Illustration 17.11. You are given below a frequency distribution.Fit a normal curve to the data.Mid value 412 417422427 432 437 442447 452 457Frequency6.34132 179 2181831466930CS Scanned with CamScanner

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Answered: Ilustration 17.11. You are given below…

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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Find the mean of the data summarized in the given frequency distribution. Compare the computed mean to the actual mean of 51.1 miles per hour. Speed (miles per hour) Frequency 42-45 23 The mean of the frequency distribution is (Round to the nearest tenth as needed.) 46 - 49 15 miles per hour. 50-53 54-57 58-61 6 4 Which of the following best discribes the relationship between the computed mean and the actual mean? O A. The computed mean is not close to the actual mean because the difference between the means is less than 5%. OB. The computed mean is close to the actual mean because the difference between the means is more than 5%. O C. The computed mean is close to the actual mean because the difference between the means is less than 5%. O D. The computed mean is not close to the actual mean because the difference between the means is more than 5%.
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n The gas mileages (in miles per gallon) for 26 cars are shown in the frequency distribution. Approximate the mean of the frequency distribution. Gas Mileage (in miles per gallon) Frequency 8 29-32 33-36 13 37-40 1 41-44 4 ... The approximate mean of the frequency distribution is (Round to one decimal place as needed.)
The gas mileages (in miles per gallon) for 38 cars are shown in the frequency distribution. Approximate the mean of the frequency distribution. The approximate mean of the frequency distribution is (Round to one decimal place as needed.) Gas Mileage (in miles per gallon) 26-30 31-35 36-40 41-45 Frequency 13 14 4 7 0
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Use the Histogram tool to develop a frequency distribution and histogram for the number of months as a customer of the bank in the data below. Compute the relative and cumulative relative frequencies and use a line chart to construct an ogive. Complete the frequency distribution for the months as a customer, and compute the relative and cumulative relative frequencies. (Type integers or decimals rounded to three decimal places as needed.) Bins Frequency Relative Frequency Cumulative Relative Frequency 10 enter your response here enter your response here enter your response here 20 enter your response here enter your response here enter your response here 30 enter your response here enter your response here enter your response here 40 enter your response here enter your response here enter your response here 50 enter your response here enter your response here enter your…
Find the mean for the set of grouped data. The mean is (Round to two decimal places if needed.) Interval 0.5-2.5 2.5-4.5 4.5-6.5 6.5-8.5 Frequency 1 8 5 1
The coefficient of variation CV describes the standard deviation as a percent of the mean. Because it has no units, you can use the coefficient of variation to compare data with different units. Find the coefficient of variation for each sample data set. What can you conclude? Standard deviation CV = • 100% Mean E Click the icon to view the data sets. Data Table C/heights = 5.0 % (Round to the nearest tenth as needed.) Heights Weights CVweights =% (Round to the nearest tenth as needed.) 68 181 77 169 72 167 72 207 71 177 70 175 78 185 69 186 67 221 70 216 67 209 73 206
The numbers of courses taught per semester by a random sample of university professors are shown in the histogram. Make a frequency distribution for the data. Then use the table to estimate the sample mean and the sample standard deviation of the data set. 25- 24 20- 17 16 15 10- 5- Number of courses taught per semester f 1 2 16 24 4 17 The sample mean is x = 2.9 (Round to one decimal place as needed.) The sample standard deviation is s = (Round to one decimal place as needed.) Number of professors
Find the mean of the data summarized in the given frequency distribution. Compare the computed mean to the actual mean of 52.9 degrees. Low Temperature (F) 40-44 45-49 50 - 54 Frequency 1 6 10 The mean of the frequency distribution is (Round to the nearest tenth as needed.) degrees. 55-59 5 60-64 3
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Don't drink and drive: A highway safety council reported that there were 3902 fatalities among drivers in auto accidents in a particular year. Following is a frequency distribution of their ages. Approximate the mean age. Round your answer to one decimal place. Age 11-20 21-30 31-40 41-50 51-60 61-70 Number of Fatalities 331 1524 821 669 426 131 Send data Excel The mean is approximately X S

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Ilustration 17.11. You are given below a frequency distribution. Fit a normal curve to the data. Mid value 412 417 422 427 432 437 442 447 452 457 Frequency 6. 34 132 179 218 183 146 69 30