A mechanic sells a brand of automobile tire that has a lfe expectancy that is no

Question

A mechanic sells a brand of automobile tire that has a lfe expectancy that is normally distributed, with a mean life of 35,000 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? Click to view page 1of the table. Click to view page 2 of the table. Tires that wear out by miles will be replaced free of charge. Round to the nearest mile as needed.)

Explanation / Answer

let the guarantee shall be for A miles

Mean = 35000 miles

Standard deviation = 2500 miles

P(X < A) = 0.1

P(X < A) = P(Z < (A - mean)/standard deviation) = 0.1

P(Z < (A - 35000)/2500) = 0.1

From standard normal distribution table,

(A - 35000)/2500 = -1.28

A = 31800 miles

Tires that wear out by 31800 miles will be replaced free of charge

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