Answer in Statistics and Probability for Shema Derrick #205142
.In 1970, 11% of Americans completed four years of college, 43% of them were women. In 1990, 22% of Americans completed four years of college; 53% of them were women. (Time, Jan.19, 1996). a) Given that a person completed four years of college in 1970, what is the probability that the person was a woman? b) What is the probability that a woman would finish four years of college in 1990? c) What is the probability that in 1990 a man would not finish college?
S= randomly chosen American completed four years of college in 1970s
N=randomly chosen American completed four years of college in 1990s
W= randomly chosen American who completed four years of college is a woman
M=randomly chosen American who completed four years of college is a man
P(S)=0.11
P(W|S)=0.43
P(N)=0.22
P(W|N)=0.53
a) P(W|S)=0.43
b)
“P(W|N) = frac{P(W cap N)}{P(N)} \nn0.53 = frac{P(W cap N)}{0.22} \nnP(W cap N) = 0.53 times 0.22 = 0.1166”
c)
“(M cap N)” = man finished college in 1990s
“(M cap N)u2019” = man had not finished college in 1990s
The events “(M cap N)” and “(M cap N)u2019” are complementary:
“P(M cap N)u2019 = 1 -(M cap N) \nnP(M|N)= 0.47 \nnP(M|N)= frac{P(M cap N)}{P(N)} \nn0.47 = frac{P(M cap N)}{0.22} \nnP(M cap N) = 0.47 times 0.22= 0.1034 \nnP(M cap N)u2019 = 1 -0.1034 \nn= 0.8966”
