Mean and variance of made free throws

Question

A basketball player takes 25 free throws, making each independently with probability p = 0.8. Let M be the number made. What are the mean and variance of M?

Accepted Answer

Independent make/miss shots with a common make probability give a binomial: M \sim 	ext{Binomial}(n=25,\ p=0.8). Binomial mean is np and variance is np(1-p): \mathbb{E}[M] = 25 	imes 0.8 = 20, \qquad \operatorname{Var}(M) = 25 	imes 0.8 	imes 0.2 = 4. Standard deviation \sqrt{4} = 2. A good shooter sinks about 20 of 25, give or take 2. Takeaway. Binomial mean np = 20 and variance np(1-p) = 4; note that p = 0.8 and p = 0.2 give the same variance because p(1-p) is symmetric around one-half.

Mean and variance of made free throws

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