PRACTICE OF STATISTICS F/AP EXAM
PRACTICE OF STATISTICS F/AP EXAM
6th Edition
ISBN: 9781319113339
Author: Starnes
Publisher: MAC HIGHER
Question
Book Icon
Chapter 5.3, Problem 81E

(a)

To determine

Represent this chance process with Tree diagram.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given information:

Fuel purchased by automobile drivers:

88% bought regular gasoline.

2% bought midgrade gas.

10% bought premium gas.

Customers paid with Credit card:

28% bought regular gasoline.

34% bought midgrade gas.

42% bought premium gas.

First level:

First level consists of three types of gasoline:

Regular, midgrade and premium

Thus,

The first level requires three children:

Regular, midgrade and premium

Second level:

In this level, there are two types of payment methods:

Credit card or No Credit card (other payment method)

Thus,

The second level consists of two children per child at the first level i.e.; Credit card and No Credit Card.

The required tree diagram can be drawn as:

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 5.3, Problem 81E

(b)

To determine

Probability for the customer paid with a credit card.

(b)

Expert Solution
Check Mark

Answer to Problem 81E

Probability that the customer paid with a credit card is 0.2952.

Explanation of Solution

Given information:

Fuel purchased by automobile drivers:

88% bought regular gasoline.

2% bought midgrade gas.

10% bought premium gas.

Customers paid with Credit card:

28% bought regular gasoline.

34% bought midgrade gas.

42% bought premium gas.

Calculations:

According to the general multiplication rule,

  P(AandB)=P(AB)=P(A)×P(B|A)=P(B)×P(A|B)

According to the addition rule for mutually exclusive events:

  P(AB)=P(AorB)=P(A)+P(B)

Let

R: Regular gasoline

M: Midgrade gas

P: Premium gas

C: Credit card

N: No Credit card

Now,

The corresponding probabilities:

Probability for the customer purchased regular gasoline,

  P(R)=0.88

Probability for the customer purchased midgrade gas,

  P(M)=0.02

Probability for the customer purchased premium gas,

  P(P)=0.10

Probability for the customer purchased regular gasoline paid with Credit card,

  P(C|R)=0.28

Probability for the customer purchased midgrade gas paid with Credit card,

  P(C|M)=0.34

Probability for the customer purchased premium gas paid with Credit card,

  P(C|P)=0.42

Apply general multiplication rule:

Probability for the customer paid with Credit card and purchased regular gasoline,

  P(CandR)=P(R)×P(C|R)=0.88×0.28=0.2464

Probability for the customer paid with Credit card and purchased midgrade gas,

  P(CandM)=P(M)×P(C|M)=0.02×0.34=0.0068

Probability for the customer paid with Credit card and purchased premium gas,

  P(CandP)=P(P)×P(C|P)=0.10×0.42=0.0420

Since the vehicles cannot be filled up with two types of gas at same time.

Apply the addition rule for mutually exclusive events:

  P(C)=P(CandR)+P(CandM)+P(CandP)=0.2464+0.0068+0.0420=0.2952

Thus,

Probability for customer paid with a Credit card is 0.2952.

(c)

To determine

Conditional probability for the customer paid with credit card bought premium gas.

(c)

Expert Solution
Check Mark

Answer to Problem 81E

Probability that the customer paid with credit card bought premium gas is approx. 0.1423.

Explanation of Solution

Given information:

Fuel purchased by automobile drivers:

88% bought regular gasoline.

2% bought midgrade gas

10% bought premium gas

Customers paid with Credit card:

28% bought regular gasoline.

34% bought midgrade gas

42% bought premium gas

Calculations:

According to the conditional probability,

  P(B|A)=P(AB)P(A)=P(AandB)P(A)

From Part (b),

We have

Probability for the customer paid with a credit card,

  P(C)=0.2952

Probability for the customer paid with credit card and purchased premium gas,

  P(CandP)=0.0420

Apply the conditional probability:

  P(P|C)=P(CandP)P(C)=0.04200.2952=352460.1423

Thus,

The probability for customer paid with credit card purchased premium gas is approx. 0.1423.

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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