Statistics

$\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 $=\frac{1}{N} \Sigma f_i\left|x_i-\bar{X}\right|=\frac{20}{13}$

To find the variance of the data set $7, 8, 9, 7, 8, 7, \lambda, 8$ with a mean of 8, follow these steps:

First, solve for $\lambda$:

$ \frac{7 + 8 + 9 + 7 + 8 + 7 + \lambda + 8}{8} = 8 $

Simplify the equation:

$ \frac{54 + \lambda}{8} = 8 $

Multiply both sides by 8:

$ 54 + \lambda = 64 $

Solve for $\lambda$:

$ \lambda = 64 - 54 = 10 $

Next, calculate the variance using the formula:

$ \text{Variance} = \frac{\Sigma x_i^2}{8} - \bar{x}^2 $

Where $\Sigma x_i^2$ is the sum of squares of each data point:

$ \begin{aligned} & \Sigma x_i^2 = 7^2 + 8^2 + 9^2 + 7^2 + 8^2 + 7^2 + 10^2 + 8^2 \\ & = 49 + 64 + 81 + 49 + 64 + 49 + 100 + 64 \\ & = 520 \end{aligned} $

Calculate the variance:

$ \frac{520}{8} - 8^2 = 65 - 64 = 1 $

Thus, the variance of the data set is 1.

Sum of first $n$ even natural number

$ \begin{aligned} = & 2+4+6+\ldots . \\ = & 2(1+2+3+\ldots .) \\ = & 2 \times \frac{n(n+1)}{2}=n(n+1) \\ \therefore \quad \text { Mean } & =\frac{n(n+1)}{n}=(n+1) \end{aligned} $

We know that variance $=\frac{1}{n}\left(\Sigma x_1\right)^2-(\bar{x})^2$

$ \begin{aligned} & =\frac{1}{n}\left(2^2+4^2+6^2+\ldots\right)-(n+1)^2 \\ & =\frac{4}{n}\left(1^2+2^2+3^2+\ldots . .\right)-(n+1)^2 \\ & =\frac{4}{n} \times \frac{n(n+1)(2 n+1)}{6}-(n+1)^2 \\ & =\frac{n+1}{3}[2(2 n+1)-2(n+1)]=\frac{n^2-1}{3} \end{aligned} $

$\therefore$ Difference between the variance of the first $n$ even natural numbers any arithmetic mean $=\frac{399}{3}-21=112$

So, statement I is false and II is true.

Given the equations:

$ \Sigma(x_i + 2)^2 = 28n $

$ \Sigma(x_i - 2)^2 = 12n $

Let's start by expanding and simplifying each equation.

First equation:

$ \begin{aligned} \Sigma(x_i + 2)^2 & = \Sigma(x_i^2 + 4x_i + 4) \\ & = \Sigma x_i^2 + 4\Sigma x_i + 4n \\ & = 28n \end{aligned} $

Second equation:

$ \begin{aligned} \Sigma(x_i - 2)^2 & = \Sigma(x_i^2 - 4x_i + 4) \\ & = \Sigma x_i^2 - 4\Sigma x_i + 4n \\ & = 12n \end{aligned} $

Let's introduce the following notation:

$ S = \Sigma x_i $

$ S_2 = \Sigma x_i^2 $

Using these, we can set up the system of equations:

$ S_2 + 4S + 4n = 28n $

$ S_2 - 4S + 4n = 12n $

Subtract the second equation from the first:

$ \begin{aligned} (S_2 + 4S + 4n) - (S_2 - 4S + 4n) &= 28n - 12n \\ 8S &= 16n \\ S &= 2n \end{aligned} $

Substituting $ S = 2n $ back into the first equation:

$ \begin{aligned} S_2 + 4(2n) + 4n &= 28n \\ S_2 + 8n + 4n &= 28n \\ S_2 &= 28n - 12n \\ S_2 &= 16n \end{aligned} $

