Question

According to government data, 68% of employed women have never been married. Rounding to 4 decimal places, if 15 employed women are randomly selected:

a. What is the probability that exactly 2 of them have never been married?

b. That at most 2 of them have never been married?

c. That at least 13 of them have been married?

Answer 1

Solution:

For the given binomial distribution, we are given

n = 15

p = 0.68

q = 1 – p = 1 – 0.68 = 0.32

The binomial formula is given as below:

P(X=x) = nCx*p^x*q^(n- x)

Part a

Here, we have to find P(X=2)

P(X=2) = 15C2*0.68^2*0.32^(15 – 2)

P(X=2) = 105*0.68^2*0.32^13

P(X=2) = 0.00002

Required probability = 0.0000

Part b

Here, we have to find P(X≤2)

P(X≤2) = P(X=0) + P(X=1) + P(X=2)

P(X=0) = 15C0*0.68^0*0.32^(15 – 0)

P(X=0) = 0.0000

P(X=1) = 15C1*0.68^1*0.32^(15 – 1)

P(X=1) = 0.0000

P(X=2) = 15C2*0.68^2*0.32^(15 – 2)

P(X=2) = 0.0000

P(X≤2) = P(X=0) + P(X=1) + P(X=2)

P(X≤2) = 0.0000 + 0.0000 + 0.0000

P(X≤2) = 0.0000

Required probability = 0.0000

Part c

Here, we have to find P(X≥13)

P(X≥13) = P(X=13) + P(X=14) + P(X=15)

P(X=13) = 15C13*0.68^13*0.32^(15 – 13)

P(X=13) = 0.071467

P(X=14) = 15C14*0.68^14*0.32^(15 – 14)

P(X=14) = 0.021695

P(X=15) = 15C15*0.68^15*0.32^(15 – 15)

P(X=15) = 0.003074

P(X≥13) = P(X=13) + P(X=14) + P(X=15)

P(X≥13) = 0.071467 + 0.021695 + 0.003074

P(X≥13) = 0.096236

Required probability = 0.096236