The Bermuda department of tourism estimates that 80% of visitors who attend a co

ID: 2506218 • Letter: T

Question

The Bermuda department of tourism estimates that 80% of visitors who attend a conference on

the island bering a guest or family member along. Suppose we would like to use a binomial random variable

X to describe how many of the 200 attendees of the Architectural Coating Manufacturer's Conference will

bring somebody with them.

a. What are the mean and variance of X?

b. What is P(X = 160)?

c. What is P(X = 180)?

d. What is P(X greater than or equal to 155)?

e. What is P(X less than or equal to 165)?

1) mean = np = 200*.80 = 160
variance = npq = 200*.80*.20 = 32

SD = sqrt (32) = 5.6568

2) P(X = 160) = 200C160 * .8^160 * .2^40 = 0.070369599

3) P(X = 180) = 200C180 * .8^180 * .2^20 = 0.00006089619681

4) P(X >= 155 ) = 1 - P(X <= 155)

Z = (155 - 160)/5.6568 = -0.88

Ans = 1- P(Z) = 1- 0.1894 = 0.8106 = 81.06 %

5) P(X <=165) = p(Z)

Z = (165-160)/5.6568 = 0.88

P(Z) = 0.8106 = 81.06 %

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