Now, to find the variance $\sigma^2$, use the formula:

$ \begin{aligned} \sigma^2 &= \frac{1}{n} \Sigma(x_i - \mu)^2 \\ &= \frac{1}{n} \Sigma x_i^2 - \mu^2 \end{aligned} $

Since $ \Sigma x_i = S = 2n $, the mean $\mu$ is:

$ \mu = \frac{\Sigma x_i}{n} = \frac{2n}{n} = 2 $

Therefore, the variance is:

$ \begin{aligned} \sigma^2 &= \frac{1}{n} \times 16n - 2^2 \\ &= 16 - 4 \\ &= 12 \end{aligned} $

$ \begin{aligned} &\text { Given frequency distribution }\\ &\begin{array}{llll} \hline \boldsymbol{x}_{\boldsymbol{i}} & 4 & 3 & 1 \\ \hline \boldsymbol{f}_{\boldsymbol{L}} & 1 & 3 & 5 \\ \hline \end{array} \end{aligned} $

$ \begin{array}{c|c|c|c|c|c} \hline \boldsymbol{x}_i & \boldsymbol{f}_i & \boldsymbol{f}_i \boldsymbol{x}_i & \left(x_i-\bar{x}\right) & \left(x_i-\bar{x}\right)^2 & f_i\left(x_i-\bar{x}\right)^2 \\ \hline 4 & 1 & 4 & 2 & 4 & 4 \\ \hline 3 & 3 & 9 & 1 & 1 & 3 \\ \hline 1 & 5 & 5 & -1 & 1 & 5 \\ \hline & \Sigma_i=9 & \Sigma \sum_i x_i=18 & & & 12 \\ \hline \end{array} $

$ \therefore \quad \text { Mean } \bar{x}=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{18}{9}=2 $

Now, standard deviation

$ \sigma=\sqrt{\frac{\Sigma f_i\left(x_i-\bar{x}\right)^2}{\Sigma f_i}}=\sqrt{\frac{12}{9}}=\frac{2}{\sqrt{3}} $

So, coefficient of variance $=\frac{\sigma}{\bar{x}} \times 100$

$ =\frac{2}{\sqrt{3}} \times \frac{1}{2} \times 100=\frac{100}{\sqrt{3}} $

$ \text { Mean } A=15, h=6 $

$ \begin{array}{c|c|c|c|c|c|c} \hline \begin{array}{l} \text { Class } \\ \text { Interval } \end{array} & \begin{array}{l} \text { Mid } \\ \text { value } x_i \end{array} & \text { Frequency } & d_{\boldsymbol{i}}=\frac{x_{\boldsymbol{i}}-15}{\mathbf{6}} & f \boldsymbol{d}_{\boldsymbol{i}} & x_{\boldsymbol{i}}-\bar{x} & \frac{\boldsymbol{f}_{\boldsymbol{i}}}{\boldsymbol{x}_{\boldsymbol{i}}-\boldsymbol{x}} \\ \hline 0-6 & 3 & 1 & -2 & -2 & 12 & 12 \\ \hline 6-12 & 9 & 2 & -1 & -2 & 6 & 12 \\ \hline 12-18 & 15 & 3 & 0 & 0 & 0 & 0 \\ \hline 18-24 & 21 & 2 & 1 & 2 & 6 & 12 \\ \hline 24-30 & 27 & 1 & 2 & 2 & 12 & 12 \\ \hline & & N=9 & & 0 & 36 & 48 \\ \hline \end{array} $

Mean, $\bar{X}=A+h\left(\frac{1}{N} \Sigma f_i d_i\right)=15+6\left[\frac{1}{9}(0)\right]$

$ \bar{X}=15 $

Mean-deviation about mean

$ \begin{aligned} & =\frac{1}{N} \sum f_i\left|x_i-\bar{X}\right|=\frac{1}{9} \times 48 \\ & =\frac{1}{9} \times 48=\frac{16}{3} \end{aligned} $

