Ie 230: Probability and Statistics in Engineering I

IE 230: PROBABILITY AND STATISTICS IN ENGINEERING I

FALL 1999

HOMEWORK #6 SOLUTIONS

10/6/99

Problem 1:

First, think about the number of integer outcomes we have. There are outcomes. To convince yourself this is correct, think about a sample space consisting of the integer elements 1, 2, 3 and 4. We have 4 elements or 4 – 1 + 1 = 4 elements. Since this is a uniform discrete distribution, the probability of any of the outcomes is . So,

Problem 2:

(a) Uniform Discrete Distribution (If we assume the die is fair.) We know parameter p = 1/n, but we do not know what parameter n is, because we do not know the number of sides on the die.

(b) Binomial Distribution. We know that parameter n = 100, the total number of calls made, but we do not know the parameter p, the probability that a given person answers the call.

(c) Geometric Distribution. We need only know p, the probability that a telephone call is answered, but we are not given this information.

(d) Negative Binomial Distribution. We know that parameter r = 5, the total number of people we want to contact, but we do not know parameter p, the probability that a phone call is answered.

(e) Hyper-Geometric Distribution. We know that N = 2000 people, the total population, we know that n = 800 people, the subset of N we are selecting and using as our sample and we know that K = 15, the number of people interested in the entire population. Therefore, in this instance, we know all the parameters.

Problem 3:

(a)

(b) Hypergeometric Distribution with N = 15, r = 3, K = 2.

(c) No, because the experimenter samples three random samples from the shipments and has no way of having information regarding the the totes in the rest of the shipment.

(d)

(e) P(X > 0) = 1 - P(X = 0) = 0.37143

(f) E(X) = 0*P(X = 0) + 1*P(X = 1) + 2*P(X = 2) = 0.34286 + 2*0.02857 = np = 0.4

Problem 4:

(a) Poisson with P(X=0) = 1 - e-0.1 = 0.095

(b) P(X>0) = 1 - P(X=0) = 0.905

= 0.002521

(d)