First slide

Introduction to statistics

Question

$If a variable x takes values 0, 1, 2, \dots, n {with frequencies proportional to the binomial coefficients}^{n} C_{0},^{n} C_{1},^{n} C_{2}, \dots$ $^{n} C_{n}, then var (x) is$

Difficult

A

$\frac{n^{2} - 1}{12}$
B

$\frac{n}{2}$
C

$\frac{n}{4}$
D

None of these

Solution

We have
$\begin{array}{l} \bar{x} = \frac{0^{n} C_{0} + 1^{n} C_{1} + 2^{n} C_{2} + \dots + n^{n} C_{n}}{^{n} C_{0} +^{n} C_{1} +^{n} C_{2} + \dots +^{n} C_{n}} = \frac{\sum_{r = 0}^{n} r^{n} C_{r}}{\sum_{r = 0}^{n}^{n} C_{r}} \\ = \frac{1}{2^{n}} \sum_{r = 1}^{n} r {\frac{n}{r}}^{n - 1} C_{r - 1} [∵ \sum_{r = 0}^{n}^{n} C_{r} = 2^{n};^{n} C_{r} = {\frac{n}{r}}^{n - 1} C_{r - 1}] \\ = \frac{n}{2^{n}} \sum_{r = 1}^{n}^{n - 1} C_{r - 1} = \frac{n}{2^{n}} 2^{n - 1} = \frac{n}{2} [∵ \sum_{r = 1}^{n}^{n - 1} C_{r - 1} = 2^{n - 1}] \\ and \frac{1}{N} \sum f_{i} x_{i}^{2} = \frac{1}{2^{n}} \sum_{r = 0}^{n} {r^{2}}^{n} C_{r} = \frac{1}{2^{n}} \sum_{r = 0}^{n} [r (r - 1) + r]^{n} C_{r} \\ = \frac{1}{2^{n}} \{\sum_{r = 0}^{n} r {(r - 1)}^{} C_{r} + \sum_{r = 0}^{n} r^{} C_{r}\} \\ = \frac{1}{2^{n}} \{\sum_{r = 2}^{n} r (r - 1) \frac{n}{r} {\frac{n - 1}{r - 1}}^{n - 2} C_{r - 2} + \sum_{r = 1}^{n} r {\frac{n}{r}}^{n - 1} C_{r - 1}\} \\ = \frac{1}{2^{n}} \{n (n - 1) 2^{n - 2} + n 2^{n - 1}\} = \frac{n (n - 1)}{4} + \frac{n}{2} \\ ∴ Var (X) = \frac{1}{N} \sum f_{i} x_{i}^{2} - {\bar{x}}^{2} = \frac{n (n - 1)}{4} + \frac{n}{2} - \frac{n^{2}}{4} = \frac{n}{4} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App