Question

Suppose that each time Diana Taurasi takes a 3-point shot, she has a 37% probability of success, independent of all other attempts. (Success yields 3 points; failure yields 0.) IfTaurasi takes seven 3-point shots in a game, what is the variance of the total number of points she scores from these shots?Please use the concept of independent random variables and/or variance and/or Linearity of Expectation and/or Binomial Distribution.

Answer 1

Answer:

The variance of the total number of points she scores from these shots is 4.8951.

Step-by-step explanation:

For each 3-point shot that she takes, there are only two possible outcomes. Either she makes it, or she misses. This means that we can solve this problem using concepts of the binomial probability distribution.

Binomial probability distribution:

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

Has the variance given by:

$Var(X) = np(1-p)$

In this problem, we have that:

She takes 7 shots, each one with a 37% probability of success. So $n = 7, p = 0.37$. So:

$Var(X) = np(1-p) = 7*0.37*0.63 = 1.6317$

This is the Variance of the number of shots that she makes. Each shot is worth three points. So the variance of the total number of points she scores from these shots is 3*1.6317 = 4.8951.