Question

Pharmaceutical companies advertise for the pill an efficacy of 99.9% in preventing pregnancy. However, under typical use the real efficacy is only about 96%. That is, 4% of women taking the pill for a year will experience an unplanned pregnancy that year. A gynecologist looks back at a random sample of 540 medical records from patients who had been prescribed the pill one year before.

(a) What are the mean and standard deviation of the distribution of the sample proportion who experience unplanned pregnancies out of 540? μp^ = (Use 2 decimal places) σp^ = (Use 3 decimal places)

(b) Suppose the gynecologist finds that 20 of the women had become pregnant within 1 year while taking the pill. How surprising is this finding? Give the probability of finding 20 or more pregnant women in the sample. (Use 3 decimal places)

(c) What is the probability of finding 22 or more pregnant women in the sample? (Use 3 decimal places)

(d) What is the probability of finding 27 or more pregnant women in the sample? (Use 3 decimal places)

Answer 1

a) p = 0.04

    n = 540

= p = 0.04

= sqrt(p(1 - p)/n)

     = sqrt(0.04 * (1 - 0.04)/540)

     = 0.008

b) = 20/540 = 0.037

P(> 0.037)

= P(( - )/> (0.037 - )/)

= P(Z > (0.037 - 0.04)/0.008)

= P(Z > -0.38)

= 1 - P(Z < -0.38)

= 1 - 0.352

= 0.648

c) = 22/540 = 0.041

P(> 0.041)

= P(( - )/> (0.041 - )/)

= P(Z > (0.041 - 0.04)/0.008)

= P(Z > 0.13)

= 1 - P(Z < 0.13)

= 1 - 0.5517

= 0.4483 = 0.448

c) = 27/540 = 0.05

P(> 0.05)

= P(( - )/> (0.05 - )/)

= P(Z > (0.05 - 0.04)/0.008)

= P(Z > 1.25)

= 1 - P(Z < 1.25)

= 1 - 0.8944

= 0.1056 = 0.106