Question

Suppose approximately 80% of all marketing personnel are extroverts, whereas about 60% of all computer programmers are introverts. (For each answer, enter a number. Round your answers to three decimal places.) (a) At a meeting of 15 marketing personnel, what is the probability that 10 or more are extroverts? What is the probability that 5 or more are extroverts? What is the probability that all are extroverts? (b) In a group of 4 computer programmers, what is the probability that none are introverts? What is the probability that 2 or more are introverts? What is the probability that all are introverts?

Answer 1

A) n = 15

P = 0.8

It is a binomial distribution

P(X = x) = nCx * p^x * (1 - p)^{n - x}

I) P(X > 10) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)

= 15C10 * (0.8)^10 * (0.2)^5 +  15C11 * (0.8)^11 * (0.2)^4 +  15C12 * (0.8)^12 * (0.2)^3 +  15C13 * (0.8)^13 * (0.2)^2 +  15C14 * (0.8)^14 * (0.2)^1 +  15C15 * (0.8)^15 * (0.2)^0 = 0.939

ii) P(X > 5) = 1 - P(X < 5)

= 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

= 1 - ( 15C0 * (0.8)^0 * (0.2)^15 +  15C1 * (0.8)^1 * (0.2)^14 +  15C2 * (0.8)^2 * (0.2)^13 +  15C3 * (0.8)^3 * (0.2)^12 +  15C4 * (0.8)^4 * (0.2)^11 = 1 - 0.000 = 1

iii) P(X = 15) =  15C15 * (0.8)^15 * (0.2)^0 = 0.035

B) n = 4

P = 0.6

i) P(X = 0) = 4C0 * (0.6)^0 * (0.4)^4 = 0.026

ii) P(X > 2) = P(X = 2) + P(X = 3) + P(X = 4)

= 4C2 * (0.6)^2 * (0.4)^2 + 4C3 * (0.6)^3 * (0.4)^1 + 4C4 * (0.6)^4 * (0.4)^0 = 0.821

iii) P(X = 4) = 4C4 * (0.6)^4 * (0.4)^0 = 0.130