Given data is

Now, $\bar{x}=\frac{4+72+140+112+72}{80}$

$ =\frac{400}{80}=5 $

Thus, mean deviation about mean is $m=\frac{4 \times 4+24 \times 2+28 \times 0+16 \times 2+8 \times 4}{80}$ $=\frac{8}{5}$

and $\sigma^2=\frac{4 \times 4^2+24 \times 2^2+28 \times 0^2+16 \times(-2)^2+8 \times(-4)^2}{80}=\frac{22}{5}$

Now, $m+\sigma^2=\frac{8}{5}+\frac{22}{5}=\frac{30}{5}=6$

Given that $\bar{x}$ and $\bar{y}$ are the arithmetic means of the runs for batsmen $A$ and $B$, respectively, and $\sigma_A$ and $\sigma_B$ represent the standard deviations of their runs.

Since Batsman $A$ is more consistent than Batsman $B$, we have:

$ \sigma_A < \sigma_B \Rightarrow \frac{\sigma_A}{\sigma_B} < 1 $

Additionally, we know that if Batsman $A$ is a higher run scorer than Batsman $B$, then:

$ \bar{x} > \bar{y} \Rightarrow \frac{\bar{x}}{\bar{y}} > 1 $

Combining both conditions, we get:

$ 0 < \frac{\sigma_A}{\sigma_B} < \frac{\bar{x}}{\bar{y}} \quad \text{and} \quad \frac{\bar{x}}{\bar{y}} > 1 $

This indicates that Batsman $A$ scores more runs only when the ratio of the standard deviations is less than the ratio of the arithmetic means, with the latter being greater than 1.

We have, $20,5,15,2,7,3,11$.

$ \begin{aligned} \text { Mean } & =\frac{20+5+15+2+7+3+11}{7} \\ & =\frac{63}{7}=9 \end{aligned} $

Data in ascending order $=2,3,5,7,11$,

15, 20

Median = 7

$m=$ Mean deviations about means

$ \begin{aligned} & |20-9|+|5-9|+|15-9|+|2-9| \\ = & \frac{+|7-9|+|3-9|+|11-9|}{7} \\ = & \frac{11+4+6+7+2+6+2}{7}=\frac{38}{7} \end{aligned} $

$M=$ Mean deviation about median

$ \begin{aligned} & M=\frac{|20-7|+|5-7|+|15-7|+|2-7|}{+|7-7|+|3-7|+|11-7|} \\ & =\frac{13+2+8+5+0+4+4}{7}=\frac{36}{7} \end{aligned} $

Mean of $m$ and $M=\left(\frac{38}{7}+\frac{36}{7}\right) \frac{1}{2}=\frac{37}{7}$

MD (Mean of $m$ and $M$ )

$ \begin{aligned} & =\frac{1}{2}\left\{\left|\frac{36}{7}-\frac{37}{7}\right|+\left|\frac{38}{7}-\frac{37}{7}\right|\right\} \\ & =\frac{1}{2} \times \frac{2}{7}=\frac{1}{7} \end{aligned} $

To find the variance given the coefficient of variation (CV) and the mean, we can use the following relationship:

The formula for the coefficient of variation (CV) is given by:

$ \text{CV} = \frac{\sigma}{\mu} \times 100 $

where $\sigma$ is the standard deviation and $\mu$ is the mean.

In this problem, the coefficient of variation is 25, and the mean ($\mu$) is 44. Substituting these values into the equation, we have:

$ \frac{\sigma}{44} \times 100 = 25 $

Solving for $\sigma$, we get:

$ \sigma = \frac{25 \times 44}{100} = 11 $

The variance ($\sigma^2$) is the square of the standard deviation:

$ \sigma^2 = 11^2 = 121 $

Therefore, the variance is 121.

