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Chapter 6: Problem 44

The transmission on a model of a specific car has a warranty for 40,000 miles. It is known that the life of such a transmission has a normal distribution with a mean of 72,000 miles and a standard deviation of 13,000 miles. a. What percentage of the transmissions will fail before the end of the warranty period? b. What percentage of the transmissions will be good for more than 100,000 miles?

Short Answer

Expert verified
a) Approximately 6.94% of transmissions will fail before the end of the warranty period. b) Approximately 1.58% of transmissions will be good for more than 100,000 miles.

Step by step solution

01

Calculate the Z-Score for 40,000 miles

The Z-score is calculated by subtracting the mean from the data value and dividing the result by the standard deviation. For a warranty value of 40,000 miles, the calculation is as follows: \(Z = (40,000 - 72,000) / 13,000 = -2.46 \)
02

Find the Percentage Corresponding to the Z-Score

Consult a Z-table or use a calculator with a normal distribution function to find the percentage of data that falls below a Z-score of -2.46. This value is approximately 0.0694, or 6.94%.
03

Calculate the Z-Score for 100,000 miles

Again, subtract the mean from the data value and divide by the standard deviation. For a value of 100,000 miles, the Z-score is: \(Z = (100,000 - 72,000) / 13,000 = 2.15 \)
04

Find the Percentage Corresponding to the Z-Score

Using the Z-table or calculator again, find the percentage of data that falls below a Z-score of 2.15. This value is approximately 0.9842, or 98.42%. However, the question asks for the percentage above 100,000 miles. To find this, subtract the obtained value from 1 (or 100%) because in a normal distribution, all data falls within 100%: \(Percentage = 1 - 0.9842 = 0.0158, or 1.58%\).

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