Purdue Pete loves to have a presence on campus. Often, he is walking around camp

Question

Purdue Pete loves to have a presence on campus. Often, he is walking around campus. He looks to give and receive high fives from anyone. When he offers a high five, a random Purdue student will high five him back with probability 0.85.

a) Purdue Pete sees a student near the Stewart Center and offers a high five. It is not yet known what this particular student will do. Let the random variable count whether or not this student high fives Purdue Pete back. What distribution, parameter(s), and support of this random variable?

b) Now, Purdue Pete sees 10 students walking out the new Wilmeth Active Learning Center (WALC). He offers each of them a high five. Let H be the number of students who high five Pete back from these students outside of WALC. What distribution, parameter(s), and support of this random variable?

c) What is the probability that 7 or 8 of these students outside WALC will high five Pete back?

d) What is the expected number of high fives Purdue will receive of the group of 10 students. What is the standard deviation?

e) Out of the ten students, it is known that least one student will always high five anyone. What is the probability 3 students high five Purdue Pete back?

f) How many students would Purdue Pete need to offer a high five to in order to be at least 99% sure that at least one student high fives him back?

Explanation / Answer

(a) Here number of student is one so here the random variable can take only two values 0 or 1 that means either he will do a hi- five or he will not.

and, p(x) = 0.85 ; x = 1

= 0.15 ; x = 0

here the given distribution is a discrete distribution. Support is 0 & 1.

(b) Now the number of students are 10. So if H = number of students who did hi- five to pete

then Support of H are the integers between 0 and 10 (both inclusive)

The distribution is a binomal one with parameter n = 10 and p = 0.85.

(c) Pr(X = 7,8) = Pr( X = 7,8 ; 10 ; 0.85) = Pr( X = 7) + Pr( X = 8) =

= 10C7 (0.85)7 (0.15)3 + 10C8 (0.85)8 (0.15)2  = 0.1298 + 0.2759 = 0.4057

(d) E(H) = 10 * 0.85 = 8.5

(e) Now here we know that one student always do high- five. So, now we are left 9 students and we have to find out of these two 2 will do highfive with pete.

Pr(3 will do high five l at least one student will always do highfive) = 9C2 (0.85)2 (0.15)7  = 4.44 * 10-3

(f) lets say there are n students needed to offer at least one high five. That means the probability of no students giving him high five shall be less than 0.01.

so for n number of students.

nC0 (0.85)0(0.15)n > 0.01

(0.15)n > 0.01

n ln (0.15) > ln (0.01)

n > ln (0.01)/ ln (0.15)

n < 2.427

so n = 3

at least 3 person shall be required to make it at lest 99% probable that at least one student will high five him back.

so

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