EBK PRACTICE OF STAT.F/AP EXAM,UPDATED
EBK PRACTICE OF STAT.F/AP EXAM,UPDATED
6th Edition
ISBN: 9781319287573
Author: Starnes
Publisher: MPS PUB
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Chapter 5.2, Problem 36E

(a)

To determine

Discuss the probability model as a valid probability model.

(a)

Expert Solution
Check Mark

Answer to Problem 36E

The probability model is valid.

Explanation of Solution

Given information:

Distribution of responses:

  EBK PRACTICE OF STAT.F/AP EXAM,UPDATED, Chapter 5.2, Problem 36E , additional homework tip  1

Criteria for valid probability model:

  • For each outcome, sum of probabilities should be equal to 1.
  • All probabilities should be between 0 and 1 (including both).

Note that

All the probabilities shown in the table are between 0 and 1.

Thus,

First requirement has been satisfied.

Now,

Sum of the probabilities shown in the table:

  Sumofprobabilities=P(E)+P(F)+P(A/P)+P(Oth)=0.63+0.22+0.06+0.09=1

The sum of probabilities shown in the table is equal to 1.

Thus,

Second requirement has also been satisfied.

We can say

This is a valid probability model.

(b)

To determine

Probability for chosen person’s mother tongue is not English.

(b)

Expert Solution
Check Mark

Answer to Problem 36E

Probability,

  P(Ec)=0.37

Explanation of Solution

Given information:

Distribution of responses:

  EBK PRACTICE OF STAT.F/AP EXAM,UPDATED, Chapter 5.2, Problem 36E , additional homework tip  2

According to complement rule,

  P(Ac)=P(notA)=1P(A)

Note that

The probability of English:

  P(E)=0.63

Apply complement rule:

  P(Ec)=P(notE)=10.63=0.37

Thus,

Probability that the chosen person’s mother tongue is not English is 0.37.

(c)

To determine

Probability for chosen person’s mother tongue is one of the Canada’s official languages.

(c)

Expert Solution
Check Mark

Answer to Problem 36E

Probability,

  P(EF)=0.85

Explanation of Solution

Given information:

Distribution of responses:

  EBK PRACTICE OF STAT.F/AP EXAM,UPDATED, Chapter 5.2, Problem 36E , additional homework tip  3

If the two events cannot occur at the same time, then two events are mutually exclusive or disjoint.

According to addition rule for mutually exclusive or disjoint events:

  P(AB)=P(AorB)=P(A)+P(B)

Note that

Probability of English,

  P(E)=0.63

Probability of French,

  P(F)=0.22

Since both English and French are two mutually exclusive events, it is impossible to have both the languages as mother tongue.

Apply the addition rule for disjoint or mutually exclusive events:

  P(EF)=P(EorF)=P(E)+P(F)=0.63+0.22=0.85

Thus,

Probability that the chosen person’s mother tongue is one of the Canada’s official languages is 0.85.

Chapter 5 Solutions

EBK PRACTICE OF STAT.F/AP EXAM,UPDATED

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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