Only parts b, c, and d The average number of miles (in thousands) that acar's tire will function before needing replacement is72 and the standard deviation is 18. Suppose that 50randomly selected tires are tested. Round allanswers to 4 decimal places where possible andassume a normal distribution.a. What is the distribution of X? X - N( 7218b. What is the distribution of æ? a - NGc. If a randomly selected individual tire is tested,find the probability that the number of miles(in thousands) before it will need replacementis between 73.9 and 76.9.d. For the 50 tires tested, find the probabilitythat the average miles (in thousands) beforeneed of replacement is between 73.9 and76.9.

Name: Only parts b, c, and d The average number of miles (in thousands) that
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Description: Only parts b, c, and d The average number of miles (in thousands) that acar's tire will function before needing replacement is72 and the standard deviation is 18. Suppose that 50randomly selected tires are tested. Round allanswers to 4 decimal places

b. From the given information, X~N72,182Therefore,X~N72,18502 c. The required probability will be…

Answered: The average number of miles (in…

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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Only parts b, c, and d

The average number of miles (in thousands) that a
car's tire will function before needing replacement is
72 and the standard deviation is 18. Suppose that 50
randomly selected tires are tested. Round all
answers to 4 decimal places where possible and
assume a normal distribution.
a. What is the distribution of X? X - N( 72
18
b. What is the distribution of æ? a - NG
c. If a randomly selected individual tire is tested,
find the probability that the number of miles
(in thousands) before it will need replacement
is between 73.9 and 76.9.
d. For the 50 tires tested, find the probability
that the average miles (in thousands) before
need of replacement is between 73.9 and
76.9.

Expert Solution

Step 1

b. From the given information,

$X ~ N (72, 18^{2}) T h e r e f o r e, X ~ N (72, {(\frac{18}{\sqrt{50}})}^{2})$

c. The required probability will be given by as follows:

$P (73.9 < X < 76.9) = P (\frac{73.9 - μ}{σ} < \frac{X - μ}{σ} < \frac{76.9 - μ}{σ}) = P (\frac{73.9 - 72}{18} < Z < \frac{76.9 - 72}{18}) = P (0.11 < Z < 0.27) = P (Z < 0.27) - P (Z < 0.11) = 0.6064 - 0.5438 = 0.0626$

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