Question

6.   According to the NBA, 12% of all 7-footers in the world will eventually play professional basketball. Suppose that you selected a sample of 20 7-footers. Use this information to answer the following questions.

a.   Is this a binomial distribution? Justify your answer.
b.   What is the mean and standard deviation of this distribution be?
c.   Would a sample of 20 7-footers with 6 people playing professional basketball be considered unusual? Justify your reasoning.
d.   What is the probability that exactly 3 seven-footers play professional basketball?
e.   What is the probability that between 1 and 4 seven-footers play professional basketball?
f.   What is the probability that at least 1 7-footer plays basketball?
g.   What is the probability that at most 2 7-footers play basketball?
h.   What is the mean and standard deviation of the distribution

Answer 1

a) Yes this is a binomial distribution.

b) n = 20

p = 0.12

Mean = n * p = 20 * 0.12 = 2.4

Standard deviation = sqrt(n * p * (1 - p)) = sqrt(20 * 0.12 * 0.88) = 1.45

c) P(X = 6) = 20C6 * 0.12⁶ * 0.88¹⁴ = 0.0193

Since probability is less than 0.05, this is unusual.

d) P(X = 3) = 20C3 * 0.12³ * 0.88¹⁷ = 0.2242

e) P(1 < X < 4) = P(X = 2) + P(X = 3) = 20C2 * 0.12² * 0.88¹⁸ + 20C3 * 0.12³ * 0.88¹⁷ = 0.4982

f) P(X > 1) = 1 - P(X = 0) = 1 - 20C0 * 0.12⁰ * 0.88²⁰ = 1 - 0.0776 = 0.9224

g) P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2) = 20C0 * 0.12⁰ * 0.88²⁰ + 20C1 * 0.12¹ * 0.88¹⁹ + 20C2 * 0.12² * 0.88¹⁸ = 0.5631

h) Mean = n * p = 20 * 0.12 = 2.4

Standard deviation = sqrt(n * p * (1 - p)) = sqrt(20 * 0.12 * 0.88) = 1.45