Pharmaceutical companies advertise for the pill an efficacy of 98.8% in preventi

Question

Pharmaceutical companies advertise for the pill an efficacy of 98.8% in preventing pregnancy. However, under typical use the real efficacy is only about 94%. That is, 6% of women taking the pill for a year will experience an unplanned pregnancy that year. A gynecologist looks back at a random sample of 510 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 510?

(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 28 or more pregnant women in the sample? (Use 3 decimal places)

Explanation / Answer

a)

Here,          
n =    510
p =    0.06

u = mean = p =    0.06 [ANSWER]

sigma = standard deviation = sqrt(p(1-p)/n) =    0.010516094 [ANSWER]

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b)

We first get the z score for the critical value:          
          
x = critical value =    20.5      
u = mean = np =    30.6      
          
s = standard deviation = sqrt(np(1-p)) =    5.363207995      
          
Thus, the corresponding z score is          
          
z = (x-u)/s =    -1.88320125      
          
Thus, the right tailed area is          
          
P(z >   -1.88320125   ) =    0.97016345 [ANSWER]

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c)

We first get the z score for the critical value:          
          
x = critical value =    24.5      
u = mean = np =    30.6      
          
s = standard deviation = sqrt(np(1-p)) =    5.363207995      
          
Thus, the corresponding z score is          
          
z = (x-u)/s =    -1.137378973      
          
Thus, the right tailed area is          
          
P(z >   -1.137378973   ) =    0.872310053 [ANSWER]

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d)

We first get the z score for the critical value:          
          
x = critical value =    27.5      
u = mean = np =    30.6      
          
s = standard deviation = sqrt(np(1-p)) =    5.363207995      
          
Thus, the corresponding z score is          
          
z = (x-u)/s =    -0.578012265      
          
Thus, the right tailed area is          
          
P(z >   -0.578012265   ) =    0.718372081 [ANSWER]

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Hi! I used the actual distribution (real efficacy) to answer the questions here. If you need the "claimed efficacy" instead, please resubmit this question indicating that you need to use the 98.8% instead. That way we can continue helping you! Thanks!

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