Question

In the book A Guide to the Development and Use of the Myers-Briggs Type Indicators by Myers and McCaully, it was reported that approximately 45% of all university professors are extroverted. Suppose you have classes with four different professors. (a) What is the probability that all four are extroverts? (Round your answer to three decimal places.) .041 (b) What is the probability that none of your professors is an extrovert? (Round your answer to three decimal places.) .091 (c) What is the probability that at least two of your professors are extroverts? (Round your answer to three decimal places.) .060 (d) In a group of four professors selected at random, what is the expected number of extroverts? (Round your answer to two decimal places.) extroverts What is the standard deviation of the distribution? (Round your answer to two decimal places.) extroverts (e) Suppose you were assigned to write an article for the student newspaper and you were given a quota (by the editor) of interviewing at least three extroverted professors. How many professors selected at random would you need to interview to be at least 89% sure of filling the quota? professors

Answer 1

Solution:- Given that n = 4, p = 0.45

(a) P(X = 4) = 4C4*0.45^4*0.55^0 = 0.0410 = 0.041 (rounded)

(b) P(X = 0) = 4C0*0.45^0*0.55^4 = 0.0915 = 0.092 (rounded)

(c) P(X >= 2) = P(2) + P(3) + P(4) = 0.3675+0.2005+0.0410 = 0.609 (rounded)
=> P(X = 2) = 4C2*0.45^2*0.55^2 = 0.3675
=> P(X = 3) = 4C3*0.45^3*0.55^1 = 0.2005
=> P(X = 4) = 4C4*0.45^4*0.55^0 = 0.0410

(d) Mean = n*p = 4*0.45 = 1.80

standard deviation = sqrt(npq) = sqrt(4*0.45*0.55) = 0.9949 = 0.99

(e) 10