Given that the mean of the observations ${x, y, 1, 2, 4, 5}$ is 5, we get the equation:

$x + y + 1 + 2 + 4 + 5 = 6 \cdot 5 \Rightarrow x + y = 18$ ..........$(1)$

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:

$10 = \frac{\Sigma x_i^2}{6} - 5^2 = \frac{\Sigma x_i^2}{6} - 25 \Rightarrow \Sigma x_i^2 = 210$.

Here, $\Sigma x_i^2$ is the sum of the squares of all the observations, thus:

$x^2 + y^2 + 1 + 4 + 16 + 25 = 210 \Rightarrow x^2 + y^2 = 164$ .........$(2)$

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:

$MD(5) = \frac{1}{6} [|1-5| + |2-5| + |4-5| + |5-5| + |8-5| + |10-5|] = \frac{16}{6} = \frac{8}{3}$.

So, the mean deviation about the mean is $\frac{8}{3}$, and the correct answer is Option B, $\frac{8}{3}$.

To solve this problem, let's break it down step by step :

Step 1 : Determine the mean of set C

The mean of set A = $5$

The mean of set B = $8$

After subtracting 3 from each element in set A, the new mean becomes $5 - 3 = 2$. After adding 2 to each element in set B, the new mean becomes $8 + 2 = 10$.

When we combine both modified sets to form set C, the mean of set C is the weighted average of the means of these modified sets:

Mean of C = $\frac{5 \times 2 + 5 \times 10}{10} = \frac{10 + 50}{10} = 6$

Step 2 : Determine the variance of set C

Variance is defined as the expectation of the squared deviation of a random variable from its mean. One property of variance is that, if you add (or subtract) a constant from each data point in a set, the variance of the set does not change.

Thus, the variance of the elements in set A remains $12$ even after subtracting 3 from each element, and the variance of the elements in set B remains $20$ even after adding 2 to each element.

Now, when combining variances from two datasets into one :

Variance of C

= $\frac{n_1 \times \text{Variance of A} + n_1 \times (\text{Mean of modified A} - \text{Mean of C})^2 + n_2 \times \text{Variance of B} + n_2 \times (\text{Mean of modified B} - \text{Mean of C})^2}{n_1 + n_2}$

Given :

$n_1 = n_2 = 5$

Variance of A = $12$, Variance of B = $20$

Mean of modified A = $2$, Mean of modified B = $10$, Mean of C = $6$

Plugging in the values, we get :

Variance of C = $\frac{5 \times 12 + 5 \times (2 - 6)^2 + 5 \times 20 + 5 \times (10 - 6)^2}{10}$

Variance of C = $\frac{60 + 80 + 100 + 80}{10} = \frac{320}{10} = 32$

Step 3 : Sum of the mean and variance of set C

Sum = Mean of C + Variance of C = $6 + 32 = 38$

So, the correct answer is :

Option C : 38.

Let ${a_1} = a \Rightarrow {a_2} = a + 1,\,.....\,{a_{100}} = a + 90$

$\mu = {{{{100a + (99 \times 100)} \over 2}} \over {100}} = a + {{99} \over 2}$

M.D $ = {{\sum {|{a_1} - \mu |} } \over {100}} = {{{{\left( {{{99} \over 2} + {{97} \over 2}\, + \,...\, + \,{1 \over 2}} \right)}^2}} \over {100}} = {{2500} \over {100}} = 25$

$\therefore$ $a \to z$

3 rotten apples are mixed with 7 good apples.

$\therefore$ Total apples = 10

Among those 10 apples 4 are chosen randomly.

${x_i}$ = Number of rotten apples drawn.

${p_i}$ = Probability of rotten apple.

