A mechanic sells a brand of automobile tire that has a life expectancy that is n

Question

A mechanic sells a brand of automobile tire that has a life expectancy that is normally distributed, with a mean life of 25,000 miles and a standard deviation of 2100 miles. He wants togive a guarantee for free replacement of tires that don't wear well. How should he word his guarantee if he is willing to replave approximately 10% if the tires.

miles and a standard deviation of

21002100

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?

Explanation / Answer

mean = 25000 , sd =2100

P(X y) = 0.1
P([X-]/ ([y-]/) = 0.1
P(Z ([Y-]/) = 0.1 = P(Z -1.28155)
[Y-]/ = -1.28155
(Y-25000)/2100 = -1.28155

y = 22308.7

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