Question

The lives of a premium sports car's brakes are normally distributed with a mean of 60,000 miles and a standard deviation of 4,000 miles. What is the probability that such brakes last between 54,000 and 66,000 miles? Group of answer choices

Answer 1

Answer:

86.64% probability that such brakes last between 54,000 and 66,000 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:

$\mu = 60000, \sigma = 4000$

What is the probability that such brakes last between 54,000 and 66,000 miles?

This is the pvalue of Z when X = 66000 subtracted by the pvalue of Z when X = 54000.

X = 66000

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

$Z = \frac{66000 - 60000}{4000}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

X = 54000

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

$Z = \frac{54000 - 60000}{4000}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668

0.9332 - 0.0668 = 0.8664

86.64% probability that such brakes last between 54,000 and 66,000 miles