Fast male cross-country runners have 5K times that are approximately Normal as shown in the image. 15 15.5 16 16.5 17 5K time for male cross-country runners (minutes) 17.5 18 What is the mean and standard deviation of this distribution? Mean 18 minutes Standard deviation = 3 minutes Mean 16.5 minutes Standard deviation = 0.5 minutes Mean 16.5 minutes Standard deviation = 1.5 minutes Mean 15 minutes Standard deviation = 0.5 minutes Mean = 0.5 minutes Standard deviation = 16.5 minutes
Fast male cross-country runners have 5K times that are approximately Normal as shown in the image. 15 15.5 16 16.5 17 5K time for male cross-country runners (minutes) 17.5 18 What is the mean and standard deviation of this distribution? Mean 18 minutes Standard deviation = 3 minutes Mean 16.5 minutes Standard deviation = 0.5 minutes Mean 16.5 minutes Standard deviation = 1.5 minutes Mean 15 minutes Standard deviation = 0.5 minutes Mean = 0.5 minutes Standard deviation = 16.5 minutes
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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Question
![Fast male cross-country runners have 5K times that are approximately Normal as shown in the image.
15
15.5
16
16.5
17
5K time for male cross-country runners (minutes)
17.5
18
What is the mean and standard deviation of this
distribution?
O Mean = 18 minutes Standard deviation
Mean = 16.5 minutes Standard deviation = 0.5
minutes
Mean = 16.5 minutes Standard deviation = 1.5
minutes
Mean = 15 minutes Standard deviation = 0.5
minutes
Mean
minutes
=
= : 3 minutes
0.5 minutes Standard deviation
= 16.5](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fec8216c7-9603-414d-880e-1b2c9e8c3a03%2F0b3dd81e-1643-45ef-800f-89725bb5e539%2F9kkcnqa_processed.png&w=3840&q=75)
Transcribed Image Text:Fast male cross-country runners have 5K times that are approximately Normal as shown in the image.
15
15.5
16
16.5
17
5K time for male cross-country runners (minutes)
17.5
18
What is the mean and standard deviation of this
distribution?
O Mean = 18 minutes Standard deviation
Mean = 16.5 minutes Standard deviation = 0.5
minutes
Mean = 16.5 minutes Standard deviation = 1.5
minutes
Mean = 15 minutes Standard deviation = 0.5
minutes
Mean
minutes
=
= : 3 minutes
0.5 minutes Standard deviation
= 16.5
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