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Stat 315 - Recommended Homework 2 - Chapter 3: Statistics of X

This document provides homework problems and solutions for Chapter 3 of a statistics textbook. It includes questions on probability mass functions, cumulative distribution functions, expected values, variances, standard deviations, linear transformations of random variables, binomial and Poisson distributions. Key concepts covered are probability, random variables, expectations, transformations, binomial and Poisson distributions. Solutions are provided using statistical formulas and a spreadsheet template for random variables.

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Dipesh Jain
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31 views3 pages

Stat 315 - Recommended Homework 2 - Chapter 3: Statistics of X

This document provides homework problems and solutions for Chapter 3 of a statistics textbook. It includes questions on probability mass functions, cumulative distribution functions, expected values, variances, standard deviations, linear transformations of random variables, binomial and Poisson distributions. Key concepts covered are probability, random variables, expectations, transformations, binomial and Poisson distributions. Solutions are provided using statistical formulas and a spreadsheet template for random variables.

Uploaded by

Dipesh Jain
Copyright
© Attribution Non-Commercial (BY-NC)
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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Stat 315 Recommended Homework 2 Chapter 3

Chapter 3
3.1.

a.

P( x) = 1.0

b.

x
F(x)
0
0.3
1
0.5
2
0.7
3
0.8
4
0.9
5
1.0
c. P(X > 2) = 1 F(2) = 0.3
3-2.

a.

P( x) = 1.0

b. x
F(x)
0
0.01
1
0.10
2
0.40
3
0.60
4
0.80
5
0.90
6
1.00
c. P ( X 4) = F(4) = 0.80
d. P ( X 2) = 1 F(1) = 0.90
3-10.

a.

P( x) = 1.0

b. 0.85, found by summing the probabilities for 0.75 0.78:


P(0.75) + P(0.76) + P(0.77) + P(0.78) = 0.25 + 0.40 + 0.15 + 0.05 = 0.85
c. 0.80 found by: 1 P(0.77) P(0.78) = 1 0.15 0.05 = 0.80
d. 0.36 found by: P(price >0.75) = P(0.76) + P(0.77) + P(0.78) = 0.40 + 0.15 + 0.05 =
0.60
for the price to be above 0.75 for two days in a row: (0.60)2 = 0.36
3-11.

For the random variable in Problem 3-1:


E(X) = 1.8
E(X 2) = 6
V(X) = 2.76
[Using the template: Random Variables.xls]

SD(X) = 1.661

Statistics of X

Mean
1.8
Variance 2.76
Std. Devn. 1.66132

3-16.

For Problem 3-6:


E(X) = 2.2
P(X > E(X)) = P(X > 2.2) = P(3) + P(4) + P(5) = 0.3

3-19 Excluded
3.21.We use Equation (3-6) for a linear function of a random variable.
E(aX + b) = aE(X) + b
Hence: E(net income of a manager)
= 0.95E(X) - 300 = 0.95(1230) - 300 = $868.5M
[Using the template: Random Variables.xls)]
Enter the formula for h(X) in cell G12 as: = -300 +0.95*x, [omitting the marks]
Definition of h(X)
h(X) =

1315

Statistics of h(X)
Mean
Variance
Std. Devn.
Skewness
(Relative) Kurtosis

868.5
123733
351.757
-0.50362
-0.54598

3.25.In Problem 3-2, penalty = X 2.


Find E(penalty). From Problem 3-11:

x
1700

E(X 2) = $12.39

3.26.Mean = 0.03, var. = 0.9291


[Using the template: Random Variables.xls]

Descriptive Statistics of a Random Variable


x
-1
0
1

P(x)
0.45
0.07
0.48

F(x)
0.45
0.52
1

Statistics of X
Mean
Variance
Std. Devn.
Skewness
(Relative) Kurtosis

0.03
0.9291
0.9639
-0.0599
-1.921

3.33.X is binomial if sales calls are independent of each other.


3.39.Using the Binomial Template, we get the following answers:

a) 0.8889
b) 11
c) 0.55
3.40.Using the Binomial Template, we get the following answers:
a) 0.5769
b) I) 27
ii) 37
iii) 0.9576

3-41.
a)
b)
c)
d)

0.9981
0.889, 0.935
4, 5
increase the reliability of each engine.

3-49. Excluded

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