Question

1. what is the standard deviation of the following probability distribution?

x 0 2 4 6 8
p(x) 0.25 0.1 0.3 0.25 0.1
a. 4.4 b. 2.6 c. 3.7 d. 6.0
a. 1.8 b. 2.2 c. 1.9 d. 2.0

Explanation:

Step1: Calculate the mean $\mu$

$\mu=\sum_{i}x_{i}P(x_{i})=0\times0.25 + 2\times0.1+4\times0.3 + 6\times0.25+8\times0.1=0 + 0.2+1.2 + 1.5+0.8 = 3.7$

Step2: Calculate $\sum_{i}(x_{i}-\mu)^{2}P(x_{i})$

$(0 - 3.7)^{2}\times0.25+(2 - 3.7)^{2}\times0.1+(4 - 3.7)^{2}\times0.3+(6 - 3.7)^{2}\times0.25+(8 - 3.7)^{2}\times0.1$
$=( - 3.7)^{2}\times0.25+( - 1.7)^{2}\times0.1+(0.3)^{2}\times0.3+(2.3)^{2}\times0.25+(4.3)^{2}\times0.1$
$=13.69\times0.25 + 2.89\times0.1+0.09\times0.3+5.29\times0.25+18.49\times0.1$
$=3.4225+0.289 + 0.027+1.3225+1.849=6.91$

Step3: Calculate the standard - deviation $\sigma$

$\sigma=\sqrt{\sum_{i}(x_{i}-\mu)^{2}P(x_{i})}=\sqrt{6.91}\approx 2.6$

Answer:

b. 2.6