Question

An aspiring professional basketball player is practicing for his tryout with the L.A. Lakers. He wants to better understand what the probabilities are related to his shooting. The aspiring basketball player makes a free throw with probability .75. Let X be a binomial random variable with p = .75, where p represents the probability of a successfully shot free throw, and n = 20, where n represents the number of shots the player shoots.

What is the expected value of X?

What is the variance of X?

What is the probability the player makes exactly 18 shots?

What is the probability the player makes at least 18 shots?

Answer 1

a)  expected value of X?

E(x)= μ=n⋅p

=(20)⋅(0.75)

=15

b)  variance of X

Variance σ^2=n⋅p⋅(1−p) = 20 * 0.75(1 -0.75) = 3.75

c)

d)

P(X≥18) Probability of at least 18 successes: 0.09126

To calculate any remaining probabilities, we need to either use the binomial probability distribution function on a calculator or the binomial probability formula. We'll add these remaining probababilities to P(18)for our final answer.P(18)+P(19)+P(20)

0.066947807599718 +0.021141412926227 +0.003171211938934

= 0.09126