First slide

Introduction to statistics

Question

If the mean deviation of the numbers $1, 1 + d, 1 + 2 d, . . ., 1 + 100 d$ from their mean is $255,$ then $d$ is equal to

Moderate

A

$10.0$
B

$20.0$
C

$10.1$
D

$20.2$

Solution

Let $X$ denote the mean of the given numbers.
Then,
$\bar{X} = \frac{1 + (1 + d) + (1 + 2 d) + \dots + (1 + 100 d)}{101}$
$\Rightarrow \bar{X} = \frac{\frac{101}{2} {1 + (1 + 100 d))\}}{101} = 1 + 50 d$
$∴$ Mean deviation $= \frac{1}{101} \{\sum_{r = 0}^{100} | (1 + r d) - (1 + 50 d) |\}$
$\Rightarrow$ Mean deviation $= \frac{1}{101} \sum_{r = 0}^{100} | r - 50 |$
$\Rightarrow$ Mean deviation $= \frac{d}{101} \times 2 \sum_{r = 1}^{50} r$
$\Rightarrow$ Mean deviation $= \frac{2 d}{101} \times \frac{50 \times 51}{2} = \frac{50 \times 51}{101} d .$
It is given that the mean deviation is $255 .$
$∴ 255 = \frac{50 \times 51}{101} d \Rightarrow d = 10.1$

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