Questions: Your best friend was really into March Madness this year. He is quite
ID: 3220926 • Letter: Q
Question
Questions:
Your best friend was really into March Madness this year. He is quite sad that the tournament is over, but has decided to do some research on college basketball. You being the loyal friend you are, helps him calculate different probabilities for scenarios he has researched.
He found that OSU’s free throw percentage as a team is 78.9%. Assuming that each free throw is independent, your friend needs help calculating the probability of certain outcomes occurring for the next 100 free throws.
State the probability distribution your friend should use to calculate the probabilities and explain why.
What is the probability OSU will score at least 80 of the free throws?
What is the probability OSU will score between 60 and 90 of the free throws (exclusively)?
What is the probability OSU will score only 18 of their free throws?
I need help putting this into r. Can you show me the codes please.
Explanation / Answer
State the probability distribution your friend should use to calculate the probabilities and explain why.
Answer:
Here, we have to use the normal approximation to binomial distribution. We are given n = 100 and p = 0.789. So,
We have
Mean = n*p = 100*0.789 = 78.9,
q = 1 – p = 1 – 0.789 = 0.211 and
Standard deviation given as below:
SD = sqrt(n*p*q) = sqrt(100*0.789*0.211) = 4.08018
What is the probability OSU will score at least 80 of the free throws?
Answer:
Here, we have to find P(X80)
P(X80) = 1 – P(X<80)
Z = (X – mean)/SD
Z = (80 – 78.9)/4.08018 = 0.2696
P(X<80) = P(Z<0.2696) = 0.606264
P(X80) = 1 – P(X<80) = 1 - 0.606264 = 0.393736
Required probability = 0.393736
What is the probability OSU will score between 60 and 90 of the free throws (exclusively)?
Answer:
P(60<X<90) = P(X<90) – P(X<60)
First we have to find P(X<90)
Z = (90 – 78.9)/4.08018 = 2.72047
P(Z<2.72047) = 0.996741
Now, we have to find P(X<60)
Z = (60 – 78.9)/4.08018 = -4.6321
P(Z<-4.6321) = 0.000002
P(60<X<90) = P(X<90) – P(X<60) = 0.996741 - 0.000002 = 0.996739
Required probability = 0.996739
What is the probability OSU will score only 18 of their free throws?
Answer:
We have to find P(X<18)
Z = (18 – 78.9)/4.08018 = -14.9258
P(Z<-14.9258) = 0.00
Required probability = 0.00
Required R codes
> n=100
> p=0.789
> mean=n*p
> SD=sqrt(n*p*(1-p))
> Prob1 = dnorm(80,mean,SD)
> Prob2 = 1 - Prob1
> Prob3=dnorm(90,mean,SD)
> Prob4=dnorm(60,mean,SD)
> Prob5=Prob3 - Prob4
> Prob6 = dnorm(18,mean,SD)
> Prob2
[1] 0.9057139
> Prob5
[1] 0.002414086
> Prob6
[1] 4.11422e-50
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