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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Question
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Chapter 5, Problem R5.4RE

(a)

To determine

Probability for the vehicle is a crossover.

(a)

Expert Solution
Check Mark

Answer to Problem R5.4RE

Probability that the vehicle is a crossover is 0.24.

Explanation of Solution

Given information:

Probability distribution for the type of vehicle chosen:

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 5, Problem R5.4RE , additional homework tip  1

Calculations:

Let the missing probability be x.

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 5, Problem R5.4RE , additional homework tip  2

For a valid probability model:

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

Now,

From the table,

Sum of all probabilities needs to be equal to 1.

  0.46+0.15+0.10+x+0.05=1

Combine like terms:

  0.76+x=1

Subtract 0.76 from both sides:

  x=0.24

Thus,

The probability that a vehicle is a crossover is 0.24.

(b)

To determine

Probability for the vehicle is not an SUV or minivan.

(b)

Expert Solution
Check Mark

Answer to Problem R5.4RE

Probability that the vehicle is not an SUV or minivan is 0.85.

Explanation of Solution

Given information:

Probability distribution for the type of vehicle chosen:

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 5, Problem R5.4RE , additional homework tip  3

Calculations:

From Part (a) result,

Table becomes:

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 5, Problem R5.4RE , additional homework tip  4

It is understood that

If a vehicle is a minivan or an SUV, then it needs to be a Passenger car, Pickup truck or Crossover.

Assume that the vehicle cannot belong to more than one vehicle type.

Apply addition rule for mutually exclusive events:

  P(notanSUVoraminivan)=P(Passengercar,pickuptruckorcrossover)=P(Passengercar)+P(Pickuptruck)+P(Crossover)=0.46+0.15+0.24=0.85

Thus,

Probability for the vehicle is not an SUV or minivan is 0.85.

(c)

To determine

Probability for vehicle not a passenger car is a pickup truck.

(c)

Expert Solution
Check Mark

Answer to Problem R5.4RE

Probability that vehicle not a passenger car is a pickup truck is approx. 0.2778.

Explanation of Solution

Given information:

Probability distribution for the type of vehicle chosen:

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 5, Problem R5.4RE , additional homework tip  5

Calculations:

Conditional probability definition:

  P(B|A)=P(AB)P(A)=P(AandB)P(A)

Complement rule:

  P(Ac)=P(notA)=1P(A)

From Part (a) result,

Table becomes:

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 5, Problem R5.4RE , additional homework tip  6

Now,

The corresponding probabilities from the table:

Probability for a Passenger car,

  P(Passengercar)=0.46

Probability for a pickup truck,

  P(Pickuptruck)=0.15

Use complement rule:

  P(NotaPassengercar)=1P(Passengercar)=10.46=0.54

From the definition of conditional probability:

Thus,

Probability for vehicle not a passenger car is a pickup truck is approx. 0.2778.

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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