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A mechanic sells a brand of automobile tire that has a life expectancy that is normally​ distributed, with a mean life of 27 comma 000 miles and a standard deviation of 2100 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?

Respuesta :

Answer:

Tires that wear out by approximately 24307.8 miles will be replaces free of charge.        

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 27,000 miles

Standard Deviation, σ = 2100  miles

We are given that the distribution of life expectancy is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

We have to find the value of x such that the probability is 0.10

[tex]P( X < x) = P( z < \displaystyle\frac{x - 27000}{2100})=0.10[/tex]  

Calculation the value from standard normal z table, we have,  

[tex]\displaystyle\frac{x - 27000}{2100} = -1.282\\\\x = 24307.8[/tex]

Tires that wear out by approximately 24307.8 miles will be replaces free of charge.

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