Question

Superstitions survey. Are Americans superstitious? A Harris (Feb. 2013) poll of over 2,000 adult Americans was designed to answer this question. One survey item concerned the phrase “see a penny, pick it up, all day long you’ll have good luck.” The poll found that about 30 percent of Americans believe finding and picking up a penny is good luck. Consider a random sample of 20 U.S. adults and let x represent the number who believe finding and picking up a penny is good luck. This is a binomial random variable.

A) Find P(x<10)

B) Find P(x>5)

C) What is the expected number (the mean) who believe picking up a penny is good luck.

Answer 1

n = 20

p = 0.3

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

A) P(X < 10) = 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)

                    = 20C0 * (0.3)^0 * (0.7)^20 + 20C1 * (0.3)^1 * (0.7)^19 + 20C2 * (0.3)^2 * (0.7)^18 + 20C3 * (0.3)^3 * (0.7)^17 + 20C4 * (0.3)^4 * (0.7)^16 + 20C5 * (0.3)^5 * (0.7)^15 + 20C6 * (0.3)^6 * (0.7)^14 + 20C7 * (0.3)^7 * (0.7)^13 + 20C8 * (0.3)^8 * (0.7)^12 + 20C9 * (0.3)^9 * (0.7)^11 = 0.9520

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

                  = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5))

                  = 1 - ( 20C0 * (0.3)^0 * (0.7)^20 + 20C1 * (0.3)^1 * (0.7)^19 + 20C2 * (0.3)^2 * (0.7)^18 + 20C3 * (0.3)^3 * (0.7)^17 + 20C4 * (0.3)^4 * (0.7)^16 + 20C5 * (0.3)^5 * (0.7)^15)

                  = 1 - 0.4164 = 0.5836

C) Expected value = n * p = 20 * 0.3 = 6