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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Question
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Chapter 5.3, Problem 96E

(a)

To determine

Probability for randomly chosen all 7 references still work two years later.

(a)

Expert Solution
Check Mark

Answer to Problem 96E

Probability that randomly selected all 7 references still work two years later is approx. 0.3773.

Explanation of Solution

Given information:

87% of the Internet sites still work within two years of publication.

7 Internet references from scientific journal are chosen at random.

Calculations:

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

According to multiplication rule for independent events,

  P(AandB)=P(AB)=P(A)×P(B)

Let

A: One reference still works two years later

B: 7 references still work two years later

Now,

Probability for the reference still works two years later,

  P(A)=87%=0.87

Since the references are selected at random, it would be more convenient to assume that references are independent of each other.

Thus,

For probability that 7 references still work two years later, apply multiplication rule for independent events:

  P(B)=P(A)×P(A)×...×P(A)7references=(P(A))7=(0.87)70.3773

Thus,

Probability for the randomly selected all 7 references still work two years later is approx. 0.3773.

(b)

To determine

Probability for at least 1 of the 7 references doesn’t work two years later.

(b)

Expert Solution
Check Mark

Answer to Problem 96E

Probability that at least 1 of the 7 references doesn’t work two years later is 0.6227.

Explanation of Solution

Given information:

87% of the Internet sites still work within two years of publication.

7 Internet references from scientific journal are chosen at random.

Calculations:

According to complement rule,

  P(Ac)=P(notA)=1P(A)

Let

B: 7 references still work two years later

Bc: None of the 7 references still work two years later

From Part (a),

We have

Probability for randomly selected all 7 references still work two years later,

  P(B)0.3773

We have of find the probability for at least 1 of the 7 references does not work two years later.

That means

None of the 7 references works two years later.

Apply the complement rule:

  P(Bc)=1P(B)=10.3773=0.6227

Thus,

Probability that at least 1 of the 7 references does not work two years later is 0.6227.

(c)

To determine

Part (a) calculations may not be valid if 7 Internet references are chosen from one issue of the same journal.

(c)

Expert Solution
Check Mark

Answer to Problem 96E

It is not necessary that references are independent of each other.

The multiplication rule for independent events cannot be applied.

Explanation of Solution

Given information:

87% of the Internet sites still work within two years of publication.

7 Internet references from scientific journal are chosen at random.

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

According to multiplication rule for independent events,

  P(AB)=P(AandB)=P(A)×P(B)

In Part (a),

Multiplication rule for independent events has been used.

When 7 references are chosen from one issue of same journal, we are more likely to select some references from the same website.

That means

If one of the 7 references no longer works, it is possible that other references also no longer work.

This implies

The references will be no longer independent.

Thus,

Use of the multiplication for independent events would be inappropriate.

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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