The mean height of an adult giraffe is 17 feet. Suppose thatthe distribution is normally distributed with standarddeviation 1 feet. Let X be the height of a randomly selectedadult giraffe. Round all answers to 4 decimal places wherepossible.a. What is the distribution of X? X - N(b. What is the median giraffe height?ft.c. What is the Z-score for a giraffe that is 20 foot tall?d. What is the probability that a randomly selected giraffewill be shorter than 16.5 feet tall?e. What is the probability that a randomly selected giraffewill be between 17.2 and 18 feet tall?f. The 90th percentile for the height of giraffes isft.

Solution Given That, The mean height of an adult giraffe is 17 feet. Suppose that the distribution…

Answered: the distribution is normally…

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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Solve A through F
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.
The heights of Crepe Myrtle saplings at a local nursery are measured and the data recorded. The data is then plotted and the distribution is found to be bell-shaped. The mean of the set is 67 inches and the standard deviation is 3 inches. A graph of the situation is shown below: 58 61 64 67 70 73 76 a) What percentage of the saplings is more than 70 inches tall? b) What percentage of the saplings is between 61 and 73 inches tall?
ssume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 15 (as on the Wechsler test). Find the following: The Z-value that separates the bottom 18% from the others. The Z-Value for the 95 percentile. What is the IQ score that separates the bottom 18% from the others? The 95 percentile is? A. 113.725 B. -.92 C. 75.325 D. -.915 E. 1.645 F. 86.275 G. 124.675 H. .915
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
A. Which one is greater the mean or the median? Which one is better measured of the center the median or the mean? Show work B. Find the mode C. Find the median D. Find the mean (even though we must round to the one decimal place to make computations easy round you’re mean to the nearest minute)
Please answer B, C, & D.
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
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During an experiment, the following times (in seconds) were recorded. Find the mean. 6.3, 5.4, 6, 3.2, 5.4, 5.5 ... The mean is (Round to one decimal place as needed.) ...
Find the z-score that has 70.54% of the distribution's area to its right. Click to view page 1 of the table. Click to view page 2 of the table. O A. - 0.82 В. - 0.54 С. 0.82 D. 0.54
Find the mean of the data summarized in the given frequency distribution. Compare the computed mean to the actual mean of 57.4 degrees. The mean of the frequency distribution is (Round to the nearest tenth as needed.)

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