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Section 4-3/Cumulative Distribution Functions 113
Example 4-4 Hole Diameter For the drilling operation in Example 4-2, F x ( ) consists of two expressions.
.
F x ( ) = 0 for x < 12 5
.
and for 12 5 ≤ x,
x
F x ( ) = ∫ 20 e − ( u −12 5 . ) du = − e − ( x −12 5 )
20
20
.
1
12 5
.
Therefore,
<
f x ( ) = 0 x 12 5 .
20
e { − 1 − ( x − 12 5 . ) 12 5 ≤ x
.
Figure 4-7 displays a graph of F x ( ).
Practical Interpretation: The cumulative distribution function enables one to easily calculate the probability a diam-
eter in less than a value (such as 12.60 mm). Therefore, the probability of a scrapped part can be easily determined.
The probability density function of a continuous random variable can be determined from the
cumulative distribution function by differentiating. The fundamental theorem of calculus states that
d ∫ x f u du = (
( )
f x)
dx −∞
Probability Density
Function from the Then, given F x ( ),
Cumulative dF x ( )
(
Distribution Function f x) =
dx
as long as the derivative exists.
Example 4-5 Reaction Time The time until a chemical reaction is complete (in milliseconds) is approximated by
the cumulative distribution function
F x ( ) = 0 x < 0
− {1 e − 0 01. x Ð 0 x
Determine the probability density function of X. What proportion of reactions is complete within 200 milliseconds?
Using the result that the probability density function is the derivative of F x ( ), we obtain
f x ( ) = 0 x < 0
.
{0 01 e − 0 01. x 0 ≤ x
The probability that a reaction completes within 200 milliseconds is
P X < 200) = ( 1 e − 2 = .
(
F 200) = −
0 8647
Exercises FOR SECTION 4-3
Problem available in WileyPLUS at instructor’s discretion.
Tutoring problem available in WileyPLUS at instructor’s discretion
4-17. Suppose that the cumulative distribution function of Determine the following:
(
(
.
the random variable X is (a) P X < 2 8. ) (b) P X >1 5)
(
(
⎧0 x < 0 (c) P X < − ) 2 (d) P X > 6)
⎪
F x ( ) = ⎨ 0 .25 x 0 ≤ x < 5
⎪ ≤ 4-18. Suppose that the cumulative distribution function of
⎩ 1 5 x the random variable X is