Statistics Question 6
Question 6 - 30 January - Shift 2
Let $S$ be the set of all values of $a_1$ for which the mean deviation about the mean of 100 consecutive positive integers $a_1, a_2, a_3, \ldots \ldots, a _{100}$ is 25 . Then $S$ is
(1) $\phi$
(2) ${99}$
(3) $\mathbb{N}$
(4) ${9}$
Show Answer
Answer: (3)
Solution:
Formula: Mean deviation about mean, Arithmetic mean of individual series (ungrouped data)
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 ’ S \\ & \bar{x}=\frac{a_1+(a_1+1)+(a_1+2)+\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 $=\frac{\sum _{1=1}^{100}|x_i-\bar{x}|}{100}$
$=\frac{2(\frac{99}{2}+\frac{97}{2}+\frac{95}{2}+\ldots+\frac{1}{2})}{100}$
$=\frac{1+3+\ldots .+99}{100}$
$=\frac{\frac{50}{2}[1+99]}{100}$
$=25$
So, it is true for every natural no. ‘a1’.
