Question

According to Harper’s index 55% of all federal inmates are serving time for drug dealing. A random sample of 20 federal inmates is selected. What is the probability that 8 or more are serving time for drug dealing? What is the probability that 2 or fewer are serving time for drug dealing? What is the expected number of inmates serving time for drug dealing?

Answer 1

Let , X number of inmates serving time for drug dealing.

According to Harper’s index 55% of all federal inmates are serving time for drug dealing.

p = 0.55 , n = 20

X follows Binomial distribution with n = 20 and p = 0.55

a )

We have to find probability that 8 or more are serving time for drug dealing, i. e P( x >= 8 )

P( x >= 8 ) = 1 - P( x <= 7 )

Using Excel function ,   =BINOMDIST( x , n , p ,1 )

P( x <= 7 ) = BINOMDIST( 7 , 20, 0.55, 1 ) = 0.058034

So, P( x >= 8 ) = 1 - 0.058034 = 0.94197

The probability that 8 or more are serving time for drug dealing is 0.94197

b)

We have to find probability that 2 or fewer are serving time for drug dealing, i.e P( x <= 2 )

Using Excel function ,   =BINOMDIST( x , n , p ,1 )

P( x <= 2 ) = BINOMDIST( 2 , 20, 0.55, 1 ) = 0.00004

The probability that 2 or fewer are serving time for drug dealing is 0.00004

c)

Expected number of inmates serving time for drug dealing is ,

E( x ) = n*p = 20*0.55 = 11