Question

According to a survey by TD Ameritrade, one out of four investors has exchange-traded funds in their portfolios (USA Today, January 11, 2007). Consider a sample of 40 investors.

Compute the probability of exactly 7 investors having exchange-traded funds in their portfolio.

Compute the probability that at least 5 of the investors have exchange-traded funds in their portfolio.

Compute the probability that at most 4 of the investors have exchange-traded funds in their portfolio.

Find the mean and standard deviation.

If you found that exactly 10 of the investors have exchange-traded funds in the

portfolio would you doubt the accuracy of the survey and why?

Answer 1

n = 40

p = 1/4 = 0.25

It is a binomial distribution.

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

a) P(X = 7) = 40C7 * (0.25)^7 * (0.75)^33 = 0.0857

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

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

                  = 1 - ( 40C0 * (0.25)^0 * (0.75)^40 + 40C1 * (0.25)^1 * (0.75)^39 + 40C2 * (0.25)^2 * (0.75)^38 + 40C3 * (0.25)^3 * (0.75)^37 + 40C4 * (0.25)^4 * (0.75)^36)

                 = 1 - 0.0160

                 = 0.9840

c) P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

                   = 40C0 * (0.25)^0 * (0.75)^40 + 40C1 * (0.25)^1 * (0.75)^39 + 40C2 * (0.25)^2 * (0.75)^38 + 40C3 * (0.25)^3 * (0.75)^37 + 40C4 * (0.25)^4 * (0.75)^36

                = 0.0160

d) mean = n * p = 40 * 0.25 = 10

Standard deviation = sqrt(np(1 - p))

                               = sqrt(40 * 0.25 * 0.75)

                               = 2.7386

e) P(X = 10) = 40C10 * (0.25)^10 * (0.75)^30 = 0.1444

Since the probability is greater than 0.05, so it is not unusual value. So we will not doubt the accuracy of the survey.