Question

The average lifespan of a set of tires is 38,000 miles, with a standard deviation of 1500 miles. What is the probability that the lifespan of a set of tires will be between 32,000 miles and 44,000 miles

Answer 1

`P(32000<X<44000)=P((32000-\mu)/(\sigma)<(X-\mu)/(\sigma)<(44000-\mu)/(\sigma))=P((32000-38000)/(1500)<Z<(44000-38000)/(1500))=P(-4<z<4)`

And we can find this probability with this difference:

`P(-4<z<4)=P(z<4)-P(z<-4)`

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

`P(-4<z<4)=P(z<4)-P(z<-4)=0.999968-0.0000317=0.999937`

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 lifespan of a population, and for this case we know the following parameters

Where
`\mu=65.5` and
`\sigma=2.6`

We are interested on this probability

`P(32000<X<44000)`

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

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

If we apply this formula to our probability we got this:

`P(32000<X<44000)=P((32000-\mu)/(\sigma)<(X-\mu)/(\sigma)<(44000-\mu)/(\sigma))=P((32000-38000)/(1500)<Z<(44000-38000)/(1500))=P(-4<z<4)`

And we can find this probability with this difference:

`P(-4<z<4)=P(z<4)-P(z<-4)`

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

`P(-4<z<4)=P(z<4)-P(z<-4)=0.999968-0.0000317=0.999937`

Answer 2

100% probability that the lifespan of a set of tires will be between 32,000 miles and 44,000 miles

Explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

`\mu = 38000, \sigma = 1500`

What is the probability that the lifespan of a set of tires will be between 32,000 miles and 44,000 miles

This is the pvalue of X when X = 44000 subtracted by the pvalue of Z when X = 32000. So

X = 44000

`Z = (X - \mu)/(\sigma)`

`Z = (44000 - 38000)/(1500)`

`Z = 4`

`Z = 4` has a pvalue of 1.

X = 32000

`Z = (X - \mu)/(\sigma)`

`Z = (32000 - 38000)/(1500)`

`Z = -4`

`Z = -4` has a pvalue of 0.

1 - 0 = 100%

100% probability that the lifespan of a set of tires will be between 32,000 miles and 44,000 miles