PRACTICE OF STATISTICS F/AP EXAM
PRACTICE OF STATISTICS F/AP EXAM
6th Edition
ISBN: 9781319113339
Author: Starnes
Publisher: MAC HIGHER
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Chapter 5.3, Problem 97E

(a)

To determine

Fill the two − way table such that A and B are mutually exclusive.

(a)

Expert Solution
Check Mark

Answer to Problem 97E

Two − way table:

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 5.3, Problem 97E , additional homework tip  1

Explanation of Solution

Given information:

Data on gender and eye color of the students summarized in two − way table:

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 5.3, Problem 97E , additional homework tip  2

A: Student is male

B: Student has blue eyes

Two events are disjoint or mutually exclusive when both events cannot occur at same time.

In this part,

Events A and B are mutually exclusive.

This implies

No male student has blue eyes.

Thus,

In the table, put 0 in the column “Male” and row “Blue”.

Also,

Put the remaining counts according to the total counts of the rows and columns.

Thus,

The two − way table becomes:

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 5.3, Problem 97E , additional homework tip  3

(b)

To determine

Fill the two − way table such that A and B are independent.

(b)

Expert Solution
Check Mark

Answer to Problem 97E

Two − way table:

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 5.3, Problem 97E , additional homework tip  4

Explanation of Solution

Given information:

Data on gender and eye color of the students summarized in two − way table:

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 5.3, Problem 97E , additional homework tip  5

A: Student is male

B: Student has blue eyes

Two events are independent, when the probability of occurrence of one event does not affect the probability of occurrence of other event.

Then

The counts will be the product of the row total and the column total, divided by the table total provided in the bottom left corner of the table.

Calculate the counts in the two − way table:

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 5.3, Problem 97E , additional homework tip  6

Thus,

The two − way table becomes:

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 5.3, Problem 97E , additional homework tip  7

(c)

To determine

Fill the two − table such that A and B are not mutually exclusive and not independent.

(c)

Expert Solution
Check Mark

Answer to Problem 97E

Two – way table:

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 5.3, Problem 97E , additional homework tip  8

Explanation of Solution

Given information:

Data on gender and eye color of the students summarized in two − way table:

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 5.3, Problem 97E , additional homework tip  9

A: Student is male

B: Student has blue eyes

For two − way table, where A and B are not mutually exclusive and not independent as well.

In this part, the count for male with blue eyes should be different from the other two parts (Part (a) and Part (b)).

Suppose, if we choose the count 10 for male with blue eyes.

Then

Put 10 in the column “Male” and the row “Blue”.

And

Put the remaining counts according to the total counts of the rows and columns.

Thus,

The two − way table becomes:

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 5.3, Problem 97E , additional homework tip  10

Chapter 5 Solutions

PRACTICE OF STATISTICS F/AP EXAM

Ch. 5.1 - Prob. 11ECh. 5.1 - Prob. 12ECh. 5.1 - Prob. 13ECh. 5.1 - Prob. 14ECh. 5.1 - Prob. 15ECh. 5.1 - Prob. 16ECh. 5.1 - Prob. 17ECh. 5.1 - Prob. 18ECh. 5.1 - Prob. 19ECh. 5.1 - Prob. 20ECh. 5.1 - Prob. 21ECh. 5.1 - Prob. 22ECh. 5.1 - Prob. 23ECh. 5.1 - Prob. 24ECh. 5.1 - Prob. 25ECh. 5.1 - Prob. 26ECh. 5.1 - Prob. 27ECh. 5.1 - Prob. 28ECh. 5.1 - Prob. 29ECh. 5.1 - Prob. 30ECh. 5.2 - Prob. 31ECh. 5.2 - Prob. 32ECh. 5.2 - Prob. 33ECh. 5.2 - Prob. 34ECh. 5.2 - Prob. 35ECh. 5.2 - Prob. 36ECh. 5.2 - Prob. 37ECh. 5.2 - Prob. 38ECh. 5.2 - Prob. 39ECh. 5.2 - Prob. 40ECh. 5.2 - Prob. 41ECh. 5.2 - Prob. 42ECh. 5.2 - Prob. 43ECh. 5.2 - Prob. 44ECh. 5.2 - Prob. 45ECh. 5.2 - Prob. 46ECh. 5.2 - Prob. 47ECh. 5.2 - Prob. 48ECh. 5.2 - Prob. 49ECh. 5.2 - Prob. 50ECh. 5.2 - Prob. 51ECh. 5.2 - Prob. 52ECh. 5.2 - Prob. 53ECh. 5.2 - Prob. 54ECh. 5.2 - Prob. 55ECh. 5.2 - Prob. 56ECh. 5.2 - Prob. 57ECh. 5.2 - Prob. 58ECh. 5.2 - Prob. 59ECh. 5.2 - Prob. 60ECh. 5.3 - Prob. 61ECh. 5.3 - Prob. 62ECh. 5.3 - Prob. 63ECh. 5.3 - Prob. 64ECh. 5.3 - Prob. 65ECh. 5.3 - Prob. 66ECh. 5.3 - Prob. 67ECh. 5.3 - Prob. 68ECh. 5.3 - Prob. 69ECh. 5.3 - Prob. 70ECh. 5.3 - Prob. 71ECh. 5.3 - Prob. 72ECh. 5.3 - Prob. 73ECh. 5.3 - Prob. 74ECh. 5.3 - Prob. 75ECh. 5.3 - Prob. 76ECh. 5.3 - Prob. 77ECh. 5.3 - Prob. 78ECh. 5.3 - Prob. 79ECh. 5.3 - Prob. 80ECh. 5.3 - Prob. 81ECh. 5.3 - Prob. 82ECh. 5.3 - Prob. 83ECh. 5.3 - Prob. 84ECh. 5.3 - Prob. 85ECh. 5.3 - Prob. 86ECh. 5.3 - Prob. 87ECh. 5.3 - Prob. 88ECh. 5.3 - Prob. 89ECh. 5.3 - Prob. 90ECh. 5.3 - Prob. 91ECh. 5.3 - Prob. 92ECh. 5.3 - Prob. 93ECh. 5.3 - Prob. 94ECh. 5.3 - Prob. 95ECh. 5.3 - Prob. 96ECh. 5.3 - Prob. 97ECh. 5.3 - Prob. 98ECh. 5.3 - Prob. 99ECh. 5.3 - Prob. 100ECh. 5.3 - Prob. 101ECh. 5.3 - Prob. 102ECh. 5.3 - Prob. 103ECh. 5.3 - Prob. 104ECh. 5.3 - Prob. 105ECh. 5.3 - Prob. 106ECh. 5.3 - Prob. 107ECh. 5.3 - Prob. 108ECh. 5 - Prob. R5.1RECh. 5 - Prob. R5.2RECh. 5 - Prob. R5.3RECh. 5 - Prob. R5.4RECh. 5 - Prob. R5.5RECh. 5 - Prob. R5.6RECh. 5 - Prob. R5.7RECh. 5 - Prob. R5.8RECh. 5 - Prob. T5.1SPTCh. 5 - Prob. T5.2SPTCh. 5 - Prob. T5.3SPTCh. 5 - Prob. T5.4SPTCh. 5 - Prob. T5.5SPTCh. 5 - Prob. T5.6SPTCh. 5 - Prob. T5.7SPTCh. 5 - Prob. T5.8SPTCh. 5 - Prob. T5.9SPTCh. 5 - Prob. T5.10SPTCh. 5 - Prob. T5.11SPTCh. 5 - Prob. T5.12SPTCh. 5 - Prob. T5.13SPTCh. 5 - Prob. T5.14SPT
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