Question

Each year a company selects a number of employees for a management training program. On average, 50 percent of those sent complete the program. Out of the 24 people sent, what is the probability that 13 or more complete the program?

a) 0.7293

b) 0.8293

c) 0.5805

d) 0.2706

e) 0.4194

f) None of the above

Answer 1

n = 24

p = 0.5

It is a binomial distribution.

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

P(X > 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) + P(X = 21) + P(X = 22) + P(X = 23) P(X = 24)

            = 24C13 * (0.5)^13 * (0.5)^11 + 24C14 * (0.5)^14 * (0.5)^10 + 24C15 * (0.5)^15 * (0.5)^9 + 24C16 * (0.5)^16 * (0.5)^8 + 24C17 * (0.5)^17 * (0.5)^7 + 24C18 * (0.5)^18 * (0.5)^6 + 24C19 * (0.5)^19 * (0.5)^5 + 24C20 * (0.5)^20 * (0.5)^4 + 24C21 * (0.5)^21 * (0.5)^3 + 24C22 * (0.5)^22 * (0.5)^2 + 24C23 * (0.5)^23 * (0.5)^1 + 24C24 * (0.5)^24 * (0.5)^0

           = 0.4194

Option - e is correct.