Supplementary Problem 5.39 In a study by Peter D. Hart Research Associates for t

Question

Supplementary Problem 5.39 In a study by Peter D. Hart Research Associates for the Nasdaq Stock Market, it was determined that 20% of all stock investors are retired people. In addition, 40% of all US. adults invest in mutual funds. Suppose a random sample of 25 stock investors is taken. a. What is the probability that exactly seven are retired peoplet? b. What is the probability that 10 or more are retired people? c. How many retired people would you expect to find in a random sample of 25 stock investors? d. Suppose a random sample of 20 U.s. adults is taken. What is the probability that exactly eight aduts ivested in mutual funds? e. Suppose a random sample of 20 U.S. adults is taken. What is the probability that fewer than six adults inwested in mutual funds? f. Suppose a random sample of 20 U.S. adults is taken. What is the probability that none of the adults invested in mutual funds? g. Suppose a random sample of 20 U.S. adults is taken. What is the probability that 12 or more adults invested in mutual funds? h. For parts e-g, what exact number of adults would produce the highest probability? How does this compare to the expected number? (Round your answers to 3 decimal places when calculating using Table A2) "(Round your answer to 1 decimal places) (Round your answer to 0 decimal places.) 017 c. Expected Value 013 Expected Numberp

Explanation / Answer

P(X = x) = nCx * px * (1 - p)n - x

a) P(X = 7) = 25C7 * (0.2)^7 * (0.8)^18 = 0.1108

b) P(X > 10) = 1 - P(X < 10)

                    = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9))

                   = 1 - (25C0 * (0.2)^0 * (0.8)^0 + 25C1 * (0.2)^1 * (0.8)^24 + 25C2 * (0.2)^2 * (0.8)^23 + 25C3 * (0.2)^3 * (0.8)^22 + 25C4 * (0.2)^4 * (0.8)^21 + 25C5 * (0.2)^5 * (0.8)^20 + 25C6 * (0.2)^6 * (0.8)^19 + 25C7 * (0.2)^7 * (0.8)^18 + 25C8 * (0.2)^8 * (0.8)^17 + 25C9 * (0.2)^9 * (0.8)^16)

                = 1 - 0.9827 = 0.0173

c) Expected value = n * p = 25 * 0.2 = 5

d) P(X = 8) = 20C8 * (0.4)^8 * (0.6)^12 = 0.1797

e) P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

                  = 20C0 * (0.4)^0 * (0.6)^20 + 20C1 * (0.4)^1 * (0.6)^19 + 20C2 * (0.4)^2 * (0.6)^18 + 20C3 * (0.4)^3 * (0.6)^17 + 20C4 * (0.4)^4 * (0.6)^16 + 20C5 * (0.4)^5 * (0.6)^15 = 0.1256

f) P(X = 0) = 20C0 * (0.4)^0 * (0.6)^20 = 0.00004

g) P(X > 12) = P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)

                   = 20C12 * (0.4)^12 * (0.6)^8 + 20C13 * (0.4)^13 * (0.6)^7 + 20C14 * (0.4)^14 * (0.6)^6 + 20C15 * (0.4)^15 * (0.6)^5 + 20C16 * (0.4)^16 * (0.6)^4 + 20C17 * (0.4)^17 * (0.6)^3 + 20C18 * (0.4)^18 * (0.6)^2 + 20C19 * (0.4)^19 * (0.6)^1 + 20C20 * (0.4)^20 * (0.6)^0 = 0.0565

h) Expected number = n * p = 20 * 0.4 = 8

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