   Chapter 9.1, Problem 26E ### 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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# 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 -5000 -2500 300 P(x) 0.008 0.052 0.940

To determine

To calculate: The expected value, variance and standard deviation of the probability distribution as shown in table,

 x −5000 −2500 300 P(x) 0.008 0.052 0.94
Explanation

Given Information:

The random variable and their probability are given as,

 x −5000 −2500 300 P(x) 0.008 0.052 0.94

Formula used:

In case of 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,

σ=V(x)

Calculation:

Consider the given information,

x1=5000P(x1)=0.008x2=2500P(x2)=0

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