Page 118 - Applied statistics and probability for engineers

96 Chapter 3/Discrete Random Variables and Probability Distributions

0.3

0.2
f (x)
0.1

0.0
0 1 2 3 4 5
x
Hypergeometric N = 50, n = 5, K = 25
Binomial n = 5, p = 0.5
FIGURE 3-13 0 1 2 3 4 5
Comparison of
hypergeometric and Hypergeometric probability 0.025 0.149 0.326 0.326 0.149 0.025
binomial distributions. Binomial probability 0.031 0.156 0.312 0.312 0.156 0.031

Example 3-29 Customer Sample A list of customer accounts at a large corporation contains 1000 customers.
Of these, 700 have purchased at least one of the corporation’s products in the last three months. To
evaluate a new product design, 50 customers are sampled at random from the corporate list. What is the probability that
more than 45 of the sampled customers have purchased from the corporation in the last three months?
The sampling is without replacement. However, because the sample size of 50 is small relative to the number of
customer accounts, 1000, the probability of selecting a customer who has purchased from the corporation in the last
three months remains approximately constant as the customers are chosen.
For example, let A denote the event that the i rst customer selected has purchased from the corporation in the last three
months, and let B denote the event that the second customer selected has purchased from the corporation in the last three
| (
/
/
months. Then, P A ( ) = 700 1000 = 0 7 and P B A) = 699 999 = . 0 6997. That is, the trials are approximately independent.
.
Let X denote the number of customers in the sample who have purchased from the corporation in the last three months.
Then, X is a hypergeometric random variable with N = 1 000, = 50, and K = 700. Consequently, p = K N = 0 7. The
/
.
n
,
(
requested probability is P X > 45). Because the sample size is small relative to the batch size, the distribution of X can be
.
approximated as binomial with n = 50 and p = 0 7. Using the binomial approximation to the distribution of X results in
0 7 1 0 7)
50
(
P X > 45) = ∑ ⎛ ⎜ 50⎞ ⎟ . x ( − . 50−x = 0 00017
.
x = 46 ⎝ x ⎠
The probability from the hypergeometric distribution is 0.00013, but this requires computer software to compute. The result
agrees well with the binomial approximation.
Exercises FOR SECTION 3-8
Problem available in WileyPLUS at instructor’s discretion.
Tutoring problem available in WileyPLUS at instructor’s discretion

3-141. Suppose that X has a hypergeometric distribution (c) P X( ≤ 2 ) (d) Mean and variance of X.
with N = 100, n = 4, and K = 20. Determine the following: 3-143. Suppose that X has a hypergeometric distribution
(a) P X( = ) 1 (b) P X( = ) 6 with N = 10, n = 3, and K = 4. Sketch the probability mass func-
(c) P X( = ) 4 (d) Mean and variance of X tion of X. Determine the cumulative distribution function for X.
3-142. Suppose that X has a hypergeometric distribution 3-144. A batch contains 36 bacteria cells and 12 of the cells
with N = 20, n = 4, and K = 4. Determine the following: are not capable of cellular replication. Suppose that you examine
(a) P X( = ) 1 (b) P X( = ) 4 three bacteria cells selected at random without replacement.