Question

The tread life of a particular brand of tire is a random variable best described by a normal distribution with a mean of 60,000 miles and a standard deviation of 2900 miles. What is the probability a particular tire of this brand will last longer than 57,100 miles

Answer 1

Answer:

84.13% probability a particular tire of this brand will last longer than 57,100 miles

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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:

$\mu = 60000, \sigma = 2900$

What is the probability a particular tire of this brand will last longer than 57,100 miles

This is 1 subtracted by the pvalue of Z when X = 57100. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{57100 - 60000}{2900}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587

1 - 0.1587 = 0.8413

84.13% probability a particular tire of this brand will last longer than 57,100 miles