# Answer to Question #5219 in Statistics and Probability for cash

Question #5219
Three stores have 8, 12, and 15 employees of whom 3, 8, and 7, respectively, are women. (i) A store is chosen at random and from that store an employee is chosen at random. If this employee is a woman, what is the probability she came from the store with 12 employees? (ii) If a second employee is chosen from the same store in (i), what is the probability that a woman will be chosen? Assume that the employee chosen in (i) was a woman and we don’t know which store she came from. [7 marks]
Consider the events:
A - the store with 8 employees is chosen,
B - the
store with 12 employees is chosen,
A - the store with 15 employees is
chosen,
W - employee is a woman.

i)
We need to find the conditional
probability

P(B|W) = |using Bayes&#039; theorem| = (P(W|B)*P(B)) / (P(W|A) +
P(W|B) + P(W|C));
P(B|W) = (8/12*1/3) / (3/8 + 8/12 + 7/15) =
80/543.

ii)
Let&#039;s first calculate
P(A|W) and P(C|W) like in part
i):
P(A|W) = (3/8*1/3) / (3/8 + 8/12 + 7/15) = 45/543.
P(C|W) = (7/15*1/3)
/ (3/8 + 8/12 + 7/15) = 56/543.

Using the law of total probability, we
get:
P(W) = P(W|A)*P{the first woman was from A} + P(W|B)*P{the first woman
was from B} +
P(W|C)*P{the first woman was from C} = 2/8*45/543 +
7/12*80/543 + 6/15*56/543 = 1/543*(45/4 + 140/3 + 112/5) = 4819/(543*60) =
0.148...

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Assignment Expert
13.07.15, 19:49

Dear tope.
Your comment is not related to question 5219, but it is the same as question 53385, which is in progress. Please wait for updates.

tope
13.07.15, 14:38

The following data show systolic blood pressure levels (mm Hg) of a random sample of six
patients undergoing a particular drug therapy for hypertension.
182 179 154 161 170 151
Can we conclude, on the basis of these data, that the population mean is greater than 165?
Hint: Use an appropriate parametric test, at the 5% significance level.