We know,

Mean $(\mu ) = \sum {{p_i}{x_i}} $

$ = 0 + {{105} \over {210}} + {{126} \over {210}} + {{21} \over {210}}$

$ = {{252} \over {210}} = {6 \over 5}$

Also,

Variance $({\sigma ^2}) = \left( {\sum {{p_i}{{({x_i})}^2}} } \right) - {\mu ^2}$

$ = {{105} \over {210}} + {{252} \over {210}} + {{63} \over {210}} - {{36} \over {25}}$

$ = {1 \over 2} + {{12} \over {10}} + {3 \over {10}} - {{36} \over {25}} = {{14} \over {25}}$

$\therefore$ $10(\mu^2 + {\sigma ^2})$

$ = 10\left( {({6 \over 5})^2 + {{14} \over {25}}} \right)$

$ = 10\left( {{{36 + 14} \over {25}}} \right)$

$ = 10 \times {{50} \over {25}} = 20$

${a_1},{a_2},{a_3},{a_4},{a_5},{a_6}$ are in AP.

Let

${a_1} = a$

${a_2} = a + d$

${a_3} = a + 2d$

${a_4} = a + 3d$

${a_5} = a + 4d$

${a_6} = a + 5d$

Now Mean of ${a_1},{a_2},{a_3},{a_4},{a_5}$ and ${a_6}$ is

$ = {{{a_1} + {a_2} + {a_3} + {a_4} + {a_5} + {a_6}} \over 6} = {{19} \over 2}$

$ \Rightarrow {{6a + 15d} \over 6} = {{19} \over 2}$

$ \Rightarrow 2a + 5d = 19$ ...... (1)

Also, given,

${a_1} + {a_3} = 10$

$ \Rightarrow a + a + 2d = 10$

$ \Rightarrow 2a + 2d = 10$

$ \Rightarrow a + d = 5$ ..... (2)

From equation (1) and (2), we get $a = 2$ and $d = 3$

$\therefore$ AP is 2, 5, 8, 11, 14, 17

Now, Variance $({\sigma ^2}) = {{\sum {x_i^2} } \over 6} - {\left( {\overline x } \right)^2}$

$ = {{{2^2} + {5^2} + {8^2} + {{11}^2} + {{14}^2} + {{17}^2}} \over 6} - {\left( {{{19} \over 2}} \right)^2}$

$ = {{105} \over 4}$

$\therefore$ $8{\sigma ^2} = 8 \times {{105} \over 4} = 210$

$ \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).

Median $ = {{2k + 12} \over 2} = k + 6$

Mean deviation $ = \sum {{{|{x_i} - M|} \over n} = 6} $

$ \Rightarrow {{(k + 3) + (k + 1) + (k - 1) + (6 - k) + (6 - k) + (10 - k) + (15 - k) + (18 - k)} \over 8}$

$\therefore$ ${{58 - 2k} \over 8} = 6$

$k = 5$

Median $ = {{2 \times 5 + 12} \over 2} = 11$

Given,

5 numbers are 3, 7, 12, a, 43 $-$ a

$\therefore$ Mean $(\overline x ) = {{3 + 7 + 12 + a + 43 - a} \over 5}$

$ = {{65} \over 5}$

$ = 13$

We know,

Variance $({\sigma ^2}) = {{\sum {x_1^2} } \over n} - {(\overline x )^2}$

$ = {{{3^2} + {7^2} + {{12}^2} + {a^2} + {{(43 - a)}^2}} \over 5} - {(13)^2}$

$ = {{9 + 49 + 144 + {a^2} + 1849 + {a^2} - 86a} \over 5} - 169$

$ = {{2{a^2} - 86a + 2051} \over 5} - 169$

$ = {{2{a^2} - 86a + 2051 - 845} \over 5}$

$ = {{2{a^2} - 86a + 1206} \over 5}$

$ = {2 \over 5}({a^2} - 43a + 603)$

Variance will be natural number if ${a^2} - 43a + 603$ in multiple of 5.

$\therefore$ ${a^2} - 43a + 603 = 5n$

$ \Rightarrow {a^2} - 43a + 603 - 5n = 0$ ..... (1)

$\therefore$ $a = {{43\, \pm \,\sqrt D } \over 2}$

Now "a" will be natural number if

(1) D is a perfect square and

(2) $43\, \pm \,\sqrt D $ is multiple of 2

From equation (1),

$D = {( - 43)^2} - 4\,.\,1\,.\,(603 - 5n)$

$ = 1849 - 2412 + 20n$

$ = 20n - 563$

For any value of n, unit digit of 20n is always 0. Then 20n $-$ 563 will give a number whose unit digit is 7.

