Test Index

Statistics Part 1

Section: Mathematics

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Question : 90 of 100

Marks: +1, -0

If the mean deviation of number

1, 1+d, 1+2d, \dots, 1+100d

from their mean is 255 , then the

d

$20.0$
$10.1$
$20.2$
$10.0$

Solution:

Mean

(\overline{x}) = \frac{\text{Sum of numbers}}{n}

= \frac{\frac{n}{2}(a+l)}{n}

= \frac{1}{2}(1+l+100d)

= 1+50d

Mean deviation (M.D)

= \frac{1}{n} \sum\limits_{i=1}^{101} \left| x_i - \overline{x} \right|

= \frac{1}{101}[50d + 49d + 48d + \dots + d + 0 + \dots + 50d]

= \frac{1}{101} \cdot 2d (1+2+\dots+50)

= \frac{1}{101} \cdot 2d \cdot \frac{50 \times 51}{2}

= \frac{50 \times 51 d}{101}

Given that M.D

= 255

\therefore \frac{50 \times 51 d}{101} = 255

\Rightarrow d = \frac{101 \times 255}{51 \times 50}

= 10.1

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