   Chapter 9.1, Problem 25E ### 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
1 views

# Finding Expected Value, Variance, and Standard Deviation In Exercises 23–26, find the expected value, variance, and standard deviation for the given probability distribution. See Examples 4, 5, and 6. x -3 -1 0 3 5 P(x) 1 5 1 5 1 5 1 5 1 5

To determine

To calculate: The value of expected value, variance and standard deviation for the given probability distribution,

 x -3 -1 0 3 5 P(x) 15 15 15 15 15
Explanation

Given Information:

The random variable and their probability are given as

 x -3 -1 0 3 5 P(x) 15 15 15 15 15

Formula used:

If the set of values of a discrete random variable consist in district values x1,x2,,xn then

E(x)=x1P(x1)+x2P(x2)++xnP(xn)

If mean μ=E(x) then variance

V(x)=(x1μ)2P(x1)+(x2μ)2P(x2)++(xnμ)2P(xn)

And the standard deviation is,

σ=V(x)

Calculation:

Consider the given information

x1=3P(x1)=15x2=1P(x2)=15

And,

x3=0P(x3)=15x4=3P(x4)=15

And,

x5=5P(x5)=15

Consider the primary equation and substitute the given values

E(x)=3×15+(1)×15+

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

#### Find more solutions based on key concepts 