Question

Question 7 ment a universities, which is going to be held in Fall 2019. For this, students who are interested to be a part of the team should take a test next weekend at the Recreation Center. Based on information from past, the coach, on average, is able to have 5 students tested in 20 minutes. 100 students have registered to take the test. One of the tests will be in the morning from 8:30 am to 11:30 am and the other one will be in the afternoon from 2 pm to 5 pm. Part a. What is the probability that exactly 40 students will have to wait until the second test (the one in the afternoon)? The height of the jump and the accuracy of the shot are the two important factors to pass the test. The coach knows that, from his experience, one out of five students can jump at least 22.2 inches, and seven out of 10 students can jump at least 15.4 inches. Assuming that the height of jump follows a Normal distribution. Part b. Show the above information on a Normal Distribution. Part c. Find the mean and the standard deviation of the height of the jump Part d. What is the probability that a randomly elected student jumps at least 19 inches? Part e. What is the probability that the coach observes 30 jumps of at least 19 inches at the ensd of the day? The second test is to measure students' ability of throwing the ball in the basket. Based on the coach's experience 60 percent of students who jump at least 19 inches make all of his/her shoots goal; and 45 percent of students who jump less than 19 inches make the entire shoots goal. The coach decides to elect the captain among students who make their entire shoots goal. Part f. What is the probability that the height of the captain's jump will be less than 19 inches?

Answer 1

7(a) Let X(t) be the number of students tested in t hour

X(t) follows Poisson with $\lambda t = 5*3 =15$ per hour

Now , given 40 students have to wait for second test

therefore out of 100, 60 students are already tested in first test (in 3 hours)

60!

=

e45(45)*0 60!

= 0.00536

Probability that exactly 40 students have to wait for second test = 0.00536

(b) Let X follow Normal distribution with mean = $\mu$ and variance =$\sigma ^{2}$

then $z= \frac{X-\mu }{\sigma}\sim N(0,1)$

Given P(X>22.2) =1/5=0.20

22.2-μ ) _ 0.20

$\Rightarrow \frac{22.2-\mu }{\sigma }=0.85$ (from z table)

and P(X>15.4) =0.70

15.4 1) = 0.70

$\Rightarrow \frac{15.4-\mu }{\sigma }=-0.53$

Solving two equations we get

$\mu =18.01 ,\sigma =4.93$

Thus

$X\sim N(\mu =18.01 ,\sigma =4.93)$

(c) Mean = 18.01

Standard deviation =4.93

(d)

$P(X\geq 19)=P(z>\frac{19-18.01}{4.93})$

= P(z > 0.20)

= 0.4207 (from z table)