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Statistics Part 1

Section: Mathematics

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Question : 6 of 100

Marks: +1, -0

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, a_{100}

S

[30-Jan-2023 Shift 2]

$\phi$
$\{99\}$
$\mathbb{N}$
$\{9\}$

Solution:

let

a_1

be any natural number

\; a_1, a_1+1, a_1+2, \ldots, a_1+99 \; \text{are values of}\; a_i' S

\overline{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}

\text{Mean deviation about mean} = \frac{\sum\limits_{i=1}^{100} |x_i - \overline{x}|}{100}

= \frac{2\left(\frac{99}{2}+\frac{97}{2}+\frac{95}{2}+\ldots+\frac{1}{2}\right)}{100}

= \frac{1+3+\ldots+99}{100}

= \frac{\frac{50}{2}(1+99)}{100}

= 25

So, it is true for every natural no. '

a_1'

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