Assume that a randomly selected subject is given a bone density test. Bone density test scores are normally distributed with a mean of 0 and a standard deviation of 1. Draw agraph and find P, the 11th percentile. This is the bone density score separating the bottom 11% from the top 89%.Which graph represents P ? Choose the correct graph below.O A.O B.OD.P11P11P11P11The bone density score corresponding to P1 is(Round to two decimal places as needed.)Click to select your answer(s).APRtvMacBook AirF9F10LLA20 F3F6F8F4F5F1F2&#378.%24

Obtain the bone density score separating the bottom 11% from the top 89%. The bone density score…

Answered: Assume that a randomly selected subject…

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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Use the standard normal table to find the z-score that corresponds to the given percentile. If the area is not in the table, use the entry closest to the area. If the area is halfway between two entries, use the Z-score halfway between the corresponding Z-scores. If convenient, use technology to find the z-score. P8 Click to view page 1 of the table. Click to view page 2 of the table. The z-score that corresponds to P3 is. 8. (Round to two decimal places as needed.)
Use the standard normal table to find the z-score that corresponds to the given percentile. If the area is not in the table, use the entry closest to the area. If the area is halfway between two entries, use the z-score halfway between the corresponding z-scores. If convenient, use technology to find the z-score. P78 Click to view page 1 of the table. Click to view page 2 of the table. The z-score that corresponds to P78 is (Round to two decimal places as needed.) C
Use the standard normal table to find the z-score that corresponds to the given percentile. If the area is not in the table, use the entry closest to the area. If the area is halfway between two entries, use the z-score halfway between the corresponding z-scores. If convenient, use technology to find the z-score. P80 Click to view page 1 of the table. Click to view page 2 of the table. The z-score that corresponds to P80 is (Round to two decimal places as needed.) Enter your answer in the answer box. of 1 at
One graph in the figure represents a normal distribution with mean u = 7 and standard deviation o = 1. The other graph represents a normal distribution with mean u = 12 and standard deviation o = 1. Determine which graph is which and explain how you know. 12 Choose the correct answer below. O A. Graph A has a mean of u = 12 and graph B has a mean of u =7 because larger mean shifts the graph to the left. O B. Graph A has a mean of u = 12 and graph B has a mean of u =7 because a larger mean shifts the graph to the right. O C. Graph A has a mean of u = 7 and graph B has a mean of u = 12 because a larger mean shifts the graph to the left. O D. Graph A has a mean of u = 7 and graph B has a mean of u = 12 because a larger mean shifts the graph to the right.
The true average diameter of ball bearings of a certain type is supposed to be 0.5 in. A one-sample t test will be carried out to see whether this is the case. What conclusion is appropriate in each of the following situations? In USE SALT (a) n = 19, t = 1.5, a = 0.05 Reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.5 in. Reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in. Do not reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.5 in. O Do not reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in. (b) n = 19, t = -1.5, a = 0.05 O Reject the null hypothesis. There is sufficient evidence that the true diameter differs from 0.5 in. Reject the nulI hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in. Do not reject the null hypothesis. There is sufficient evidence…
The number of hospital beds in a sample of 20 hospitals is shown below. Construct a frequency distribution and a frequency histogram for the data set using 5 classes. Describe the shape of the histogram as symmetric, uniform, negatively skewed, positively skewed, or none of these. 166 159 132 125 185 163 165 216 145 135 D 199 210 152 250 262 241 285 139 205 179 Construct a frequency distribution for the data set using 5 classes. (Type whole numbers.) Class Frequency 125-157 158–190 191-223 224-256 257-289 Construct a frequency histogram for the data set using the frequency distribution. Choose the correct answer below. O A. OB. О с. OD. Hospital Beds Hospital Beds Hospital Beds Hospital Beds 8- 8- * 6- - 2- E 2+ 0+ 108 174 240 306 0+ 108 174 240 306 0+ 108 174 240 306 0+ 108 174 240 306 Numhar of harde Numhar of hade Numher nf hede Numher of hede Aou anbai Frequency Frequency
One graph in the figure represents a normal distribution with mean u= 15 and standard deviation o = 2. The other graph represents a normal distribution with mean u=11 and standard deviation o = 2. Determine which graph is which and explain how you know. A B 11 15 ... Choose the correct answer below. O A. Graph A has a mean of u = 15 and graph B has a mean of u = 11 because a larger mean shifts the graph to the right. O B. Graph A has a mean of u = 11 and graph B has a mean of u= 15 because a larger mean shifts the graph to the left. O C. Graph A has a mean of u = 11 and graph B has a mean of u = 15 because a larger mean shifts the graph to the right. O D. Graph A has a mean of u = 15 and graph B has a mean of u = 11 because a larger mean shifts the graph to the left.
Use the Standard Normal Table to determine the z-score that corresponds to the 33rd percentile A) -0.44 b.-0.33 C. 0.33 d.0.44 Draw a graph
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 population standard deviation. Compare these results to the actual mean u = $1.727 and standard deviation o = S1.041. E Click the icon to view the frequency distribution for the tax rates. Frequency distribution of cigarette tax rates The population mean is $ Tax Rate 0.00-0.49 0.50-0.99 1.00-1,49 1.50-1.99 2.00-2.49 2.50-2.99 3.00-3.49 3.50-3.99 Frequency D (Round to three decimal places as needed.) 5 15 The population standard deviation is S (Round to three decimal places as needed.) Compare these results to the values found using the actual data. O A. The grouped mean is slightly larger, while the grouped standard deviation is slightly smaller. O B. The grouped values are both slightly larger. O C. The grouped values are both slightly smaller. O D. The grouped mean is slightly smaller, while the grouped standard…
Statistics that describe typical scores are called: Select one: a. Models b. Standard deviations c. Z scores d. Measures of central tendency
Subject: algbera
search A study was conducted that resulted in the following relative frequency histogram. Determine whether or not the histogram indicates that a normal distribution could be used as a model for the variable. Yooo Choose the correct answer below. OA. The histogram is not bell-shaped, so a normal distribution could be used as a model for the variable. OB. The histogram is not bell-shaped, so a normal distribution could not be used as a model for the variable. OC. The histogram is bell-shaped, so a normal distribution could not be used as a model for the variable. OD. The histogram is bell-shaped, so a normal distribution could be used as a model for the variable. ly 144 is 4+ 2000 3000 4000 Q Q G 5000 prt sc delete

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