The frequency distribution was obtained using a class width of 0.5 for data on cigarette tax rates. Use the frequency distribution to approximate the population mean and populationstandard deviation Compare these results to the actual mean µ = $1.667 and standard deviation o=$1.073.Click the icon to view the frequency distribution for the tax rates.The population mean is $(Round to three decimal places as needed)The population standard deviation is $(Round to three decimal places as needed.)Compare these results to the values found using the actual data.OA. The grouped values are both slightly larger.OB. The grouped values are both slightly smaller.OC. The grouped mean is slightly larger, while the grouped standard deviation is slightly smaller.OD. The grouped mean is slightly smaller, while the grouped standard deviation is slightly larger.Frequency distribution of cigarette tax ratesTax Rate0.00 0.490.50-0.991.00-1.491.50-1.992.00-2.492.50-2.993.00-3.493.50-3.994.00-4.49Frequency5158666332

Given,The actual mean, μ = 1.667The actual standard deviation, σ = 1.073Frequency distribution table…

Answered: The frequency distribution was obtained…

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 scores and their percent of the final grade for a statistics student are given. What is the student's weighted mean score? Score Percent of final grade Homework 81 20 Quiz Quiz 90 15 92 15 Project Final Exam 96 20 91 30 The student's weighted mean score is (Simplify your answer. Round to two decimal places as needed.)
Automobiles traveling on a road with a posted speed limit of 65 miles per hour are checked for speed by a state police radar system. The following is a frequency distribution of speeds. Speed (miles per hour) Frequency 45 up to 55 70 55 up to 65 360 65 up to 75 250 75 up to 85 110 (See the Excel Data File.) The mean speed of the automobiles traveling on this road is the closest to ________.
The amount of caffeine in a sample of five-ounce servings of brewed coffee is 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. W Click the icon to view the histogram. .. Complete the table. Round values to the nearest tenth as needed. Graph/chart f Midpoint x xf 70.5 Ay 30- 25- 24 20- |14 15- 10- 17 5- X 48.5 70.5 92.5 114.5 136.5 158.5
A fitness instructor measured the heart rates of the participants in a yoga class at the conclusion of the class. The data is summarized in the histogram below. There were fifteen people who participated in the class between the ages of 25 and 45. 90 100 110 120 130 140 150 160 Beats per minute (bpm) What percent of the participants had a heart rate between 130 and 150 bpm? 4. 2. Frequency
Assume that the readings on the thermometers are normally distributed with a mean of 0° and standard deviation of 1.00°C. A thermometer is randomly selected and tested. Draw a sketch and find the temperature reading corresponding to Pa1, the 91st percentile. This is the temperature reading separating the bottom 91% from the top 9%. Click to view page 1 of the table Click to view page 2 of the table. Which graph represents Pa,? Choose the correct graph below. OD. Oc. OB. OA. Ay The temperature for P91 is approximately . (Round to two decimal places as needed.) XLSX Click to select vour answerls). 33283.xls Exit Honorlock 22 MacBook Air 吕0 000 000 F4 F3 F6 F7 F8 F9 FIC #3 2$ * 3 4 5 6 7 8 E R Y U
Find the mean and median from the frequency table below.If necessary, round your answer to 2 decimal places. data frequency 29 4 30 5 32 2 35 3 36 16 Mean= Median=
Find the mean of the data summarized in the given frequency distribution. Compare the computed mean to the actual mean of 52.5 degrees. Low Temperature (o F) 40 - 44 45 - 49 50 - 54 55 - 59 60 – 64 D Frequency 2 9. ... The mean of the frequency distribution is degrees. (Round to the nearest tenth as needed.) Which of the following best describes the relationship between the computed mean and the actual mean? O A. The computed mean is close to the actual mean because the difference between the means is more than 5% of the actual mean. B. The computed mean is not close to the actual mean because the difference between the means is less than 5% of the actual mean. C. The computed mean is not close to the actual mean because the difference between the means is more than 5% of the actual mean. D. The computed mean is close to the actual mean because the difference between the means is less than 5% of the actual mean.
Use the data in the following table, which lists carbon dioxide concentrations (in parts per million) for each year from 1960 to 2009, with projected values used for the last four years. After finding the values of the mean and range for each decade, find the values of x, and R. Also find the values of LCL and UCL for an R chart, and find the values of LCL and UCL for an x chart. 1960s 318.4 319.5 321.8 322.9 324.8 325.4 326.9 328.5 330.3 331.4 1970s 333.4 334.6 335.5 337.1 339.3 340.4 341.4 343.1 344.7 346.5 1980s 348.1 348.8 351.2 352.1 353.8 355.4 356.9 358.5 359.8 360.6 1990s 361.9 363.6 365.8 366.7 368.6 369.2 371.7 372.4 373.5 375.2 2000s 377.6 378.7 379.6 381.8 382.8 384.2 386.3 387.6 388.7 390.5 x=enter your response here (Round to three decimal places as needed.) Part 2 R=enter your response here (Round to two decimal places as needed.) Compute the values of LCL and UCL for an R chart. LCL=enter your response here, UCL=enter your response here (Round to…
Assume that the readings on the thermometers are normally distributed with a mean of 0° and standard deviation of 1.00°C. A thermometer is randomly selected and tested. Draw a sketch and find the temperature reading corresponding to P34, the 84th percentile. This is the temperature reading separating the bottom 84% from the top 16%. Click to view page 1 of the table. Click to view page 2 of the table. Which graph represents P ? Choose the correct graph below. 84 A. В. OC. D. Ay X The temperature for P84 is approximately (Round to two decimal places as needed.)
The table available below shows the weights (in pounds) for a sample of vacuum cleaners. The weights are classified according to vacuum cleaner type. At a = 0.10, can you conclude that at least one mean vacuum cleaner weight is different from the others? E Click the icon to view the vacuum cleaner weight data. Let PBu, PBLU. and HTC represent the mean weights for bagged upright, bagless upright, and top canister vacuums respectively. What are the hypotheses for this test? O A. Ho: Peu = HBLU "HTC Ha: HBu * HBLU HTC Weight of Vacuum Cleaners by Type B. Ho: At least one of the means is different. Hg: PBu = HBLU =HTC O C. Ho: HBU PBLU HTC Bagged upright 21 20 18 15 23 24 16 24 Bagless upright Top canister 17 18 20 18 Hg: HBu = HBLU =HTC 23 22 24 26 20 23 O D. Ho: HBU HBLU FHTC Ha: At least one of the means is different. What is the test statistic? Print Done F= (Round to two decimal places as needed.) What is the P-value? P-value = (Round to three decimal places as needed.) What is the…

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