Statistics

$\because$ $\overline x = 6 = {{a + b + 8 + 5 + 10} \over 5} \Rightarrow a + b = 7$ ...... (i)

And ${\sigma ^2} = {{{a^2} + {b^2} + {8^2} + {5^2} + {{10}^2}} \over 5} - {6^2} = 6.8$

$ \Rightarrow {a^2} + {b^2} = 25$ ..... (ii)

From (i) and (ii) (a, b) = (3, 4) or (4, 3)

Now mean deviation about mean

$M = {1 \over 5}(3 + 2 + 2 + 1 + 4) = {{12} \over 5}$

$ \Rightarrow 25M = 60$

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} $

$\text { Mean } \bar{x}=\frac{\text { Sum of quantities }}{n}$

$=\frac{\frac{n}{2}(a+l)}{n}$ [$l$ = last team]

$\begin{aligned} & =\frac{1}{2}(a+l)=\frac{1}{2}(1+1+100 d) \\ & =1+50 d \end{aligned}$

$\begin{aligned} & \mathrm{MD}=\frac{1}{n} \Sigma\left|x_i-\bar{x}\right| \\ & \Rightarrow 255=\frac{1}{101}[50 d+49 d+48 d+ .... +d+0+d + .... + 50d]\\ & \Rightarrow 255=\frac{2 d}{101}\left[\frac{50 \times 51}{2}\right] \\ & d=\frac{255 \times 101}{50 \times 51}=10.1 \end{aligned}$

Given, data of number are $p, 6,6,7,8,11,15$ and 16.

$\begin{aligned} \text { Mean } \bar{x} & =\frac{\text { Sum of all observations }}{\text { Total number of observations }} \\ & =\frac{p+6+6+7+8+11+15+16}{8} \end{aligned}$

$\begin{aligned} & \Rightarrow \quad \frac{69+p}{8}=3 p \quad \text{(given)}\\ & \Rightarrow \quad 24 p-p=69 \Rightarrow 23 p=69 \\ & \Rightarrow \quad p=3 \\ \end{aligned}$

So given data is 3, 6, 6, 7, 8, 11, 15, 16 and mean is 9.

So, the mean deviation

$\begin{aligned} & |3-9|+|6-9|+|6-9|+|7-9|+|8-9| \\ = & \frac{+|11-9|+|15-9|+|16-9|}{8} \\ = & \frac{6+3+3+2+1+2+6+7}{8}=\frac{30}{8}=3.75 \end{aligned}$

Mean,

$\begin{aligned} \bar{x} & =\frac{5+6+7+8+6+9+13+12+15}{9} \\ & =\frac{81}{9}=9 \end{aligned}$

Deviation, $x_i-\bar{x}$ :

$\begin{aligned} & 5-9,6-9,7-9,8-9,6-9,9-9,13-9,12-9,15-9 \\ & \Rightarrow-4,-3,-2,-1,-3,0,4,3,6 \\ & \left|x_i-\bar{x}\right|: 4,3,2,1,3,0,4,3,6 \end{aligned}$

$\begin{aligned} \mathrm{MD} & =\frac{\sum_\limits{i=1}^9\left|x_i-\bar{x}\right|}{9} \\ & =\frac{4+3+2+1+3+0+4+3+6}{9} \\ & =\frac{26}{9}=2.88 \end{aligned}$

Let data be

$x_1, x_2, x_3, \ldots, x_n$

Given that,

$\begin{aligned} \frac{x_1+x_2+x_3+\ldots+x_n}{n} & =10 \\ \Rightarrow \quad x_1+\ldots+x_n & =10 n \end{aligned}$

When all observation multiplied by 2, we obtain new data as $2 x_1, 2 x_2, \ldots, 2 x_n$ then

$\begin{aligned} \text { Mean } & =\frac{2 x_1+2 x_2+\ldots+2 x_n}{n} \\ & =\frac{2\left(x_1+\ldots+x_n\right)}{n}=\frac{2(10 n)}{n}=20 \end{aligned}$

Mean $ = {{ - 1 + 0 + 4} \over 3} = 1$

Deviation $ = | - 1 - 1|,|0 - 1|,|4 - 1| = 2,1,3$

Mean deviation $ = {{2 + 1 + 3} \over 3} = 2$

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

$1014 = {{{{60}^2} + {a^2} + {b^2}} \over 3} - {60^2}$

$10242 = {a^2} + {b^2}$ ..... (i)

$a + b = 120$ .... (ii)

From Eqs. (i) and (ii), we get

a = 21 and b = 99

$\begin{aligned} & \Sigma P(X=r)=1 \\ & K+K+\frac{1}{7}+2 K+\frac{2}{5}=1 \Rightarrow 4 K=\frac{3}{5}-\frac{1}{7} \\ & 4 K=\frac{16}{35} \Rightarrow K=\frac{4}{35} \\ & \propto=\frac{\Sigma P_i X_i}{\Sigma P_i}=\Sigma P_i X_i=K+\frac{1}{7}+4 K+\frac{6}{5} \\ & =5 K+\frac{47}{35}=5\left(\frac{4}{35}\right)+\frac{47}{35}=\frac{67}{35}\end{aligned}$

Mean $\bar{x}=5$ and $\Sigma x^2=400$

$\begin{aligned} \because & \operatorname{var}(x)=0 \\ \because \operatorname{var}(x) & =\frac{1}{n} \sum_{i=1}^n\left(x_i-\bar{x}\right)^2 \\ 0 & =\frac{1}{n}\left[\sum_{i=1}^i x_i^2+\sum_{i=1}^i \bar{x}^2-2 \bar{x} \sum_{i=1}^i x_i\right] \\ 0 & =\frac{1}{n}[40+25 n-2 \times 5 \times n \bar{x}] \\ 0 & =\frac{1}{n}[400+25 n-50 n] \\ \Rightarrow \quad \frac{400}{n} & =25 \\ \Rightarrow \quad n & =16 \end{aligned}$

$\begin{aligned} & \text { Mean }=\bar{x}=\frac{605}{5}=121 \\ & \text { Variance }=\frac{\Sigma(x-\bar{x})^2}{n}=\frac{244}{5}=48.8\end{aligned}$

Clearly, difference of values from arithmetic mean is least in option (d).

$\Rightarrow$ Option (d) has least standard deviation.