The Basic Practice of Statistics
The Basic Practice of Statistics
7th Edition
ISBN: 9781464142536
Author: David S. Moore, William I. Notz, Michael A. Fligner
Publisher: W. H. Freeman
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 15, Problem 15.42E

(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 15.42E

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 15.42E

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 15.42E

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!
Students have asked these similar questions
Solve the following LP problem using the Extreme Point Theorem: Subject to: Maximize Z-6+4y 2+y≤8 2x + y ≤10 2,y20 Solve it using the graphical method. Guidelines for preparation for the teacher's questions: Understand the basics of Linear Programming (LP) 1. Know how to formulate an LP model. 2. Be able to identify decision variables, objective functions, and constraints. Be comfortable with graphical solutions 3. Know how to plot feasible regions and find extreme points. 4. Understand how constraints affect the solution space. Understand the Extreme Point Theorem 5. Know why solutions always occur at extreme points. 6. Be able to explain how optimization changes with different constraints. Think about real-world implications 7. Consider how removing or modifying constraints affects the solution. 8. Be prepared to explain why LP problems are used in business, economics, and operations research.
ged the variance for group 1) Different groups of male stalk-eyed flies were raised on different diets: a high nutrient corn diet vs. a low nutrient cotton wool diet. Investigators wanted to see if diet quality influenced eye-stalk length. They obtained the following data: d Diet Sample Mean Eye-stalk Length Variance in Eye-stalk d size, n (mm) Length (mm²) Corn (group 1) 21 2.05 0.0558 Cotton (group 2) 24 1.54 0.0812 =205-1.54-05T a) Construct a 95% confidence interval for the difference in mean eye-stalk length between the two diets (e.g., use group 1 - group 2).
An article in Business Week discussed the large spread between the federal funds rate and the average credit card rate. The table below is a frequency distribution of the credit card rate charged by the top 100 issuers. Credit Card Rates Credit Card Rate Frequency 18% -23% 19 17% -17.9% 16 16% -16.9% 31 15% -15.9% 26 14% -14.9% Copy Data 8 Step 1 of 2: Calculate the average credit card rate charged by the top 100 issuers based on the frequency distribution. Round your answer to two decimal places.
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.
Similar questions
SEE MORE QUESTIONS
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