   # Finding Expected Value, Variance, and Standard Deviation In Exercises 13–16, find the expected value, variance, and standard deviation for the given probability distribution. x 0 1 2 3 4 P ( x ) 1 10 3 10 2 10 3 10 1 10 ### 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 13RE
Textbook Problem
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## Finding Expected Value, Variance, and Standard Deviation In Exercises 13–16, find the expected value, variance, and standard deviation for the given probability distribution. x 0 1 2 3 4 P(x) 1 10 3 10 2 10 3 10 1 10

To determine

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

 x 0 1 2 3 4 P(x) 110 310 210 310 110

### Explanation of Solution

Given Information:

The probability distribution:

 x 0 1 2 3 4 P(x) 110 310 210 310 110

Formula used:

If a discrete random variable takes the values {x1,x2,,xm}, then the expected value of the random variable is defined as,

E(x)=x1P(x1)+x2P(x2)++xmP(xm)

Variance of the random variable is defined as,

V(x)=(x1μ)2P(x1)+(x2μ)2P(x2)++(xmμ)2P(xm)

The standard deviation of the random variable is defined as,

σ=V(x)

Calculation:

Consider the provided probability distribution.

 x 0 1 2 3 4 P(x) 110 310 210 310 110

First use the provided probability distribution and the expected value formula E(x)=x1P(x1)+x2P(x2)++xmP(xm) to calculate the expected value of the probability distribution. So, the expected value is calculated as,

E(x)=(0)(110)+(1)(310)+(2)(210)+(3)(310)+(4)(110)=3+4+9+410=2010=2

Thus, the expected value is E(x)=2

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