Question

The lifetime (in miles) of bicycle tires produced by a manufacturer is a normally distributed random variable with a mean of 3000 miles and a standard deviation of 150 miles. Marketing research by the manufacturer has indicated that if a tire lasts 3270 or more miles, the customer will definitely purchase another of these tires. What percentage of all tires produced by this company will have a lifetime of 3270 or more miles?

Answer 1

3.59% of all tires produced by this company will have a lifetime of 3270 or more miles

Explanation:

Normal Probability Distribution:

Problems of normal distributions can be 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.

Normally distributed random variable with a mean of 3000 miles and a standard deviation of 150 miles.

This means that
`\mu = 3000, \sigma = 150`

What percentage of all tires produced by this company will have a lifetime of 3270 or more miles?

The proportion is 1 subtracted by the pvalue of Z when X = 3270. So

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

`Z = (3270 - 3000)/(150)`

`Z = 1.8`

`Z = 1.8` has a pvalue of 0.9641

1 - 0.9641 = 0.0359

0.0359*100% = 3.59%

3.59% of all tires produced by this company will have a lifetime of 3270 or more miles