Question

A college basketball player makes 80% of his free throws. Over the course of the season he will attempt 100 free throws. Assuming free throw attempts are independent, what is the probability that he makes at least 90 of these attempts?

Answer 1

the answer of this question is 9/10

Answer 2

The probability that he makes at least 90 of these attempts is 0.0062

What is the probability that he makes at least 90 of these attempts?

From the question, we have the following parameters that can be used in our computation:

Sample size, n = 100

Proportion, p = 80%

This means that the mean value is

Mean = 80% * 100

Mean = 80

The standard deviation is

SD = √(Mean * (1 - p))

SD = √(80 * (1 - 80%))

SD = 4

The standard score of 90 attempts is calculated using

z = (x - Mean)/SD

So, we have

z = (90 - 80)/4

z = 2.5

The probability that he makes at least 90 of these attempts is represented as

P(x ≥ 90) = P(z ≥ 2.5)

Using the z-score of probabilities, we have

P(x ≥ 90) = 0.0062

Hence, the probability is 0.0062