}=2$

The numbers on a single die are 1 , $2,3,4,5$ and 6 .

$ \begin{aligned} \text { So, the mean } & =\frac{1+2+3+4+5+6}{6} \\ & =\frac{21}{6}=3.5 \end{aligned} $

When multiple dice are thrown, the mean of the sum of the numbers appearing on them = the sum of means of individual dice.

Since, each die has mean 3.5.

So, the mean of the sum of the numbers appearing on three dice

$ =3.5+3.5+3.5=10.5 $

$\bar{x}=\frac{1+2+3+5+8+13+17}{7} $

$ =\frac{49}{7}=7 $

$ \begin{aligned} \text { Variance } & =\frac{\Sigma\left(x_{1}-x\right)^{2}}{7} \\\\ & =\frac{218}{7}=3114 \end{aligned} $

To calculate the variance of the first 10 natural numbers that are multiples of 3, we use the formula for variance:

$ \sigma^2 = \frac{1}{n} \sum_{i=1}^{n}(x_i)^2 - (\bar{x})^2 $

Here, let $ x_i = 3k $ where $ k \in [1, 10] $. So the numbers are $ 3 \times 1, 3 \times 2, \ldots, 3 \times 10 $.

First, we find the sum of squares of these numbers:

$ \sigma^2 = \frac{1}{n} \times 9 \sum_{k=1}^{10} k^2 - 9(\bar{k})^2 $

Calculating step-by-step:

The formula for the sum of squares of the first $ n $ natural numbers is:

$ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} $

Substituting $ n = 10 $:

$ \sum_{k=1}^{10} k^2 = \frac{10 \times 11 \times 21}{6} $

The mean $ \bar{k} $ of the first 10 natural numbers is:

$ \bar{k} = \frac{1}{10} \sum_{k=1}^{10} k = \frac{10 \times 11}{2 \times 10} $

Plugging these into the variance formula:

$ \sigma^2 = \frac{9}{10} \times \frac{10 \times 11 \times 21}{6} - 9 \left(\frac{10 \times 11}{20}\right)^2 $

This simplifies to:

$ \sigma^2 = \frac{9}{10} \times 385 - 9 \times 30.25 $

Finally, calculating the result:

$ \sigma^2 = \frac{9}{10} \times 385 - 272.25 = 74.25 $

Thus, the variance of the first 10 multiples of 3 is $ 74.25 $.

We have, data $ 44,5,27,20,8,54,9 $, 14, 35

Data in ascending order is

$ 5,8,9,14,20,27,35,44,54 $

Mean

$ \begin{aligned} (\bar{x}) & =\frac{5+8+9+14+20+27+35+44+54}{9} \\\\ & =\frac{216}{9}=24 \end{aligned} $

Median $ (M)=20 $

$ M_{1}=\frac{1}{\eta} \Sigma\left|x_{i}-\bar{x}\right| $

$ =\frac{1}{9}[19+16+15+10+4+3+11+20+30] $ $ =\frac{128}{9} $

$ M_{2}=\frac{1}{\eta} \Sigma\left|x_{i}-M\right| $

$ =\frac{1}{9}[15+12+11+6+0+7+15+24+34] $

$ =\frac{124}{9} $

$ M_{1}-M_{2}=\frac{128}{9}-\frac{124}{9}=\frac{4}{9} $

Mean $ =1 $

$ \begin{array}{l} \Rightarrow n p=1 \Rightarrow P(X=2)=\frac{27}{128} \\\\ \Rightarrow{ }^{n} C_{2} p^{2} q^{n-2}=\frac{27}{128} \\\\ \Rightarrow \frac{n(n-1)}{2} p^{2} \cdot q^{n-2}=\frac{27}{128} \\\\ \Rightarrow \frac{n(n-1)}{2}\left(\frac{1}{n}\right)^{2} \times\left(1-\frac{1}{n}\right)^{n-2}=\frac{27}{128} \\\\ \Rightarrow\left(\frac{n-1}{2 n}\right)\left(\frac{n-1}{n}\right)^{n-2}=\frac{27}{128} \end{array} $

