Applied Statistics and Probability for Engineers

86 CHAPTER 3 DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS

Let Xequal the number of parts in the sample from the local supplier. Then, Xhas a

hypergeometric distribution and the requested probability is Consequently,

What is the probability that two or more parts in the sample are from the local supplier?

What is the probability that at least one part in the sample is from the local supplier?

The mean and variance of a hypergeometric random variable can be determined from

the trials that comprise the experiment. However, the trials are not independent, and so the

calculations are more difficult than for a binomial distribution. The results are stated as

follows.

P 1 X 12  1 P 1 X 02  1 

a

100

0

b a

200

4

b

a

300

4

b

0.804

0.298 0.098 0.01190.408

P 1 X 22 

a

100

2

b a

200

2

b

a

300

4

b

a

100

3

b a

200

1

b

a

300

4

b

a

100

4

b a

200

0

b

a

300

4

b

P 1 X 42 

a

100

4

b a

200

0

b

a

300

4

b

0.0119

P 1 X 42.

If Xis a hypergeometric random variable with parameters then

(3-14)

where .pKN

E 1 X 2 np and 2 V 1 X 2 np 11 p 2 a

Nn

N 1

b

N, K, and n,

Here pis interpreted as the proportion of successes in the set of Nobjects.

EXAMPLE 3-28 In the previous example, the sample size is 4. The random variable Xis the number of parts in

the sample from the local supplier. Then,. Therefore,

E 1 X 2  (^41100)  3002 1.33
p (^100)  300  (^1)  3
PQ220 6234F.Ch 03 13/04/2002 03:19 PM Page 86