Answer to Question #120245 in Statistics and Probability for Jospeh Camo

Question

6. At any given time about 5.5% of women (age 15-45) are pregnant. A home pregnancy test is accurate 99% of the time if the woman taking the test is actually pregnant and 99.5% accurate if the woman is not pregnant. If the test yields a positive result, what is the posterior probability of the hypothesis that the woman is pregnant?
A. 0.08
B. 0.99
C. 0.995
D. 0.92

7. Your boss is a biologist who needs wood samples from long-leaf pine trees with a fungal disease which is only visible under a microscope, and she sends you on an assignment to collect the samples. She wants at least 50 different diseased samples. She tells you that approximately 28% of long-leaf pine trees currently have the fungal disease. If you sample 160 long-leaf pine trees at random, what is the probability you’ll have at least 50 diseased samples to return to your boss?

A. 92%

B. 28%

C.13%

D. 82%


18%

Answer 1

6) D) 0.92

We are given that at any given time about 5.5% of women (age 15-45) are pregnant;

Let Probability that women is pregnant, P(A) = 0.055

Probability that women is not pregnant, P(A') = 1 - P(A) = 1 - 0.055 = 0.945

Also, Let B be the event that test is positive;

Probability that pregnancy test is positive given that the woman is actually pregnant, P(B/A) = 0.99

Probability that pregnancy test is positive given that the woman is not actually pregnant, P(B/A') = 1 - 0.995 = 0.005

Now, we have to find that if the test yields a positive result what is the posterior probability that the woman is pregnant i.e.; P(A/B)

Using Bayes' Theorem we get

P(A/B) =(P(A)*P(B|A))/((P(A)*P(B|A)+ P(A')*P(B/A') )=0.92

7) 18%

μ=np=160*0.28=44.8

σ=$\sqrt{npq}$= $\sqrt{160*0.28*(1-0.72)}$ =5.7

P(x≥50)=1-P(x<50)

P(x<50), z=(50-44.8)/5.7=0.91, using table P(x<50)=0.81859

P(x≥50)=1-0.81859=0.18