Question

A basketball player participates in a one-and-one free throw situation. The player has a 70% free throw percentage. What is the average number of points this player will get in 100 one-and-one free throw situations?

Answer 1

Answer:

I don't know, I just need points for a question.

Step-by-step explanation:

Answer 2

Answer:

The average number of points this player will get in 100 one-and-one free throw situations is 70.

Step-by-step explanation:

For each free throw, there are only two possible outcomes. Either the player makes it, or he does not. The probability of the player making a free throw is independent of other free throws. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

$E(X) = np$

70% free throw percentage.

This means that $p = 0.7$

What is the average number of points this player will get in 100 one-and-one free throw situations?

This is E(X) when n = 100. So

$E(X) = np = 100*0.7 = 70$

The average number of points this player will get in 100 one-and-one free throw situations is 70.