A college basketball player makes 80% of his free throws. At the end of a game, his team is down one point. He is fouled attempting a three-pointer and hence awarded three free throws. Note that one free throw is worth one point. Assume each free throw is independent.

What is the standard deviation of the number of free throws he would make?

Question

A college basketball player makes 80% of his free throws. At the end of a game, his team is down one point. He is fouled attempting a three-pointer and hence awarded three free throws. Note that one free throw is worth one point. Assume each free throw is independent.

What is the standard deviation of the number of free throws he would make?

goformit100 · Answer

Take Probabily in use..

anonymous · Answer

I know missing all of them is 0.008 and i don't know how to get the probability for the rest

goformit100 · Answer

You can do, I belive on you.

anonymous · Answer

hint? :(

jim_thompson5910 · Answer

This is a binomial distribution problem where

n = 3 (he attempts 3 free throws)
p = 0.80 (the probability he makes any given free throw)

jim_thompson5910 · Answer

the mean of the binomial distribution is

xbar = n*p

the standard deviation is 

sigma = sqrt(n*p*(1-p))

anonymous · Answer

How do i get the probability of making 1 free throw?

jim_thompson5910 · Answer

by using the formula

P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)

where in this case

n = 3
p = 0.8
k = 1

anonymous · Answer

oh :/ didn't learn that yet

jim_thompson5910 · Answer

you haven't learned about the binomial distribution yet?

anonymous · Answer

Nope, i don't know if we will