   Chapter 9.3, Problem 49E ### 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 The daily demand x for water (in millions of gallons) in a town is a random variable with the probability density function f ( x ) = 1 9 x e − x / 3 ,           [ 0 , ∞ ) (a) Find the mean and standard deviation of the demand.(b) Find the probability that the demand is greater than 4 million gallons on a given day.

(a)

To determine

To calculate: The expected value(mean) and standard deviation of the probability density function f(x)=19xex/3 on the interval [0,) for the daily demand x as a random variable for water (in millions of gallons) in a town.

Explanation

Given information:

The function is,

f(x)=19xex/3

And the interval is [0,).

Formula used:

To find the expected value or mean of x,

μ=E(x)=abxf(x)dx

The variance of x is,

V(x)=abx2f(x)dxμ2

The standard deviation of x is,

σ=V(x)

Where, f is a probability density function with a continuous random variable x in the interval [a,b]

Calculation:

Consider the provided function is,

f(x)=19xex/3

Now apply the formula μ=E(x)=abxf(x)dx to find the expected value,

μ=0x[19xex/3]dx=190[x2ex/3]dx

Now, apply by parts formula u(x)v(x)dx=uvdxu(vdx)dx,

Thus,

μ=19[3ex3x26ex3xdx]

Now, further solve,

μ=limb3[x29ex/32(x31)ex/3]0b=6

Thus, the mean value is 6

(b)

To determine

To calculate: The probability that the demand is greater than 4 million gallons on a given day, if the probability density function f(x)=19xex/3 on the interval [0,) for the daily demand x as a random variable for water (in millions of gallons) in a town.

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