You work for a consumer watchdog publication and are testing the advertising claims of a tire manufacturer. The manufacturer claims that the life spans of the tires are normally distributed, with a mean of 40,000 miles and a standard deviation of6500 miles. You test 16 tires and get the following life spans. Complete parts (a) through (c) below.47,493 39,585 28,756 36,034 30,952 38,287 42,718 34,078 D24,464 29,573 37,049 36,753 32,401 36,752 34,850 44,188(a) Draw a frequency histogram to display these data. Use five classes. Choose the correct answer below.OA.OB.Oc.Life spans of tiresLife spans of tires.Life spans of tires45,1526,45190526,7545,1905Distance (in miles)Distance (in miles)Distance (in miles)Is it reasonable to assume that the life spans are normally distributed? Why? Choose the correct answer below.O A. No, because the histogram is symmetric and bell-shaped.O B. Yes, because the histogram is neither symmetric nor bell-shaped.O C. Yes, because the histogram is symmetric and bell-shaped.neither symmetric nor bell-shaped.O D. No, because the histogram(b) Find the mean of your sample.The mean is 1. (Round to one decimal place as needed.)Noxt You work for a consumer watchdog publication and are testing the advertising claims of a tire manufacturer. The manufacturer claims that the life spans of the tires are normally distributed, with a mean of 40,000 miles and a standard deviation of6500 miles. You test 16 tires and get the following life spans. Complete parts (a) through (c) below.47,493 39,585 28,756 36,034 30,952 38,287 42,718 34,07824,464 29,573 37,049 36,753 32,401 36,752 34,850 44,188O A. No, because the histogram is symmetric and bell-shaped.O B. Yes, because the histogram is neither symmetric nor bell-shaped.OC. Yes, because the histogram is symmetric and bell-shaped.O D. No, because the histogram is neither symmetric nor bell-shaped.(b) Find the mean of your sample.The mean is . (Roundone decimal place as needed.)Find the standard deviation of your sample.The standard deviation is. (Round to one decimal place as needed.)(c) Compare the mean and standard deviation of your sample with those in the manufacturers claim. Discuss the differences. Choose the corect answer below.O A. The sample mean is greater than the claimed mean, so, on average, the tires in the sample lasted for a longer time. The sample standard deviation is greater than the claimed standard deviation, so the tires in the sample had a greatervariation in life span.O B. The sample mean is less than the claimed mean, so, on average, the tires in the sample lasted for a shorter time. The sample standard deviation is less than the claimed standard desijation, so the tires in the sample had a smallervariation in life span.The sample mean is less than the claimed mean, so, on average, the tires in the sample lasted for a shorter time. The sample standard deviationvariation in life span.Oc.greater than the claimed standard deviation, so the tires in the sample had a greater

Answered: You work for a consumer watchdog…

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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Given the following data: 8,5,6,10,8,3,6 find the A. Median B.Sample Mean C.Sample standard deviation Explain your answers.
Consider a data set containing the following values: 80, 94, 93, 106, 91, 90, 96, 99The mean of the preceding values is 93.625. The deviations from the mean have been calculated as follows: -13.625, 0.375, -0.625, 12.375, -2.625, -3.625, 2.375, 5.375If this is sample data, what is the sample variance? What is the sample standard deviation? If this is population data, what is the population variance? What is the population standard deviation?Suppose the largest value of 106 in the data was misrecorded as 100. If you were to recalculate the variance and standard deviation with the 100 instead of the 106, your new values for the variance and standard deviation would be...
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In the 1992 presidential election, Alaska's 40 election districts averaged 1980 votes per district for President Clinton. The standard deviation was 573. (There are only 40 election districts in Alaska.) The distribution of the votes per district for President Clinton was bell-shaped. Let X = number of votes for President Clinton for an election district. (Source: The World Almanac and Book of Facts) Round all answers except part e. to 4 decimal places.PLEASE ANSWER C, D AND E AND ROUND TO THE 4 DECIMALc. Find the probability that a randomly selected district had fewer than 2062 votes for President Clinton. d. Find the probability that a randomly selected district had between 2052 and 2347 votes for President Clinton. e. Find the first quartile for votes for President Clinton. Round your answer to the nearest whole number.
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