Question

Pharmaceutical companies advertise for the pill an efficacy of 99.4% in preventing pregnancy. However, under typical use the real efficacy is only about 97%. That is, 3% of women taking the pill for a year will experience an unplanned pregnancy that year. A gynecologist looks back at a random sample of 530 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 530?

(b) Suppose the gynecologist finds that 18 of the women had become pregnant within 1 year while taking the pill. How surprising is this finding? Give the probability of finding 18 or more pregnant women in the sample. (Use 3 decimal places)

(c) What is the probability of finding 23 or more pregnant women in the sample? (Use 3 decimal places)

(d) What is the probability of finding 28 or more pregnant women in the sample? (Use 3 decimal places)

Answer 1

Success =pregnant women

X =Number of successes

p =proportion of success for each observation =0.03

P =Probability

n =sample size =530

Distribution: Binomial

Formula: P(X =x) =C(n, x).p^x.(1-p)^n-x

(a)

Mean, =np =530*0.03 =15.9

Standard deviation, s = = =3.93

(b)

It's not surprising because 18 is close to the expected value (mean) of 15.9 and 18 is less than one std.deviation away from the mean. 15.9+3.93 =19.83 and 18 < 19.83.

P(X 18) =0.33​​​​0

(c)

P(X 23) =0.052

(d)

P(X 28) =0.003