A mechanic sells a brand of automobile tire that has a life expectancy that is normally distributed, with a mean life of 34 , 000 miles and a standard deviation of 2500 miles. He wants to give a guarantee for free replacement of tires that don't wear well. How should he word his guarantee if he is willing to replace approximately 10% of the tires?

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A1peakenbe A1peakenbe · Answer 1 · 2019-07-31T10:12:39+02:00

Answer:

He should word his statement as "Free replacement for tires that wear before 30875 miles."

Step-by-step explanation:

If he is willing to replace 10% of tires he should find the life of tires that gives an area of 10% in the normal distribution graph.

Now for 10% of area standard normal deviate Z can be obtained from normal distribution table

Using normal distribution table for 10% area we have Z = -1.28

Thus we have [tex]Z=\frac{X-\overline{X}}{\sigma }\\\\\therefore X=\sigma Z+\overline{X}[/tex]

Applying given values we get

[tex]X=-1.28\times 2500+34000\\\\X=30875miles[/tex]

raphealnwobi raphealnwobi · Answer 2 · 2022-03-28T12:23:21+02:00

The mechanic would replace the tires that spoil without covering up to 30800 miles.

Z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:

z = (raw score - mean) / standard deviation

For a mean of 34000 miles and standard deviation of 2500. The probability of 10% correspond with a z score of -1.28. Hence:

-1.28 = (x - 34000)/2500

x = 30800

The mechanic would replace the tires that spoil without covering up to 30800 miles.

Find out more on z score at: https://brainly.com/question/25638875