   # Demand Repeat Exercise 59 for the probability density function f ( x ) = 6 125 x ( 5 − x ) , [ 0 , 5 ] . Demand The daily demand x for a certain product (in hundreds of pounds) is a random variable with the probability density function f ( x ) = 3 256 x ( 8 − x ) , [ 0 , 8 ] . (a) Find the mean and standard deviation of the demand. (b) Find the median of the demand. (c) Find the probability that the demand is within one standard deviation of the mean. ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
Publisher: Cengage Learning
ISBN: 9781305860919

#### Solutions

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

10th Edition
Ron Larson
Publisher: Cengage Learning
ISBN: 9781305860919
Chapter 9, Problem 60RE
Textbook Problem
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## Demand Repeat Exercise 59 for the probability density function f ( x ) =   6 125 x ( 5 − x ) , [ 0 ,   5 ] . Demand The daily demand x for a certain product (in hundreds of pounds) is a random variable with the probability density function f ( x ) = 3 256 x   (   8   − x ) ,       [ 0 ,   8 ] . (a) Find the mean and standard deviation of the demand.(b) Find the median of the demand.(c) Find the probability that the demand is within one standard deviation of the mean.

(a)

To determine

To calculate: The mean and standard deviation of the random variable x which is the daily demand (in hundreds of pounds) for a certain product with the probability density function f(x)=6125x(5x) over the interval [0,5].

### Explanation of Solution

Given Information:

The daily demand x for a certain product is defined by the probability density function,

f(x)=6125x(5x) over the interval [0,5].

Formula used:

For any probability density function f of a continuous random variable x over the interval [a,b], the expected value of x is defined as,

μ=abxf(x) dx

For any probability density function f of a continuous random variable x over the interval [a,b], the standard deviation of x is defined as,

σ=V(x)

Here, V(x)=ab(xμ)2f(x) dx

Calculation:

Consider the provided probability density function,

f(x)=6125x(5x) over the interval [0,5].

Use the formula E(x)=abxf(x) dx for the provided probability density function to calculate the mean,

μ=05x(6125x(5x)) dx=612505(5x2x3) dx=6125(5x33x44)05=6125[20x33x412]05

Further solve,

μ=1250[(20(53)3(54))(20(03)3(04))]=1250(25001875)=625250=2.5

Thus, the mean of the demand is 2.5 pounds.

Now, calculate the standard deviation of the demand. So, first calculate variance of the demand.

Now use the formula V(x)=ab(xμ)2f(x) dx for the provided probability density function to calculate the variance. So,

V(x)=05(x2.5)2(6125x(5x)) dx=612505(x2

(b)

To determine

To calculate: The median of the random variable x which is the daily demand (in hundreds of pounds) for a certain product with the probability density f(x)=6125x(5x) over the interval [0,5].

(c)

To determine

To calculate: The probability that the demand is within one standard deviation of the mean of the daily demand (in hundreds of pounds) for a certain product with the probability density function f(x)=6125x(5x) over the interval [0,5].

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