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.

A)    

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

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

Now x can take values 0, 1 , 2

Putting values, we get,

B) Mean of X

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

b

step-by-step explanation:

the total is 23.97. 0.25% is off so you have to subtracting 25% from 100% and you will get 75% if you convert it to decimal you will get 0.75. you have to find the original because the formula is % times original = total.so you have the total and the % so you just have to plug it in.