## Gwen, a friend of Mary from the previous question, is also practicing free throws. However, she is trying to score 3 points in a single set.

Question

Gwen, a friend of Mary from the previous question, is also practicing free throws. However, she is trying to score 3 points in a single set. She will keep shooting sets until she has three successful shots in a single set. Gwen is more confident in her abilities, and believes that she can successfully make any single shot with a probability of 0.8.

a) Given the information above, how many sets does Gwen expect to make?

b)b) Given the information above, what is the variance for the number of sets Gwen will make?

c) Given the information above, how many shots does Gwen expect to make?

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1 week 2021-10-11T13:18:20+00:00 1 Answer 0

a. Gwen expect to make 3.75 sets

b. The variance for the number of sets Gwen will make is 0.9375

c. Gwen expect to make 2 shots

Step-by-step explanation:

a. According to the given data we have the following:

Here this follows negative binomial distribution with parameter r =3 and p=0.8

To calculate how many sets does Gwen expect to make we have to calculate the following formula:

expected number of sets =r/p

expected number of sets =3/0.8=3.75

Gwen expect to make 3.75 sets.

b. In order to calculate the variance for the number of sets Gwen will make we have to use the following formula:

variance for the number of sets=σ∧2=r(1-p)/p∧2

variance for the number of sets=3*(1-0.8)/0.8^2

variance for the number of sets=0.9375

The variance for the number of sets Gwen will make is 0.9375

c. To calculate how many shots does Gwen expect to make, we have to calculate first the probability she shoots all the three in the set as follows:

probability she shoots all the three in the set=0.8∧3=0.512

if E(X)=1/p, therefore, 1/p=1/0.512=1.95

Gwen expect to make 2 shots