Question

A mechanic sells a brand of automobile tire that has a life expectancy that is normally distributed, with a mean life of 29,000 miles and a standard deviation od 2400 miles. He wants to give a guarantee for free replacement of tires that don't wear weel. How should he work his guarantee if he is willing to replace approximately 10% of the tires?

Tires that wear out by _____ miles will be replaces free of charge. Round to the nearest mile as needed.

Answer 1

`a=29000 -1.28*2400=25928`

Tires that wear out by 25928 miles will be replaces free of charge

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

`X \sim N(29000,2400)`

Where
`\mu=29000` and
`\sigma=2400`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

He wants to give a guarantee for free replacement of tires that don't wear weel so then we need to find a value a, such that we satisfy this condition:

`P(X>a)=0.90` (a)

`P(X<a)=0.10` (b)

Because we are interested in the lower tail.

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.10 of the area on the left and 0.90 of the area on the right it's z=-1.28. On this case P(Z<-1.28)=0.10 and P(z>-1.28)=0.90

If we use condition (b) from previous we have this:

`P(X<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.1`

`P(z<(a-\mu)/(\sigma))=0.1`

But we know which value of z satisfy the previous equation so then we can do this:

`z=-1.28=<(a-29000)/(2400)`

And if we solve for a we got

`a=29000 -1.28*2400=25928`

So the value of height that separates the bottom 10% of data from the top 90% is 25928 mi.

Tires that wear out by 25928 miles will be replaces free of charge