This is a very difficult question that I have been working on for 2 hours now.. Can any one help?
In a survey, some of the questions concern sensitive issues (e.g., income, drug use, sexual experiences). As a result, some respondents do not answer the questions truthfully. Suppose that 30% of the members of a particular population had incomes over $100,000 last year. A random sample of n = 10 members of this population is taken, and each person in the sample is asked “Was your income over $100,000 last year?” If a person really had an income over $100,000, the probability that he or she will give a truthful answer to this question is 0.7. If a person’s income was not over $100,000, the probability that he or she will give a truthful answer is 0.9.
(a) Find the probability distribution of X, the number of people in the sample who had incomes over $100,000 last year.
(b) Find the probability distribution of Y, the number of “yes” answers to the question in the sample of size 10.
(c) Find the probability distribution of T, the number of truthful answers from the 10 respondents.
(d) Given the answer “Yes, my income was over $100,000,” what is the probability that the income of this person was, indeed, over $100,000? Similarly, given the answer “No, my income was not over $100,000,” what is the probability that the income of this person was, indeed, below $100,000?
P(Income>100000) = 0.30
P(Truth|Income>100000) = 0.70
P(Truth|Income < 100000) = 0.90
a) X is Binomial(n=10, p=0.30)
b) Person says "yes" if
i)telling truth when: Income>100000 or Income<100000
ii)lying:Income>100000 or Income<100000
P("yes") = P(yes|Income>100000).P(Income>100000) + P(yes|Income<100000).P(Income<100000)
= P(Truth|Income>100000)*P(Income>100000) + P(Lied|Income<100000)*P(Income<100000)
= 0.70*0.30+0.10*0.70
=0.28
Y is Binomial (n=10,p=0.28)
c)
P(truth) = P(Truth|Income>100000)*P(Income>100000) +
P(Truth|Income<100000)*P(Income<100000)
= 0.70*0.30 + 0.90*0.70
=0.84
T is Binomial (n=10,p=0.84)
d) Using Bayes Theorem
P(Income>100000|"yes")
= P("yes"|Income>100000).P(Income>100000)/P("yes")
=0.7*0.30/0.28
=0.75