For perfect square numbers ex : 1² = 1, 2² = 4, 3² = 9, 4² = 16, 5² = 25, 6² = 36, 7² = 49, 8² = 64, 9² = 81, 10² = 100

So, unit digit is either 1, 4, 6, 5, 6, 9, 0 it can't be 7. So D can't be perfect square.

Mean $(\overline x ) = {{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}} \over 5}$

Given, ${{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}} \over 5} = {{24} \over 5}$

$ \Rightarrow {x_1} + {x_2} + {x_3} + {x_4} + {x_5} = 24$ ...... (1)

Now, Mean of first 4 observation

$ = {{{x_1} + {x_2} + {x_3} + {x_4}} \over 4}$

Given, $ = {{{x_1} + {x_2} + {x_3} + {x_4}} \over 4} = {7 \over 2}$

$ \Rightarrow {x_1} + {x_2} + {x_3} + {x_4} = 14$ ...... (2)

From equation (1) and (2), we get

$14 + {x_5} = 24$

$ \Rightarrow {x_5} = 10$

Now, variance of first 5 observation

$ = {{\sum {x_i^2} } \over n} - {\left( {\overline x } \right)^2}$

$ = {{x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2} \over 5} - {\left( {{{24} \over 5}} \right)^2}$

Given,

${{x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2} \over 5} - {\left( {{{24} \over 5}} \right)^2} = {{194} \over {24}}$

$ \Rightarrow x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 = 5\left( {{{194} \over {25}} + {{576} \over {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$

Now, variance of first 4 observation

$ = {{x_1^2 + x_2^2 + x_3^2 + x_4^2} \over 4} - {\left( {{7 \over 2}} \right)^2}$

Given,

${{x_1^2 + x_2^2 + x_3^2 + x_4^2} \over 4} - {\left( {{7 \over 2}} \right)^2} = a$

$ \Rightarrow {{54} \over 4} - {{49} \over 4} = a$

$ \Rightarrow a = {5 \over 4}$

$\therefore$ $4a + {x_5}$

$ = 4 \times {5 \over 4} + 10 = 15$

Mean $ = {{4 + 5 + 6 + 6 + 7 + 8 + x + y} \over 8} = 6$

$\therefore$ $x + y = 12$ ..... (i)

And variance

$ = {{{2^2} + {1^2} + {0^2} + {0^2} + {1^2} + {2^2} + {{(x - 6)}^2} + {{(y - 6)}^2}} \over 8}$

$ = {9 \over 4}$

$\therefore$ ${(x - 6)^2} + {(y - 6)^2} = 8$ ..... (ii)

From (i) and (ii)

x = 4 and y = 8

$\therefore$ ${x^4} + {y^2} = 320$

Given $\overline x = 15,\,\sigma = 2 \Rightarrow {\sigma ^2} = 4$

$\therefore$ ${x_2} + {x_2} + \,\,.....\,\, + \,\,{x_{50}} = 15 \times 50 = 750$

$4 = {{x_1^2 + x_2^2 + \,\,.....\,\, + \,\,x_{50}^2} \over {50}} - 225$

$\therefore$ $x_1^2 + x_2^2 + \,\,.....\,\, + \,\,x_{50}^2 = 50 \times 229$

Let a be the correct observation and b is the incorrect observation

then $a + b = 70$

and $16 = {{750 - b + a} \over {50}}$

$\therefore$ $a - b = 50 \Rightarrow a = 60,\,b = 10$

$\therefore$ Correct variance $ = {{50 \times 229 + {{60}^2} - {{10}^2}} \over {50}} - 256 = 43$