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In Exercise 5.129 - 5.132, We have provided the probability distributions of the random variables considered in Exercise 5.109 - 5.112 of section 5.4 For each exercise, do the following tasks.

Part (a) Find the mean of the random variable.

Part (b) Obtain the standard deviation of the random variable by using one of the formulas given in definition 5.10.

Short Answer

Expert verified

Part (a) μ=3

Part (b)σ0.9

Step by step solution

01

Part (a) Step 1. Given information. 

y1234
P(Y=y)0.10.10.50.3
02

Part (a) Step 2. A random variable's mean is;

μ=x·(X=x)μ=1×0.1+2×0.1+3×0.5+(4×0.3)μ=3

03

Part (b) Step 1.The random variable's standard deviation:

We have subpart (a) μ=3

y
P(Y=y)
y2
y2P(Y=y)
10.110.1
20.140.4
30.594.5
40.3164.8
Total

9.8

The standard deviation is defined as

σ=x2P(X=x)-μ2σ=9.3-32σ=0.8σ=0.8944σ0.9

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Most popular questions from this chapter

In Exercises 5.16-5.26, express your probability answers as a decimal rounded to three places.

Cardiovascular Hospitalizations. From the Florida State Center for Health Statistics report Women and Cardiovascular Disease Hospitalization, we obtained the following table showing the number of female hospitalizations for cardiovascular disease, by age group, during one year.

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