Question Help ▼ Determine whether the graph shown could represent a variable with a normal distribution. Explain your reasoning. If the graph appears to represent a normal distribution, estimate the mean and standard deviation. Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. and its standard deviation is approximately O A. The graph could represent a variable with a normal distribution because the curve is symmetric and bell-shaped. Its mean is appraximately (Round to one decimal place as needed.) O B. The graph could not represent a variable with a normal distribution because the curve has two modes. O C. The graph could not represent a variable with a normal distribution because the graph is skewed to the right. O D. The graph could not represent a variable with a normal distribution because the curve crosses the x-axis. O E. The graph could not represent a variable with a normal distribution because the curve is constant O F. The graph could not represent a variable with a normal distribution because the graph is skewed to the left

Question Help ▼ Determine whether the graph shown could represent a variable with a normal distribution. Explain your reasoning. If the graph appears to represent a normal distribution, estimate the mean and standard deviation. Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. and its standard deviation is approximately O A. The graph could represent a variable with a normal distribution because the curve is symmetric and bell-shaped. Its mean is appraximately (Round to one decimal place as needed.) O B. The graph could not represent a variable with a normal distribution because the curve has two modes. O C. The graph could not represent a variable with a normal distribution because the graph is skewed to the right. O D. The graph could not represent a variable with a normal distribution because the curve crosses the x-axis. O E. The graph could not represent a variable with a normal distribution because the curve is constant O F. The graph could not represent a variable with a normal distribution because the graph is skewed to the left

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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Psychology students in the Izmir University of Economics completed the Dental Anxiety Scale questionnaire in 2020. In this study, scores on the scale range from 0 (no anxiety) to 20 (extreme anxiety). The mean score was 14 and the standard deviation was 3.5. Assume that the distribution of all scores on the Dental Aroxiety Scale is normal with u= 14 and g=35. Complete parts a through c A Click here to view a table of areas under the standardized normal curve O Table of Normal Curve Areas a. Suppose you score a 18 on the Dental Anxiety Scale. Find the z-value for this score. z= (Round to the nearest hundredth as needed ) b. Find the probability that someone scores between 13 and 18 on the Dental Anxiety Scale P(13sxs 18) =D (Round to four decimal places as needed) c. Find the probability that someone scores above 17 on the Dental Anxiety Scale P(x> 17)=(Round to four decimal places as needed.) Enter your answer in each of the answer boxes. |8 目國 100 |8 案 |8 業可 0 | 会效到商同我目 |
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. 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- 92.5 26 25- 114.5 136.5 20- 158.5 15- 12 Ef = Exf = 10- Find the mean of the data set. 5- X = (Round to the nearest tenth as needed.) 48.5 70.5 92.5 114.5 136.5 158.5 Complete the table. Round values to the nearest tenth as needed. Midpoint x (x-x)? (x-x)?f X-X 70.5 Print Done 92.5 114.5 136.5 158.5 (x-x)?r= ] of
instructor gives a 100-point examination in which the grades are normally distributed. The mean is 66 and the standard deviation is 5. If there are 5% A's and 5% F's, 15% B's and 15% D's, and 60% C's, find the scores that divide the distribution into those categories. Round the cutoff scores to the nearest whole number. Round intermediate z-value calculations to 2 decimal places. A:_- B: C: D: --- F: _-_
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
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 ..... CVneights = 71.2% (Round to the nearest tenth as needed.) Heights Weights 68 181 77 169 72 167 72 207 71 177 70 175 78 185 69 186 67 221 70 216 67 209 73 206
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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
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Augusta National Golf Course. Earlier in this section, we found that the population standard deviation of the lengths of the holes at the Augusta National Golf Club is 132.0 yd. In this context, is the number 132.0 a parameter or a statistic? Explain your answer.
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