Question

Chauncey Billups, a current shooting guard for the Los Angeles Clippers, has a career free-throw percentage of 89.4%. Suppose he shoots six free throws in tonight’s game. What is the standard deviation of the number of free throws that Billups will make?

Answer 1

Answer:

$\sigma = 0.754$

Step-by-step explanation:

We are given the following in the question:

We treat free throws as a success.

P(Billups make a free throw) = 89.4% = 0.894

Then the number free throw follows a binomial distribution, where

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

WE have to find the standard deviation of the number of free throws that Billups will make.

We are given n  = 6.

Standard deviation formula:

$\sigma = \sqrt{np(1-p)} = \sqrt{6(0.894)(1-0.894)} = 0.754$

0.754 is the standard deviation of the number of free throws that Billups will make.