Question

A mechanic sells a brand of automobile tire that has a life expectancy that is normally​ distributed, with a mean life of miles and a standard deviation of 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?

Answer 1

Complete question is;

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?

Answer:

The mechanics worded guarantee that the tyre will be replaced if it lasts less than or equal to 30796 miles.

Step-by-step explanation:

Let X = Life expectancy of the automobile tire

For the normal distribution, we are given;

μ = 34000

σ = 2700

Let y miles be the minimum miles of the tyres which the mechanic guarantees that he will replace if it's last less than that.

Now, since the mechanic is willing to replace approximately 10% of the tires, y would be such that:

P(X ≤ y) = 0.1

Using the z formula, we have;

P[(X - μ)/σ) ≤ [(y - μ)/σ] = 0.1

Thus;

P(Z ≤ ([y - μ]/σ) = 0.1

From z-distribution table the Z-score at 0.1 is approximately -1.2816

This is P(Z ≤ -1.2816)

Thus;

[y - μ]/σ = -1.2816

(y - 34000)/2500 = -1.2816

y - 34000 = -1.2816 × 2500

y - 34000 = -3204

y = 34000 – 3204

y = 30796

So, we conclude that the mechanic will guarantee that the tyre will be replaced if it lasts less than or equal to 30796 miles.