grear tire company has produced a new tire with an estimated mean lifetime mileage of 36,500 miles. Management

Question

believes that the standard deviation is 5000 miles and that tire mileage is normally distributed. using a worksheet simulate the miles obtained for a sample of 500 tires.
a) use excel COUNTIF function to determine the number of tires that last longer than 40,000miles. whats ur estimate of the percentage of tires that will exceed 40,000miles?
b)use COUNTIF to find the number of tires that obtain mileage less than32,000 miles then find the number with less than 30,000 miles and the number with less than 28,000 miles
c)if management would like to advertise a tire mileage guarantee such that approximately no more than 10% of the tires would obtain mileage low enough to qualify for the guarantee, what tire mileage considered in part b would u recommend for the guarantee?

Answer 1

Dear cherita,

To make the requested estimates, you need to look up numbers in normal tables, or have a way of calculating them. If you have specific questions about Excel, you should probably ask those separately, because it is not clear that you want help with Excel here.


a. If you use normal tables, they are usually standardized with a mean of 0 and standard deviation of 1. To put numbers for the new tire into standardized form, first subtract the mean from the mileage number under consideration, then divide by the standard deviation. For this part, we are considering tires lasting longer than 40000 miles, so

(40000 - 36500) / 5000
= 3500 / 5000
= 0.7 .

This standardized number tells you how many standard deviations above the mean your selected mileage number is. (A negative number would be below the mean.) The probability that your a tire gets more mileage than this is determined by the area under the normal density curve to the right of this number. (If your normal tables give you the probability less than, rather than greater than the selected number, you will need to subtract from 1 to get the complementary probability.) Thus, using the normal distribution in this way,

P(mileage of tire > 40000 miles) = 0.24196 (to five decimal places).

From this we also conclude that our best estimate of the percentage of tires that exceed 40000 miles is 24.196%. (So, in a sample of 500 tires, you would estimate that around 121 would exceed 40000 miles.)


b. Using the same method as described above we can obtain estimates for other mileage amounts.

P(mileage of tire < 32000) = 0.18406 (to five decimal places),
so our best estimate is that 0.18406 x 500 = 92.030 tires in our sample would get fewer than 32000 miles.

P(mileage of tire < 30000) = 0.09680 (to five decimal places),
so our best estimate is that 0.09680 x 500 = 48.400 tires in our sample would get fewer than 30000 miles.

P(mileage of tire < 28000) = 0.04457 (to five decimal places),
so our best estimate is that 0.04457 x 500 = 22.285 tires in our sample would get fewer than 28000 miles.


c. Notice that in the previous part of this problem, we can see that the answer is 30000 miles, since we determined that 9.68% of tires would not last that far. If you want a more precise number, then you need to work in the opposite direction. Start by finding a 0.10 probability and then determine what number of standard deviations away from the mean correspond to it. Once you do that, you will also need to convert back out of standardized units into mileage of the tires. (If you were work through this you would find that 10% of tires would get fewer than 30092 miles.)

Answer 2

i might ask them first till now doing any artwork. you may visually inspect your tire treads fantastically undemanding. Take a penny and turn it the different way up and positioned it interior the grooves of your tire. It the tread is larger than Lincolns head, you have quite a few tread left and don't need new tires!