$ \begin{array}{l} \Rightarrow \quad \frac{(n-1)^{n-1}}{(n)^{n-1} \cdot 2}=\frac{27}{128}=\left(\frac{3}{4}\right)^{3} \times \frac{1}{4} \\\\ \therefore \quad n=4 \\\\ \text { Variance }=n p q=\frac{3}{4} \end{array} $

Mean, $\bar{x}=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{56}{8}=7$

Variance, $\sigma^2=\frac{\Sigma f_i\left(x_i-\bar{x}\right)^2}{\Sigma f_i}$

$ \begin{aligned} & =\frac{2(2-7)^2+\frac{3(6-7)^2+2(10-7)^2+(1)(14-7)^2}{8}}{=\frac{50+3+18+49}{8}=\frac{120}{8}=15} \end{aligned} $

Let assumed mean $A=5$ and $h=2$

Here, $N=13, \Sigma f_i d_i=0$

Mean $\bar{X}=A+h\left(\frac{1}{N} \Sigma f_i d_i\right)=5$

$\therefore \quad$ Mean deviation about mean $=\frac{1}{N} \Sigma f_i\left|x_i-\bar{X}\right|=\frac{20}{13}$

$ \begin{aligned} & \text { } x=40 \\ & \Rightarrow \sum x=40 \times 100=4000 \Rightarrow \sigma=5.1 \\ & \Rightarrow(5 \cdot 1)^2=\frac{\sum x_i^2}{100}-(40)^2 \\ & \Rightarrow \sum x_i^2=100(26.01+1600)=162601 \end{aligned} $

After correcting 50 by 40

$ \begin{aligned} \text { Correct } \sum x_i^2 & =162601-(50)^2+(40)^2 \\ & =161701 \end{aligned} $

Variance of 50 observation $=7$

When data is multiply by 6 , then new variance $=6^2 \times$ odd variance

$ =36 \times 7=252 $

and data is subtracted by 5 , then variance remains same.

Hence, final variance $=252$

$ \begin{aligned} & \text { Mean } \\ & =\frac{1+3+5+7+11+13+17+19+23}{9}=11 \end{aligned} $

$M$ (mean deviation)

$ =\frac{|1-11|+|3-11|+|5-11|+|7-11|+|11-11|+|13-11|+|17-11|+|19-11|+|23-11|}{9} $

$ =\frac{10+8+6+4+0+2+6+8+12}{9}=\frac{56}{9} $

$ \sigma^2=\frac{\Sigma\left(x_i-\bar{x}\right)^2}{9} $

$ \frac{\begin{array}{r} (1-11)^2+(3-11)^2+(5-11)^2+(7-11)^2 +(11-11)^2+(13-11)^2+(17-11)^2 \end{array}+(19-11)^2+(23-11)^2}{9} $

$ \begin{aligned} & =\frac{100+64+36+16+0+4+36+64+144}{9} \\ & =464 / 9 \\ & \text { Hence, } 3\left(\sigma^2-M\right)=3\left(\frac{464}{9}-\frac{56}{9}\right) \\ & =\frac{408}{3}=136 \end{aligned} $

$ \begin{aligned} &\text { Given that }\\ &\begin{aligned} & \quad \begin{aligned}\because 3 P(X=2) & =P(X=4) \\ \frac{3\left(e^{-\lambda} \lambda^2\right)}{2!} & =\frac{e^{-\lambda} \lambda^4}{4!} \\ 3 e^{-\lambda} \lambda^2 & =\frac{e^{-\lambda} \lambda^4}{12} \Rightarrow \lambda^2=36 \\\therefore \lambda & =6 \\ \quad P(X=6) & =\frac{e^{-\lambda} \lambda^6}{6!}=\frac{e^{-6}(6)^6}{6 \times 5 \times 4 \times 3 \times 2 \times 1} \\ & =\frac{6^4}{e^6 \times 5 \times 4}=\frac{6 \times 6 \times 6 \times 6}{e^6 \times 5 \times 4} \\ & =\frac{324}{5 e^6} \end{aligned} \end{aligned} \end{aligned} $

