Question

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.

Answer 1

The standard deviation of the probability distribution is 0.79

Explanation:

We are given the following in the question:

x: 0 1 2 3

P(x): 0.46 0.41 0.09 0.04

We have to find the standard deviation of the probability distribution.

Formula:

`(x)/(y) E(x) = \displaystyle\sum x_iP(x_i)\\E(x) = 0(0.46 ) + 1(0.41)+2(0.09 )+3 (0.04)\\E(x) = 0.71\\E(x^2) = \sum x_i^2p(X_i)\\E(x^2) =0^2(0.46 ) + 1^2(0.41)+2^2(0.09 )+3^2 (0.04)\\E(x^2)= 1.13\\\sigma^2 = E(x^2) - (E(x))^2\\\sigma^2 = 1.13 - (0.71)^2 = 0.6259\\\sigma = √(0.6259) = 0.79`

Thus, the standard deviation of the probability distribution is 0.7911