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Chapter 18: A. Basic Statistics and Applications 245
EXAMPLE 18.30
Suppose a quality characteristic of a product is normally distributed with mean m = 18
and standard deviation s = 1.5. The specification limits furnished by the customer are
(15, 21). Determine what percentage of the product meets the specifications set by the
customer.
Solution:
Let the random variable X denote the quality characteristic of interest. Then X is normally
distributed with mean m = 18 and standard deviation s = 1.5.
We are interested in finding the percentage of product with the characteristic of
interest within the limits (15, 21), which is given by
15
18 X 18 21
18
100P( 15
X 21)=
100P
15 . 15 . 15 .
= 100P
20 . Z 20
.
= 100[ P( 20 . Z 20
. )+ P( 0Z
2.
0)
]
= 100× 2P( 0 Z
2.
0)
= 100× 2( 4. 772)=
9545
. %.
( )
In this case, the percentage of product that will meet the specifications set by the
customer is 95.45 percent.
Part IV.A.5