   Chapter 13.7, Problem 45E ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# Quality control The probability density function for the life span of an electronics part is f ( t )   = 0.08 e 0.08 t , where t is the number of months in service. Find the probability that any given part of this type lasts longer than 24 months.

To determine

To calculate: The probability that a randomly selected part lasts longer than 24 months if the probability density function for the life span of an electronic part is f(t)=0.08e0.08t.

Explanation

Given Information:

The probability density function for the life span of an electronic part is,

f(t)=0.08e0.08t

Where t is the number of months in service.

Formula used:

When f(x) is a continuous probability density function, then

Pr(axb)=abf(x)dx

According to the exponential rule of integrals,

exdx=ex+C

Calculation:

It is provided that the probability density function for the life span of an electronic part is,

f(t)=0.08e0.08t

Where t is the number of months in service.

Now, the formula for probability for continuous probability density function is,

Pr(axb)=abf(x)dx

Thus, to obtain the probability that a randomly selected part lasts longer than 24 months, substitute 24 for a, 0.08e0.08t for f(x) and for b in above formula to get,

P(x24)=240.08e0.08tdx

Now, use the formula, af(x)dx=limbabf(x)dx to rewrite the integral as,

P(x24)=limb24b0

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