Pharmaceutical companies advertise for the pill an efficacy

Pharmaceutical companies advertise for the pill an efficacy of 99.5% 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 520 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 520?

(b) Suppose the gynecologist finds that 19 of the women had become pregnant within 1 year while taking the pill. How surprising is this finding? Give the probability of finding 19 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

Solution

a)

Here, the probability of being pregnant is

up = 1 - 0.95 = 0.05 [ANSWER]

sigmap = sqrt(p(1-p)/n) = sqrt(0.05*(1-0.05)/520) = 0.009557518 [ANSWER]

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b)

Note that P(at least x) = 1 - P(at most x - 1).          
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    520      
p = the probability of a success =    0.05      
x = our critical value of successes =    19      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   18   ) =    0.059903675
          
Thus, the probability of at least   19   successes is  
          
P(at least   19   ) =    0.940096325 [ANSWER]

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c)

Note that P(at least x) = 1 - P(at most x - 1).          
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    520      
p = the probability of a success =    0.05      
x = our critical value of successes =    23      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   22   ) =    0.245420111
          
Thus, the probability of at least   23   successes is  
          
P(at least   23   ) =    0.754579889 [ANSWER]

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d)

Note that P(at least x) = 1 - P(at most x - 1).          
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    520      
p = the probability of a success =    0.05      
x = our critical value of successes =    28      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   27   ) =    0.628971528
          
Thus, the probability of at least   28   successes is  
          
P(at least   28   ) =    0.371028472 [ANSWER]