Question

USA Today reported that approximately 25% of all state prison inmates released on parole become repeat offenders while on parole. Suppose the parole board is examining five prisoners up for parole. Let x = number of prisoners out of five on parole who become repeat offenders.x 0 1 2 3 4 5P(x) 0.207 0.367 0.227 0.162 0.036 0.001(d) Compute μ, the expected number of repeat offenders out of five. (Round your answer to three decimal places.)μ =(e) Compute σ, the standard deviation of the number of repeat offenders out of five. (Round your answer to two decimal places.)σ = prisonersplease help

Answer 1

`\mu =E(X) = 0*0.207+1*0.367+ 2*0.227+ 3*0.162+ 4*0.036+ 5*0.001= 1.456`

For the variance we need to calculate first the second moment given by:

`E(X) = \sum_(i=1)^n X^2_i P(X_i)`

And replacing we got:

`E(X^2) = 0^2*0.207+1^2*0.367+ 2^2*0.227+ 3^2*0.162+ 4^2*0.036+ 5^2*0.001= 3.33`

And the variance is given by:

`\sigma^2 = E(X^2) - [E(X)]^2 = 3.334 -[1.456]^2 =1.21`

And the deviation is:

`\sigma = √(1.214) = 1.10`

Explanation:

For thi case we have the following distribution given:

X 0 1 2 3 4 5

P(X) 0.207 0.367 0.227 0.162 0.036 0.001

For this case the expected value is given by:

`E(X) = \sum_(i=1)^n X_i P(X_i)`

And replacing we got:

`\mu =E(X) = 0*0.207+1*0.367+ 2*0.227+ 3*0.162+ 4*0.036+ 5*0.001= 1.456`

For the variance we need to calculate first the second moment given by:

`E(X) = \sum_(i=1)^n X^2_i P(X_i)`

And replacing we got:

`E(X^2) = 0^2*0.207+1^2*0.367+ 2^2*0.227+ 3^2*0.162+ 4^2*0.036+ 5^2*0.001= 3.33`

And the variance is given by:

`\sigma^2 = E(X^2) - [E(X)]^2 = 3.334 -[1.456]^2 =1.21`

And the deviation is:

`\sigma = √(1.214) = 1.10`