A certain brand of automobile tire has a mean life span of 38,000 miles and a standard deviation of 2,250 miles. (Assume the life spans of the tires have a bell-shaped distribution.) Question content area bottom Part 1 The life spans of three randomly selected tires are 33,000 miles, 37,000, miles, and 32,000 miles. Find the z-score that corresponds to each life span. Note: I know how to find Z-Scores. My issue is that the second part wants me to take the results and place them in percentiles. (ex. 16th, 50th, 86th) so 33,000-38,0000/2250= -2.22 37,000-38,000/2250= -0.44 32,000-38,000= -2.67 38,000-2450= 35,550 38,000+2450= 40,450 = 38,000 How do I get the percentiles?
A certain brand of automobile tire has a mean life span of 38,000 miles and a standard deviation of 2,250 miles. (Assume the life spans of the tires have a bell-shaped distribution.) Question content area bottom Part 1 The life spans of three randomly selected tires are 33,000 miles, 37,000, miles, and 32,000 miles. Find the z-score that corresponds to each life span. Note: I know how to find Z-Scores. My issue is that the second part wants me to take the results and place them in percentiles. (ex. 16th, 50th, 86th) so 33,000-38,0000/2250= -2.22 37,000-38,000/2250= -0.44 32,000-38,000= -2.67 38,000-2450= 35,550 38,000+2450= 40,450 = 38,000 How do I get the percentiles?
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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A certain brand of automobile tire has a mean life span of
38,000 miles and a standard deviation of 2,250
miles. (Assume the life spans of the tires have a bell-shaped distribution.)Question content area bottom
Part 1
The life spans of three randomly selected tires are 33,000 miles, 37,000, miles, and 32,000 miles. Find the z-score that corresponds to each life span.
Note: I know how to find Z-Scores. My issue is that the second part wants me to take the results and place them in percentiles. (ex. 16th, 50th, 86th)
so 33,000-38,0000/2250= -2.22
37,000-38,000/2250= -0.44
32,000-38,000= -2.67
38,000-2450= 35,550
38,000+2450= 40,450
= 38,000
How do I get the percentiles?
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Follow-up Question
Explain how
For X = 33000 miles, Z = -2.22
P(Z<-2.22) = 0.0132 Because I dont see it. Divide? Multiply? What. How do you end up with 0.0132
Solution
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