BASIC PRACTICE OF STATISTICS >C<
BASIC PRACTICE OF STATISTICS >C<
8th Edition
ISBN: 9781319220280
Author: Moore
Publisher: MAC HIGHER
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 15, Problem 15.41E

(a)

To determine

To explain: The accuracy of the probability obtained in part (b) of Exercise 15.39 using normal approximation when compared to the exact probability in the given problem.

(a)

Expert Solution
Check Mark

Answer to Problem 15.41E

The probability obtained in part (b) of Exercise 15.39 using normal approximation is under estimated compared with exact answer.

Explanation of Solution

Given info:

The probability that the Joe’s average winnings between 0.5 and 0.7 is 0.4961. The number of bets is 14,000.

Justification:

Here, the exact probability that the Joe’s average winning between 0.5 and 0.7 is 0.4961. From the part (b) of Exercise 15.39 using normal approximation the Joe’s average winnings is between 0.5 and 0.7 is 0.4648.

From the above two probabilities, it can be observed that the probability obtained by using normal approximation is less than the exact probability. Thus, the probability obtained in part (b) of Exercise 15.39 by using normal approximation is under estimated.

(b)

To determine

To explain: The accuracy of the probability of x¯ is between 0.5 and 0.7 using normal approximation when compared to the exact probability in the given problem.

(b)

Expert Solution
Check Mark

Answer to Problem 15.41E

The probability of x¯ is between 0.5 and 0.7 using normal approximation is less than the exact probability.

Explanation of Solution

Given info:

The data shows that the probability that the Joe’s average winnings are between 0.5 and 0.7 is 0.4048. The number of bets is 3,500.

Calculation:

From the part (a) of the Exercise 15.39 the population mean is μ=0.60 and standard deviation is σ=18.96 .

Central limit theorem:

It states that as the sample size n becomes large the sampling distribution of x¯ follows normal distribution with mean μ and standard deviation σn .

Mean of the Sampling distribution of x¯ :

The sample mean x¯ of sample size n has mean μ . Thus, the mean of the sampling distribution is 0.60.

Standard deviation of the Sampling distribution of x¯ :

The standard deviation can be calculated by using the formula σn .

Substitute 18.96 for σ and 3,500 for n

σn=18.963,500=18.9659.1608=0.3205

Thus, the standard deviation of x¯ is 0.3205.

Probability using normal approximation:

P(0.5<x¯<0.7)=P(0.50.618.963,500<x¯μσn<0.70.618.963,500)=P(0.10.3205<Z<0.10.3205)=P(0.31<Z<0.31)=P(Z<0.31)P(Z<0.31)

                            =0.62170.3783[FromTableA:StandardNormalCumulativeProportions]=0.2434

Thus, the probability of x¯ is between 0.5 and 0.7 using normal approximation is 0.2434.

From the given problem the probability that the Joe’s average winnings are between 0.5 and 0.7 is 0.4048. From the above results it can be observed that probability 0.4048 is greater than 0.2434.

Hence, the probability of x¯ is between 0.5 and 0.7 using normal approximation is less than the exact probability.

(c)

To determine

To explain: The accuracy of the probability of x¯ is between 0.5 and 0.7 using normal approximation when compared to the exact probability in the given problem.

(c)

Expert Solution
Check Mark

Answer to Problem 15.41E

The probability of x¯ is between 0.5 and 0.7 using normal approximation is less than the exact probability.

Explanation of Solution

Given info:

The data shows that the probability that the Joe’s average winnings are between 0.5 and 0.7 is 0.9629. The number of bets is 150,000.

Calculation:

Mean of the Sampling distribution of x¯ :

The sample mean x¯ of sample size n has mean μ . Thus, the mean of the sampling distribution is 0.60.

Standard deviation of the Sampling distribution of x¯ :

The standard deviation can be calculated by using the formula σn .

Substitute 18.96 for σ and 150,000 for n

σn=18.96150,000=18.96387.2983=0.0490

Thus, the standard deviation of x¯ is 0.0490.

Probability using normal approximation:

P(0.5<x¯<0.7)=P(0.50.618.96150,000<x¯μσn<0.70.618.96150,000)=P(0.10.0490<Z<0.10.0490)=P(2.04<Z<2.04)=P(Z<2.04)P(Z<2.04)

                            =0.97930.0207[FromTableA:StandardNormalCumulativeProportions]=0.9586

Thus, the probability of x¯ is between 0.5 and 0.7 using normal approximation is 0.9586.

From the given problem the probability that the Joe’s average winnings are between 0.5 and 0.7 is 0.9629. From the above results it can be observed that probability 0.9586 is less than 0.9629.

Hence, the probability of x¯ is between 0.5 and 0.7 using normal approximation is less than the exact probability.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Knowledge Booster
Background pattern image
Statistics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
MATLAB: An Introduction with Applications
Statistics
ISBN:9781119256830
Author:Amos Gilat
Publisher:John Wiley & Sons Inc
Text book image
Probability and Statistics for Engineering and th...
Statistics
ISBN:9781305251809
Author:Jay L. Devore
Publisher:Cengage Learning
Text book image
Statistics for The Behavioral Sciences (MindTap C...
Statistics
ISBN:9781305504912
Author:Frederick J Gravetter, Larry B. Wallnau
Publisher:Cengage Learning
Text book image
Elementary Statistics: Picturing the World (7th E...
Statistics
ISBN:9780134683416
Author:Ron Larson, Betsy Farber
Publisher:PEARSON
Text book image
The Basic Practice of Statistics
Statistics
ISBN:9781319042578
Author:David S. Moore, William I. Notz, Michael A. Fligner
Publisher:W. H. Freeman
Text book image
Introduction to the Practice of Statistics
Statistics
ISBN:9781319013387
Author:David S. Moore, George P. McCabe, Bruce A. Craig
Publisher:W. H. Freeman
Statistics 4.1 Point Estimators; Author: Dr. Jack L. Jackson II;https://www.youtube.com/watch?v=2MrI0J8XCEE;License: Standard YouTube License, CC-BY
Statistics 101: Point Estimators; Author: Brandon Foltz;https://www.youtube.com/watch?v=4v41z3HwLaM;License: Standard YouTube License, CC-BY
Central limit theorem; Author: 365 Data Science;https://www.youtube.com/watch?v=b5xQmk9veZ4;License: Standard YouTube License, CC-BY
Point Estimate Definition & Example; Author: Prof. Essa;https://www.youtube.com/watch?v=OTVwtvQmSn0;License: Standard Youtube License
Point Estimation; Author: Vamsidhar Ambatipudi;https://www.youtube.com/watch?v=flqhlM2bZWc;License: Standard Youtube License