First slide

Introduction to statistics

Question

If a variable $X$ takes values $0, 1, 2, . . ., n$ with frequencies $^{n} C_{0},^{n} C_{1},^{n} C_{2}, \dots .^{n} C_{n}$ respectively, then S.D $=$

Moderate

A

$\frac{n}{4}$
B

$\frac{n}{2}$
C

$\frac{\sqrt{n}}{2}$
D

$\frac{\sqrt{n}}{4}$

Solution

We have,
$σ^{2} = \frac{1}{N} \sum_{i = 1}^{n} f_{i} x_{i}^{2} - {(\frac{1}{n} \sum_{i = 1}^{n} f_{i} x_{i})}^{2}$
$∴ σ^{2} = \frac{1}{2^{n}} \sum_{r = 0}^{n} C_{r} r^{2} - {(\frac{1}{2^{n}} \sum_{r = 0}^{n} C_{r} r)}^{2} [∵ N = \sum_{r = 0}^{n} C_{r}^{n} = 2^{n}]$
Now,
$\sum_{r = 0}^{n} r^{n} C_{r} = \sum_{r = 0}^{n} r {\frac{n}{r}}^{n - 1} C_{r - 1}$
$\Rightarrow \sum_{r = 0}^{n} r^{n} C_{r} = n \sum_{r = 1}^{n - 1} C_{r - 1} = n 2^{n - 1}$
and,
$\sum_{r = 0}^{n} {r^{2}}^{n} C_{r} = \sum_{r = 0}^{n} {r (r - 1) + r}^{n} C_{r}$
$\begin{array}{l} \Rightarrow \sum_{r = 0}^{n} {r^{2}}^{n} C_{r} = \sum_{r = 0}^{n} r {(r - 1)}^{n} C_{r} + \sum_{r = 0}^{n} r^{n} C_{r} \\ \Rightarrow \sum_{r = 0}^{n} {r^{2}}^{n} C_{r} = \sum_{r = 2}^{n} r (r - 1) \frac{n - 1}{r - 1} \frac{n - 2}{r - 1} C_{r - 2} + \sum_{r = 1}^{n} r \frac{n - n^{n - 1} C_{r - 1}}{r} \end{array}$
$\begin{array}{l} \Rightarrow \sum_{r = 0}^{n} {r^{2}}^{n} C_{r} = n (n - 1) \{\sum_{r = 2}^{n}^{n - 2} C_{r - 2}\} + n \{\sum_{r = 1}^{n}^{n - 1} C_{r - 1}\} \\ \Rightarrow \sum_{r = 0}^{n} {r^{2}}^{n} C_{r} = n (n - 1) 2^{n - 2} + n 2^{n - 1} \end{array}$
$∴ σ^{2} = \frac{n (n - 1) 2^{n - 2} + n \times 2^{n - 1}}{2^{n}} - {(\frac{n \times 2^{n - 1}}{2^{n}})}^{2}$
$\Rightarrow σ^{2} = \frac{n (n - 1)}{4} + \frac{n}{2} - \frac{n^{2}}{4} = \frac{n}{4} \Rightarrow σ = \frac{\sqrt{n}}{2}$

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