Question

according to harpers index, 40% of all federal inmates are serving time for drug dealing. a random sample of 16 federal inmates is selected.

a) what is the probability that 13 or more are serving time for drug dealing?

b) what is the probability that 5 or fewer are serving time for drug dealing?

Answer 1

Solution(a)

Given in the question

Population proportion = 0.40

Sample proportion = 13/16 =

No. Of sample = 16

We need to calculate P(pcap>13/16) = 1 - P(pcap<= 13/16)

Z = ((13/16)-0.40)/sqrt(0.40*0.60/16) = 0.4125/0.1225 = 3.37

From Z table we found p- value

P(pcap>13/16) = 1 - 0.9996 = 0.0004

Solution(b)

P(pcap<5/16)

Z = (5/16)-0.40/sqrt(0.40*0.60/16) = -0.0875/0.1225 = -0.71

P-value can be found from Z table

P(X<5/16) = 0.2389