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UPSC NDA I 2024 Mathematics Paper

Question Numbers: 117-118

Consider the following for the items that follow :

$x_{i}$	1	2	3	...	$n$
$f_{i}$	1	$2^{-1}$	$2^{-2}$	...	$2^{-(n-1)}$

Question : 118 of 120

Marks: +1, -0

What is the mean of the distribution ?

Solution:

Concept:

The mean of a distribution is the weighted average:

\bar{x} = \frac{\sum(x_i f_i)}{\sum f_i}

Explanation:

Given

x_i = i

for

i=1,2,\dots,n

Frequencies:

f_1 = 1

, and for

i \ge 2

f_i = \frac{1}{2^{i-1}}

Compute numerator:

\sum x_i f_i = 1\cdot1 + 2\cdot\frac12 + 3\cdot\frac1{2^2} + \dots + n\cdot\frac1{2^{n-1}} = S

Let

S = \frac1{1} + \frac2{2} + \frac3{2^2} + \dots + \frac{n}{2^{n-1}}

Multiply by

\frac12

\frac12 S = \frac1{2} + \frac2{2^2} + \frac3{2^3} + \dots + \frac{n}{2^n}

Subtract:

\frac12 S = \left(1 + \frac12 + \frac1{2^2} + \dots + \frac1{2^{n-1}}\right) - \frac{n}{2^n}

The geometric series sum is

\frac{1-\frac1{2^n}}{1-\frac12} = 2\left(1-\frac1{2^n}\right)

\frac12 S = 2\left(1-\frac1{2^n}\right) - \frac{n}{2^n} = 2 - \frac{2}{2^n} - \frac{n}{2^n} = 2 - \frac{n+2}{2^n}

Multiply by

2

S = 4 - \frac{2n+4}{2^n} = \frac{4\cdot2^n - (2n+4)}{2^n} = \frac{2^{n+2} - 2n -4}{2^n} = \frac{2^{n+1} - n -2}{2^{n-1}}

Denominator:

\sum f_i = 1 + \frac12 + \frac1{2^2} + \dots + \frac1{2^{n-1}} = \frac{1-\frac1{2^n}}{1-\frac12} = 2\left(1-\frac1{2^n}\right) = \frac{2^n -1}{2^{n-1}}

Thus,

\bar{x} = \frac{S}{\sum f_i} = \frac{\frac{2^{n+1} - n -2}{2^{n-1}}}{\frac{2^n -1}{2^{n-1}}} = \frac{2^{n+1} - n -2}{2^n -1}

Answer:

Option C:

\frac{2^{n+1} - n -2}{2^n -1}

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