Question

Stephen Curry's Free Throws As we see in Exercise P.37, during the 2015-16 NBA season, Stephen Curry of the Golden State Warriors had a free throw shooting percentage of 0.908. Assume that the probability Stephen Curry makes any given free throw is fixed at 0.908, and that free throws are independent. Let X be the number of free throws Stephen Curry makes in two attempts Questions (P.3) Complete the following tasks: 12. In the space provided, create the probability distribution of X. (I pt) 13. What is the mean of X? Interpret this value.

Answer 1

Answer:

A) $P(X=x) = \binom{2}{x}.(0.908)^x.(0.092)^{2-x}$    

B) Mean = 1.816

Step-by-step explanation:

We are given the following information:

We treat Stephen Curry making any given free throw as a success.

P(Stephen Curry makes any given free throw) = 0.908

Since the probability for the free throw is equal for each trial and  free throws are independent.

Then the number of free shots follows a binomial distribution.

A) Probability distribution

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

Here n = 2, p = 0.908

$P(X=x) = \binom{2}{x}.(0.908)^x.(1-0.908)^{2-x}\\\\P(X=x) = \binom{2}{x}.(0.908)^x.(0.092)^{2-x}$

Now x can take values 0, 1 , 2

Putting values, we get,

$P(x = 0) = 0.008464\\P(x = 1) = 0.167072\\P(x = 2) = 0.82446$

B) Mean of X

$\mu = np = 2(0.908) =1.816$

Thus, the mean number of free shots made by Stephen Curry is 1.816