Mike and Ike decide to have a basketball shooting contest, and they decide shoot
ID: 2923725 • Letter: M
Question
Mike and Ike decide to have a basketball shooting contest, and they decide shoot three pointers. Ike
truly makes 40% of his three point shots, while Mike truly makes 30% of his three point shots. It is safe
to assume each shot is independent from the next.
a. What is the probability the Mike makes at least one of his first 5 three pointers?
b. They decide to shoot 50 three pointers apiece. What are the mean and standard deviation for the
number of three pointers that Mike will make in 50 three point shots?
c. What is the approximate probability that Kevin will make at least 20 of his 50 three
point shots?
Please explain your steps...Thanks
Explanation / Answer
a)probability the Mike makes at least one of his first 5 three pointers=1-P( he makes none of his first 3 pointers)
=1-(1-0.3)5 =1-0.1681 =0.8319
b)from binomial distribution
mean number of three pointers that Mike will make in 50 three point shots =np =50*0.3 =15
std deviation of number of three pointers that Mike will make in 50 three point shots =(np(1-p))1/2
=(50*0.3*(1-0.3))1/2 =3.2404
c)
we can solve part c by normal approximation: but for this Kevin percentage is required while in question it is given for onlt Mike and Lke. PLease revert on this so that I may help.