Data 2, 3, 5, 8, 12

Median = 5

Mean deviation from median

$ \begin{aligned} & \qquad \begin{array}{l} =\frac{\Sigma \mid x-\text { Median } \mid}{n} \\ =\frac{3+2+0+3+7}{5} \\ M=\frac{15}{5}=3 \end{array} \\ & \text { Variance } \sigma^2=\frac{\Sigma x^2}{n}-\left(\frac{\Sigma x}{n}\right)^2 \\ & =\frac{2^2+3^2+5^2+8^2+12^2}{5}-\left(\frac{2+3+5+8+12}{5}\right)^2 \\ & =13.2 \\ & \sigma^2-M=10.2 \end{aligned} $

Assertion (A) The variance of the first ' $n$ ' odd natural number is $\frac{n^2-1}{3}$, which is correct.

Reason ( $\mathbf{R}$ ) The sum of first $n$ odd natural number is $n^2$ and the sum of the squares of the first $n$ odd natural number is $\frac{(4 n-1)}{3} n$. Which is also correct.

and Reason (R) is correct explanation of Assertion (A).

Thus, Assertion (A) and Reason (R) both true. Reason ' $R$ ' is correct explanation of Assertion (A).

Statement (I) Range depends on minimum and maximum value only. So, it will not be affected by intermediate observations.

Statement (II) It is true that the mean deviation of an ungrouped data about the median is always less than or equal to the value of the mean deviation computed about any other measure of central tendency.

Statement (III) For a grouped data, range is approximated as the difference between lowest (minimum) value and the largest (maximum) value of the given data or among all given class interval.

Statements I, II are true but III is false.

Mean deviation of $x_1 x_2, x_3, \ldots \ldots x_n$

is 10 .

We know that, mean deviation of $a x_1+b, a x_2+b \ldots . a x_n+b$ is a times Mean deviation of $x_1, x_2 \ldots x_n$

∴ Mean deviation of

$ \begin{gathered} \frac{2 x_1+5}{3}, \frac{2 x_2+5}{3} \ldots . \frac{2 x_n+5}{3} \\ =10 \times \frac{2}{3}=\frac{20}{3} \end{gathered} $

Let $x_1=-y_1, x_2=-y_2, x_3=-y_3$, …., $x_n=-y_n$ where $y_1, y_2, y_3, \ldots \ldots y_n$ are positive numbers.

Now, the series is $-y_1,-y_2,-y_3, \ldots,-y_n \left(y_1>y_2>y_3>y_4>\ldots>y_n\right)$ If the sign first and last term are changed, then the series in ascending order are

$ \begin{array}{r} -y_2,-y_3, \ldots-y_n, y_n, y_1 \\ \text { Range }=\left|y_1\right|-\left|-y_2\right|=\left|x_1\right|-x_2 \end{array} $

$ \bar{x}=\frac{1+3+5+7+11+13+17+19+23}{9} $

$ \begin{aligned} & =\frac{99}{9}=11 \\ & M D|\bar{x}|=\sum \frac{\left|x_i-\bar{x}\right|}{n} \\ & =\frac{1}{9}[|1-11|+|3-11|+|5-11| \\ & \quad+|7-11|+|11-11|+|13-11| \\ & \quad+|17-11|+|19-11|+\mid 23-11] \\ & =\frac{1}{9}[10+8+6+4+0+2+6+8+12] \\ & =\frac{56}{9}=6 \frac{2}{9} \end{aligned} $

$ \begin{aligned} &\text { Mean }\\ &\begin{gathered} (x)=\frac{1+3+4+7+11+18+29+47+78}{9} \\ =\frac{198}{9}=22 \end{gathered} \end{aligned} $

