Question

1.Stephen is a basketball player who makes 82 % of his free throws over the course of a season. Each day, Stephen shoots 70 free throws during practice. Assume that each day constitutes a simple random sample, SRS, of all free throws shot by Stephen, and that each free throw is independent of the rest. Let the random variable X equal the count of free throws that Stephen makes. Compute the probability that Stephen makes at least 56 free throws using the binomial distribution. Report your answer to at least four decimal places of precision.

2.Using the normal approximation to the binomial, compute this same probability again. First, compute μ X , the mean of X , and report to two decimal places of precision.

Answer 1

1) n = 70

p = 0.82

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

P(X > 56) = P(X = 56) + P(X = 57) + P(X = 58) + P(X = 59) + P(X = 60) + P(X = 61) + P(X = 62) + P(X = 63) + P(X = 64) + P(X = 65) + P(X = 66) + P(X = 67) + P(X = 68) + P(X = 69) + P(X = 70)

= 70C56 * (0.82)^56 * (0.18)^14 + 70C57 * (0.82)^57 * (0.18)^13 + 70C58 * (0.82)^58 * (0.18)^12 + 70C59 * (0.82)^59 * (0.18)^11 + 70C60 * (0.82)^60 * (0.18)^10 + 70C61 * (0.82)^61 * (0.18)^9 + 70C62 * (0.82)^62 * (0.18)^8 + 70C63 * (0.82)^63 * (0.18)^7 + 70C64 * (0.82)^64 * (0.18)^6 + 70C65 * (0.82)^65 * (0.18)^5 + 70C66 * (0.82)^66 * (0.18)^4 + 70C67 * (0.82)^67 * (0.18)^3 + 70C68 * (0.82)^68 * (0.18)^2 + 70C69 * (0.82)^69 * (0.18)^1 + 70C70 * (0.82)^70 * (0.18)^0

= 0.729882

2) = np = 70 * 0.82 = 57.4

= sqrt(np(1 - p))

= sqrt(70 * 0.82 * (1 - 0.82))

= 3.2143

P(X > 56)

= P((X - )/ > (55.5 - )/)

= P(Z > (55.5 - 57.4)/3.2143)

= P(Z > -0.59)

= 1 - P(Z < -0.59)

= 1 - 0.2776

= 0.7224