   Chapter 9.2, Problem 39E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Demand Using the probability density function in Exercise 38, determine the number of tons that should be ordered each week so that the demand can be met for 90% of the weeks.Demand The weekly demand x (in tons) for a certain product is described by the probability density function f ( x ) = 1 36 x e − x / 6 ,       [ 0 , ∞ ] Find and interpret each probability. ( a ) P ( x   < 6 ) ( b )   P ( 6   < x <   12 ) ( c )   P ( 12 ≤ x ≤ 24 ) ( d ) P ( x >   12 )   =   1   - P ( x ≤   12 )

To determine

To calculate: The number of tons that should be ordered each week so that demand can be met for 90% of the weeks, when the probability density function is,

f(x)=136xex/6,  [0,)

Explanation

Given Information:

The weekly demand x for a certain product is described by the probability density function,

f(x)=136xex/6,  [0,)

Formula used:

Integration by parts.

When u and v is assumed to be the differentiable functions of x then,

u dv=uvv du

Calculation:

Consider the exponential probability density function,

f(x)=136xex/6,  [0,)

In order to calculate the number of tons integrate f(x) over interval [0,b].

Thus,

P(0<x<b)=0.900b136xex/6dx=0.90

In order to solve the integral 0b136xex/6dx apply integration by parts formula.

Let u=x and dv=ex/6dx, then

dv=ex/6dx

Apply integral on both sides of the above equation as,

dv=ex/6dxv=6ex/6

Thus, v=6ex/6,

Then differentiate both sides of the equation u=x as,

du=dx

Substitute u=x, dv=ex/6dx, v=6ex/6 and du=dx in the formula u dv=uvv du as,

xex/6dx=x(6ex/6)6ex/6dx=6xex/6

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