Question

solve the problem.

1. find the standard deviation for the given probability distribution.

x	p(x)
0	0.05
1	0.28
2	0.47
3	0.11
4	0.09

a) 0.97
b) 0.94
c) 1.00
d) 2.14

Explanation:

Step1: Calculate the mean $\mu$

$\mu=\sum_{x}x\cdot P(x)=0\times0.05 + 1\times0.28+2\times0.47 + 3\times0.11+4\times0.09=0 + 0.28+0.94 + 0.33+0.36 = 1.91$

Step2: Calculate $\sum_{x}(x - \mu)^2\cdot P(x)$

$(0 - 1.91)^2\times0.05+(1 - 1.91)^2\times0.28+(2 - 1.91)^2\times0.47+(3 - 1.91)^2\times0.11+(4 - 1.91)^2\times0.09$
$=( - 1.91)^2\times0.05+( - 0.91)^2\times0.28+(0.09)^2\times0.47+(1.09)^2\times0.11+(2.09)^2\times0.09$
$=3.6481\times0.05 + 0.8281\times0.28+0.0081\times0.47+1.1881\times0.11+4.3681\times0.09$
$=0.182405+0.231868+0.003807+0.130691+0.393129 = 0.9419$

Step3: Calculate the standard deviation $\sigma$

$\sigma=\sqrt{\sum_{x}(x - \mu)^2\cdot P(x)}=\sqrt{0.9419}\approx0.97$

Answer:

A. 0.97