   # In Exercises 5 and 6, find the expected value, variance, and standard deviation for the given probability distribution. x 0 1 2 3 P ( x ) 2 10 1 10 4 10 3 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 5TYS
Textbook Problem
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## In Exercises 5 and 6, find the expected value, variance, and standard deviation for the given probability distribution. x 0 1 2 3 P(x) 2 10 1 10 4 10 3 10

To determine

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

 x 0 1 2 3 P(x) 210 110 410 310

### Explanation of Solution

Given Information:

The probability distribution:

 x 0 1 2 3 P(x) 210 110 410 310

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 P(x) 210 110 410 310

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)(210)+(1)(110)+(2)(410)+(3)(310)=0+1+8+910=1810=1

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