$ \begin{array}{cc} \hline x_i & \left|x_i-x\right| \\ \hline 1 & 21 \\ \hline 3 & 19 \\ \hline 4 & 18 \\ \hline 7 & 15 \\ \hline 11 & 11 \\ \hline 18 & 4 \\ \hline 29 & 7 \\ \hline 47 & 25 \\ \hline 78 & 56 \\ \hline \Sigma\left|x_i-x\right|=176 / 9 \\ \hline \end{array} $

Given, mean $=\bar{x}$

Mean of absolute deviation,

$ \begin{aligned} & =\frac{\Sigma|x-\bar{x}|}{2} \\ & =\text { Mean deviation of the data } \end{aligned} $

The variance of first $2 k$ natural numbers

$ \begin{aligned} S_1 & =\frac{2 k(2 k+1)(4 k+1)}{6 \times 2 k}-\left(\frac{2 k(2 k+1)}{2 \times 2 k}\right)^2 \\ & =(2 k+1)\left[\frac{4 k+1}{6}-\frac{2 k+1}{4}\right] \\ & =\frac{2 k+1}{12}[8 k+2-6 k-3] \\ & =\frac{4 k^2-1}{12} \end{aligned} $

and variance of first $k$ natural numbers

$ \begin{aligned} S_2= & \frac{k(k+1)(2 k+1)}{6 \times k}-\left(\frac{k(k+1)}{2 \times k}\right)^2 \\ & =(k+1)\left[\frac{2 k+1}{6}-\frac{k+1}{4}\right] \\ & =\frac{k+1}{12}[4 k+2-3 k-3]=\frac{k^2-1}{12} \\ \therefore \quad & \frac{S_1}{S_2}=\frac{4 k^2-1}{k^2-1}=4+\frac{3}{k^2-1},(k>1) \\ \therefore \quad & \frac{S_1}{S_2} \in(1,4] \end{aligned} $

From the given information, we can say that

$ \begin{aligned} & \sqrt{\frac{\sum_{i=1}^{100}\left(x_i-\bar{x}\right)^2}{100}}=5 \text { and } \sqrt{\frac{\sum_{i=1}^{100}\left(y_i-\bar{y}\right)^2}{100}}=6 \\ & \therefore \quad \sum_{i=1}^{100}\left(x_i-\bar{x}\right)^2=2500 \end{aligned} $

and $\sum_{i=1}^{100}\left(y_i-\bar{y}\right)^2=3600$

and it is also given that

$ \begin{gathered} \sum_{i=1}^{100}\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right)=600 \\ \because \quad a^2+b^2-2 a b=(a-b)^2 \\ \sum_{i=1}^{100}\left(x_i-\bar{x}\right)^2+\sum_{i=1}^{100}\left(y_i-\bar{y}\right)^2 \\ -\sum_{i=1}^{100}\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right) \end{gathered} $

$ \begin{aligned} & =\sum_{i=1}^{100}\left(\left(x_i-\bar{x}\right)-\left(y_i-\bar{y}\right)\right)^2 \\ \Rightarrow \quad & \sum_{i=1}^{100}\left[\left(x_i-\bar{x}\right)-\left(y_i-\bar{y}\right)\right]^2 \\ & =2500+3600-1200 \\ \Rightarrow \quad & \sum_{i=1}^{100}\left[\left(x_i-y_i\right)-(\bar{x}-\bar{y})\right]^2=4900 \end{aligned} $

$ \Rightarrow \sqrt{\frac{\sum_{i=1}^{100}\left[\left(x_i-y_i\right)-(\bar{x}-\bar{y})\right]^2}{100}}=7 $

$ \begin{aligned} & \text { Given, } \frac{1}{4} \text { th observation }=a \\ & \text { another } \frac{1}{4} \text { th observation }=-a \\ & \quad \frac{1}{4} \text { th observation }=b \\ & \Rightarrow \frac{1}{4} \text { th observation }=-b \\ & \therefore \quad \bar{x}=\frac{a-a+b-b}{n}=0 \end{aligned} $

