Question

1. a police department reports that the probabilities that 0, 1, 2, and 3 burglaries will be reported in a given day are 0.46, 0.41, 0.09, and 0.04, respectively. find the standard deviation for the probability distribution. round answer to the nearest hundredth. a) $sigma = 0.79$ b) $sigma = 0.63$ c) $sigma = 1.06$ d) $sigma = 1.04$

Explanation:

Step1: Calculate the mean ($\mu$)

$\mu=\sum xP(x)=0\times0.46 + 1\times0.41+2\times0.09 + 3\times0.04=0 + 0.41+0.18+0.12 = 0.71$

Step2: Calculate $\sum(x - \mu)^2P(x)$

$(0 - 0.71)^2\times0.46+(1 - 0.71)^2\times0.41+(2 - 0.71)^2\times0.09+(3 - 0.71)^2\times0.04$
$=( - 0.71)^2\times0.46+(0.29)^2\times0.41+(1.29)^2\times0.09+(2.29)^2\times0.04$
$=0.5041\times0.46 + 0.0841\times0.41+1.6641\times0.09+5.2441\times0.04$
$=0.231886+0.034481+0.149769+0.209764$
$=0.6259$

Step3: Calculate the standard deviation ($\sigma$)

$\sigma=\sqrt{\sum(x - \mu)^2P(x)}=\sqrt{0.6259}\approx0.79$

Answer:

A. $\sigma = 0.79$