# What should the guaranteed lifetime be if the tire company desires that no more than 5 percent of the tires will fail to meet the guaranteed lifetime? A tire company has developed a new type of...

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### 1 Answer

The best way to look at this is to consider what it means to be "normally distributed."

Normal distribution is defined by the following function:

`f(x) = 1/(sigmasqrt(2pi))e^(-(x-mu)^2/(2sigma^2))`

where `mu` is the mean of the population, `sigma` is the standard deviation of the population, and x is a possible value in the population. The area under the curve between `-oo` and `x` gives the probability that a sample will have a value less than `x`.

To find probability with the above distribution, you would need to integrate. In our case, the distribution uses the following rule because of our given mean of 65,000 and standard deviation of 6500:

`f(x) = 1/(6500*sqrt(2pi))e^(-(x-65000)^2/84500000)`

So, in order to find the the mileage where only 5% of the tires fail, we need to see at which x that the area below the curve from `-oo` to `x` is 0.05. The area is given by the following integral:

`P(x) = 1/(6500sqrt(2pi)) int_-oo^x e^-((tau-65000)^2/84500000) d tau = 0.05`

Don't worry about the `tau` in the equation. It's simply a place-holder for `x` in the integral.

You can solve this integral by first multiplying both sides of the equation by `6500sqrt(2pi)` :

`int_-oo^x e^(-(tau-65000)^2/84500000) d tau = 814.65`

Now, we're going to make a substitution here:

`u = (tau-65000)/sqrt(84500000)`

Looking at the derivative:

`du = 1/sqrt(84500000) d tau`

`sqrt(84500000) du = d tau`

So, we get a much easier integral, after taking out the constant term and adjusting the limits:

`sqrt(84500000) int_-00^((x-65000)/sqrt(84500000)) e^(-u^2) du = 814.65`

Now, we get a tricky result for our integral, the error function (see link below):

`sqrt(84500000) sqrt(pi/4)erf(u) = 814.65`

Where we evaluate the definite integral using the boundaries above:

`sqrt(21125000pi) (erf((x-65000)/sqrt(84500000)) - erf(-oo)) = 814.65`

Now, the error function of `-oo` is simply -1, which is convenient. The other term is not so. After simplifying the above equation by dividing the constant terms and approximating with decimals, we get:

`erf(x/9192.38 - 7.071) + 1 = 0.100`

`erf(x/9192.38 - 7.071) = -0.900`

There is simply no good analytical way to evaluate the error function, so we're going to use the table provided at the link to find at what value of the error function is -0.9 (it's roughly 1.15):

`x/9192.38 - 7.071 =-1.15`

`x/9192.38 = 5.921`

`x =54428.1`

So, in order to make a guarantee that is likely to have only 5% of people that it would apply to, we would have that guarantee be for up to 54428 miles.

Another way to do this is to look at a z table where you find the number of standard deviations below the mean to get 5% of the population (where the table reads 0.95, where z = -1.645 based on the second table below because we will clearly be below the mean). You can then use the following function to find x:

`x = mu - zsigma = 65000 - 1.645*6500 = 54307.5`

These two methods disagree slightly, but they are easily within 1% of ech other. So, either answer is going to be reasonable. Now that I look at your grade level, it's more likely your instructor is going after the 54307.5 miles, so I'd go with that! I hope that helps!

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