$ \begin{aligned} & \text { Variance }=a b \\ & \qquad a b=\frac{\Sigma x_i^2}{n}-(\bar{x})^2=\frac{\Sigma x_i^2}{n} \\ & \Rightarrow \quad a b=\frac{\frac{n}{4}\left(a^2+a^2\right)+\frac{n}{4}\left(b^2+b^2\right)}{n} \end{aligned} $

$ \begin{array}{llrl} \Rightarrow & a b=\frac{a^2+b^2}{2} & \\ \Rightarrow & a^2+b^2-2 a b =0 \\ \Rightarrow & (a-b)^2 =0 \\ \Rightarrow & a =b \end{array} $

Let variance of $x_1, x_2, x_3, \ldots, x_n$ be $\sigma^2$, then variance of $4 x_1, 4 x_2, \ldots, 4 x_n$ is $4^2 \sigma^2$ is $16 \sigma^2$ And variance dependent on change of scale.

$ \begin{aligned} &\text { Mean of first five prime numbers }\\ &\begin{aligned} & =\frac{2+3+5+7+11}{5} \\ & =\frac{28}{5}=5.6 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { ∴ Mean deviation about mean }\\ &\begin{aligned} & =\frac{|2-5.6|+|3-5.6|+|5-5.6|+|7-5.6|}{+|11-5.6|} \\ & =\frac{3.6+2.6+0.6+1.4+5.4}{5} \\ & =\frac{13.6}{5}=2.72 \\ & \text { Variance }=\frac{\Sigma x i^2}{n}-\left(\frac{\Sigma x i}{n}\right)^2 \\ & \quad=\frac{4+9+25+49+121}{5}-(5.6)^2 \\ & \quad=41.6-31.36=10.24 \end{aligned} \end{aligned} $

$ \begin{array}{c|c|c|c|c|c} \hline \mathbf{C l} & \boldsymbol{f}_{\boldsymbol{i}} & \boldsymbol{x}_{\boldsymbol{i}} & \boldsymbol{f}_{\boldsymbol{i}} \boldsymbol{x}_{\boldsymbol{i}} & \boldsymbol{x}_{\boldsymbol{i}}^2 & \boldsymbol{f}_{\boldsymbol{i}} \boldsymbol{x}_{\boldsymbol{i}}^2 \\ \hline 0-5 & 4 & 2.5 & 10 & 6.25 & 25 \\ \hline 5-10 & 1 & 7.5 & 7.5 & 56.25 & 56.25 \\ \hline 10-15 & 10 & 12.5 & 125 & 156.25 & 1562.5 \\ \hline 15-20 & 3 & 17.5 & 52.5 & 306.25 & 918.75 \\ \hline 20-25 & 2 & 22.5 & 45 & 506.25 & 1012.50 \\ \hline \text { Total } & 20 & & 240 & & 3575 \\ \hline \end{array} $

$ \begin{gathered} \text { } \begin{aligned} Now,\,\,\, & \operatorname{var}(x)=\frac{1}{N} \Sigma f_i x_i^2-\left(\frac{1}{N} \Sigma f_i x_i\right)^2 \\ = & \frac{1}{20} \times 3575-\left(\frac{1}{20} \times 240\right)^2 \\ = & 178.75-144=34.75 \end{aligned} \end{gathered} $

We have,

$ \begin{aligned} & \text { Mean }=6 \\ & \therefore \quad \frac{8+9+6+5+x+4+6+5}{8}=6 \\ & \Rightarrow \quad 43+x=48 \Rightarrow x=5 \end{aligned} $

So, data are $8,9,6,5,5,4,6,5$

$ \begin{array}{lc} \therefore & x_i=8,9,6,5,5,4,6,5 \\ \Rightarrow & x_i^2=64,81,36,25,25,16,36,25 \\ \therefore & \Sigma x_i^2=64+81+36+25+25+16+36+25 \\ & =308 \\ \therefore & \text { SD }(x)=\sqrt{\frac{1}{N} \Sigma x_i^2-(\bar{x})^2} \\ & =\sqrt{\frac{1}{8} \times 308-(6)^2}=\sqrt{38.5-36} \\ & =\sqrt{2.5} \quad=1.58 \end{array} $