Pharmaceutical companies advertise for the pill an efficacy of 99% in preventing

Question

Pharmaceutical companies advertise for the pill an efficacy of 99% 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 500 medical records from patients who had been prescribed the pill one year before.

(b) 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)

(c) What is the probability of finding 25 or more pregnant women in the sample? (Use 3 decimal places)

(d) What is the probability of finding 26 or more pregnant women in the sample? (Use 3 decimal places)

Explanation / Answer

a.
possibile chances (x)=21
sample size(n)=500
success rate ( p )= x/n = 0.042 = 4.2%
it is surprise in findig , since this fall under typical use the real efficacy is about 4%
b.
NORMAL APPROXIMATION TO BINOMIAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
mean ( np ) = 500 * 0.04 = 20
standard deviation ( npq )= 500*0.04*0.96 = 4.3818
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
P(X < 21) = (21-20)/4.3818
= 1/4.3818= 0.2282
= P ( Z <0.2282) From Standard NOrmal Table
= 0.5903
P(X > = 21) = (1 - P(X < 21))
= 1 - 0.5903 = 0.4097 ~ 0.41
c.
P(X < 25) = (25-20)/4.3818
= 5/4.3818= 1.1411
= P ( Z <1.1411) From Standard NOrmal Table
= 0.8731
P(X > = 25) = (1 - P(X < 25))
= 1 - 0.8731 = 0.127
d.
P(X < 26) = (26-20)/4.3818
= 6/4.3818= 1.3693
= P ( Z <1.3693) From Standard NOrmal Table
= 0.9145
P(X > = 26) = (1 - P(X < 26))
= 1 - 0.9145 = 0.086

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