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. (If you are unfamiliar with a “high five” –please ask!)

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

Answer to part a)

this variable follows Binomial distribution

there are only two possible outcomes : either the students returns hi five , or not

the probability is fixed to 0.85

.

Answer to part b)

n = 10 , p = 0.85

H denotes the number of hi fives he gets

it follows Binomial distribution

.

Answer to part c)

P(7 or 8) = P(x=7) + P(x=8)

We make use of binomial formula

P(X=x) = nCx * p^x * (1-p)^(n-x)

P(X=7) = 10C7 * 0.85^7 * (0.15)^3 = 0.1298

P(x=8) = 10C8 * (0.85)^8 * (0.15)^2 = 0.2759

P(7 or 8) = 0.1298 + 0.2759 = 0.4057

.

Answer to part d)

Expected value = n * p

Expected value = 10*0.85 = 8.5

.

Standard deviation = sqrt(n*p*q)

Standard deviation= sqrt(10 *0.85 *0.15)

Standard deviation = 1.1292

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