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Suppose that each time Michael Jordan takes a 3-point shot, he has a 37% probability of success, independent of all other attempts. If Jordan takes seven 3-point shots in a game, what is the variance of the total number of points he scores from these shots?

0.37 chance of making the shot. So the variance of 7 shots being made is 7 * p (1-p) = 7 * (0.37) (1-0.37) = 1.63

How do you account for the 3 points that each shot is worth?

Wayne L
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1 Answers1

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If $S$ is a random variable representing the number of three-pointers he makes, then $S$ has a binomial distribution with success probability $p=0.37$ across $n=7$ independent trials, with PMF

$$\mathrm{Pr}(S=s)=\begin{cases}\dbinom7s0.37^s(1-0.37)^{7-s}&\text{for }s\in\{0,1,2,\ldots,7\}\\[1ex]0&\text{otherwise}\end{cases}$$

Each successful shot carries a value of $3$ points, so you can describe the number of points MJ gets by a new random variable, $P=3S$, whose variance would be $\mathrm{Var}[P]=3^2\mathrm{Var}[S]$.

user170231
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