Based on all student records at Camford University, students spend an average of 5.80 hours per week playing organized sports. Thepopulation's standard deviation is 3.50 hours per week. Based on a sample of 64 students, Healthy Lifestyles Incorporated (HLI) wouldlike to apply the central limit theorem to make various estimates.Required:a. Compute the standard error of the sampling distribution of sample means. (Round your answer to 2 decimal places.)Standard errorb. What is the chance HLI will find a sample mean between 5 and 6.6 hours? (Round your z and standard error values to 2 decimalplaces. Round your intermediate and final answer to 4 decimal places.)Chancec. Calculate the probability that the sample mean will be between 5.4 and 6.2 hours. (Round your z and standard error values to 2decimal places. Round your intermediate and final answer to 4 decimal places.)Probability d. How strange would it be to obtain a sample mean greater than 7.80 hours?O This is very unlikely.This is very likely.

Since you have posted a question with multiple sub parts, we will provide the solution only to the…

Answered: Based on all student records at Camford…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 50 ounces and a standard deviation of 4 ounces. Use the Standard Deviation Rule, also known as the Empirical Rule (see image below). Do not use normalcdf on your calculator. Suggestion: sketch the distribution in order to answer these questions. a) 95% of the widget weights lie between and b) What percentage of the widget weights lie between 46 and 58 ounces? c) What pagcentage of the widget weights lie below 62 ? % 34% 34% 2.35% 13.5% 2.35% 0.15% 13.5% -35 25 0.15% -1s 1s 25
The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 46 ounces and a standard deviation of 9 ounces. Use the Standard Deviation Rule, also known as the Empirical Rule (see image below). Do not use normalcdf on your calculator. Suggestion: sketch the distribution in order to answer these questions. a) 99.7% of the widget weights lie between b) What percentage of the widget weights lie between 37 and 73 ounces? c) What percentage of the widget weights lie above 28? 0.15% -3s 2.35% -25 13.5% -1s 34% 34% 1s 13.5% 2s 2.35% 3s and 0.15% % %
A bottled water distributor wants to determine whether the mean amount of water contained in 1-gallon bottles purchased from a nationally known water bottling company is actually 1 gallon. You know from the water bottling company specifications that the standard deviation of the amount of water is 0.03 gallon. You select a random sample of 45 bottles, and the mean amount of water per 1-gallon bottle is 0.994 gallon. Complete parts (a) through (d) below. 1.96,-1.96 (Round to two decimal places as needed. Use a comma to separate answers as needed.) What is the final conclusion? A. Fail to reject Ho. There is not sufficient evidence that the mean amount is different from 1.0 gallon. OB. Reject Ho. There is not sufficient evidence to warrant rejection of the claim that the mean amount is different from 1.0 gallon. OC. Reject Ho. There is sufficient evidence to warrant rejection of the claim that the mean amount is different from 1.0 gallon. D. Fail to reject Ho. There is sufficient…

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