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Mathematical Applications for the Management, Life, and Social Sciences11th Edition

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

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

Solutions

Chapter
Section
Problem 1CP
Problem 2CP
Problem 1E
Problem 2E
Problem 3E
Problem 4E
Problem 5E
Problem 6E
Problem 7E
Problem 8E
Problem 9E
Problem 10E
Problem 11E
Problem 12E
Problem 13E
Problem 14E
Problem 15E
Problem 16E
Problem 17E
Problem 18E
Problem 19E
Problem 20E
Problem 21E
Problem 22E
Problem 23E
Problem 24E
Problem 25E
Problem 26E
Problem 27E
Problem 28E
Problem 29E
Problem 30E
Problem 31E
Problem 32E
Problem 33E
Problem 34E
Problem 35E
Problem 36E
Problem 37E
Problem 38E
Problem 39E
Problem 40E
Problem 41E
Problem 42E
Problem 43E
Problem 44E
Problem 45E
Problem 46E
Problem 47E
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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)=0.08e−0.08t.

Explanation

Given Information:

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

f(t)=0.08e−0.08t

Where t is the number of months in service.

Formula used:

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

Pr(a≤x≤b)=∫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)=0.08e−0.08t

Where t is the number of months in service.

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

Pr(a≤x≤b)=∫abf(x)dx

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

P(x≥24)=∫24∞0.08e−0.08tdx

Now, use the formula, ∫a∞f(x)dx=limb→∞∫abf(x)dx to rewrite the integral as,

P(x≥24)=limb→∞∫24b0

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