Pharmaceutical companies advertise for the pill an efficacy of 99.6% in preventi

Question

Pharmaceutical companies advertise for the pill an efficacy of 99.6% in preventing pregnancy. However, under typical use the real efficacy is only about 95%. That is, 5% 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.

Suppose the gynecologist finds that 21 of the women had become pregnant within 1 year while taking the pill. How surprising is this finding? Give the probability of finding 21 or more pregnant women in the sample. (Use 3 decimal places)

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

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

Explanation / Answer

mean ( np ) = 540 * 0.05 = 27
standard deviation ( npq )= 540*0.05*0.95 = 5.0646
a.
P(X < 21) = (21-27)/5.0646
= -6/5.0646= -1.1847
= P ( Z <-1.1847) From Standard NOrmal Table
= 0.1181
P(X > = 21) = (1 - P(X < 21))
= 1 - 0.1181 = 0.8819
the proportion is 88.19%, which is normal

b.
P(X < 23) = (23-27)/5.0646
= -4/5.0646= -0.7898
= P ( Z <-0.7898) From Standard NOrmal Table
= 0.2148
P(X > = 23) = (1 - P(X < 23))
= 1 - 0.2148 = 0.7852
c.
P(X < 28) = (28-27)/5.0646
= 1/5.0646= 0.1974
= P ( Z <0.1974) From Standard NOrmal Table
= 0.5783
P(X > = 28) = (1 - P(X < 28))
= 1 - 0.5783 = 0.4217

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