2. A specially designed car tire has an average life of 38,500 miles with a standard deviation of 2,500 ms isapproximately normally distributed.(a) Let X = number of miles of a single tire. Find the probability that in the sample above the mean will beless than 38,000 miles. (Use 4 decimals)rox< 38000) =nomacr (IEa, 3,0o, 38,500, 2,50).5442PCXbobnpt.ou vnio(b) Use the software tool below (or any other) to graph and shade the probability from part (a).Do not graph manually. The graph should show the appropriate area from part (a).Tool link: https://mathcracker.com/normal-probability-calculator-sampling-distributions(c) The tire manufacturer wants to set the tire warranty to 39,100 miles. Is this a good decision? What arethe possible consequences to the manufacturer? Explain in detail.

Given: Mean = 38500 Standard deviation = 2500 n = 60 Let X = number of miles of a single tire.

Answered: The manufacturer tests 60 such tires to…

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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