Question

The distribution of lifetimes of a particular brand of car tires has a mean of 51,200 miles and a standard deviation of 8,200 miles. Assuming that the distribution of lifetimes is approximately normally distributed and rounding your answers to the nearest thousandth, find the probability that a randomly selected tire lasts: A) Between 55,000 and 65,000 miles B) Less than 48,000 miles C) At least 41,000 miles D) A lifetime that is within 10,000 miles of the mean

Answer 1

a) 0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.

b) 0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.

c) 0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.

d) 0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean

Explanation:

Problems of normally distributed distributions 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 question, we have that:

`\mu = 51200, \sigma = 8200`

Probabilities:

A) Between 55,000 and 65,000 miles

This is the pvalue of Z when X = 65000 subtracted by the pvalue of Z when X = 55000. So

X = 65000

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

`Z = (65000 - 51200)/(8200)`

`Z = 1.68`

`Z = 1.68` has a pvalue of 0.954

X = 55000

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

`Z = (55000 - 51200)/(8200)`

`Z = 0.46`

`Z = 0.46` has a pvalue of 0.677

0.954 - 0.677 = 0.277

0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.

B) Less than 48,000 miles

This is the pvalue of Z when X = 48000. So

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

`Z = (48000 - 51200)/(8200)`

`Z = -0.39`

`Z = -0.39` has a pvalue of 0.348

0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.

C) At least 41,000 miles

This is 1 subtracted by the pvalue of Z when X = 41,000. So

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

`Z = (41000 - 51200)/(8200)`

`Z = -1.24`

`Z = -1.24` has a pvalue of 0.108

1 - 0.108 = 0.892

0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.

D) A lifetime that is within 10,000 miles of the mean

This is the pvalue of Z when X = 51200 + 10000 = 61200 subtracted by the pvalue of Z when X = 51200 - 10000 = 412000. So

X = 61200

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

`Z = (61200 - 51200)/(8200)`

`Z = 1.22`

`Z = 1.22` has a pvalue of 0.889

X = 41200

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

`Z = (41200 - 51200)/(8200)`

`Z = -1.22`

`Z = -1.22` has a pvalue of 0.111

0.889 - 0.111 = 0.778

0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean