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Refer to Exercise 1.a. Find the marginal probability mass

Statistics for Engineers and Scientists | 4th Edition | ISBN: 9780073401331 | Authors: William Navidi ISBN: 9780073401331 38

Solution for problem 2E Chapter 2.6

Statistics for Engineers and Scientists | 4th Edition

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Statistics for Engineers and Scientists | 4th Edition | ISBN: 9780073401331 | Authors: William Navidi

Statistics for Engineers and Scientists | 4th Edition

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Problem 2E

Refer to Exercise 1.

a. Find the marginal probability mass function pX(x).

b. Find the marginal probability mass function pY(y).

c. Find μX.

d. Find μγ.

e. Find σX.

f. Find σγ.

g. Find Cov(X, Y).

h. Find ρX,Y .

i. Are X and Y independent? Explain.

Exercise 1: In a certain community, levels of air pollution may exceed federal standards for ozone of for particulate matter on some days. For a particular summer season, let X be the number of days on which the ozone standard is exceeded and let Y be the number of days on which the particulate matter standard is exceeded. Assume that the joint probability mass function of X and Y is given in the following table

Step-by-Step Solution:

Solution :

Step 1 of 9:

The number of days on which the ozone standard is exceeded is X and

The number of days on which the particulate matter standard is exceeded is Y.

Then the joint probability mass function of X and Y is given in the table below.

                                 Y

 X

 0

1

2

0

0.10

0.11

0.05

1

0.17

0.23

0.08

2

0.06

0.14

0.06

Our goal is:

a). We need to find the marginal probability function .

b). We need to find the marginal probability function .

c). We need to find mean .

d). We need to find mean .

e). We need to find mean .

f). We need to find mean .

g). We need to find Cov(X,Y).

h). We need to find .

i). We need to find find X and Y are independent? Explain.

a).

Now we have to find the marginal probability function .

Here we are summing all the rows to find a probability mass function.

Then the  table is given below.

                                 Y

Total

 X

 0

1

2

0

0.10

0.11

0.05

0.10+0.11+0.05

1

0.17

0.23

0.08

0.17+0.23+0.08

2

0.06

0.14

0.06

0.06+0.14+0.06

Then,

                                 Y

Total

 X

 0

1

2

0

0.10

0.11

0.05

0.26

1

0.17

0.23

0.08

0.48

2

0.06

0.14

0.06

0.26

Therefore the marginal probability function =0.26,=0.48 and= 0.26.


Step 2 of 2

Chapter 2.6, Problem 2E is Solved
Textbook: Statistics for Engineers and Scientists
Edition: 4
Author: William Navidi
ISBN: 9780073401331

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Refer to Exercise 1.a. Find the marginal probability mass