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During the 2015-16 NBA season, Steven Adams of the Oklahoma City Thunder had a free throw shooting percentage of 0.502 . Assume that the probability Steven Adams makes any given free throw is fixed at 0.502 , and that free throws are independent.

Required:
a. If Steven Adams shoots 8 free throws in a game, what is the probability that he makes at least 7 of them?
b. If Steven Adams shoots 80 free throws in the playoffs, what is the probability that he makes at least 70 of them?
c. If Steven Adams shoots 8 free throws in a game, what are the mean and standard deviation for the number of free throws he makes during the game?
d. If Steven Adams shoots 80 free throws in the playoffs, what are the mean and standard deviation for the number of free throws he makes during the playoffs?

1 Answer

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Answer:

Explanation:

Since the probability that Steven Adams makes any given free throw is fixed at 0.502, then this is a binomial probability. It is either he makes a free throw or he doesn't. The probability of success, p = 0.502

The probability of failure, q = 1 - p

q = 1 - 0.502 = 0.498

a) We want to determine P(x ≥ 7)

n = 8

Using the binomial probability calculator,

P(x ≥ 7) = 0.036

b) We want to determine P(x ≥ 70)

n = 80

Using the binomial probability calculator,

P(x ≥ 70) < 0.000001

c) Mean = np = 8 × 0.502 = 4.016

Standard deviation = √npq = √(8 × 0.502 × 0.498) = 1.41

d) Mean = 80 × 0.502 = 40.16

Standard deviation = √npq = √(80 × 0.502 × 0.498) = 4.47

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