Question

The tread life of tires mounted on light-duty trucks follows the normal probability distribution with a mean of 60,000 miles and a standard deviation of 4,000 miles. Suppose you bought a set of four tires, what is the likelihood the mean tire life of these four tires is between 57,000 and 63,000 miles

Answer 1

Answer:

86.64% probability that the mean tire life of these four tires is between 57,000 and 63,000 miles

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples 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:

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

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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$, the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 60000, \sigma = 4000, n = 4, s = \frac{4000}{\sqrt{4}} = 2000$

Suppose you bought a set of four tires, what is the likelihood the mean tire life of these four tires is between 57,000 and 63,000 miles

This is the pvalue of Z when X = 63000 subtracted by the pvalue of Z when X = 57000. So

X = 63000

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

By the Central Limit Theorem

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

$Z = \frac{63000 - 60000}{2000}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

X = 57000

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

$Z = \frac{57000 - 60000}{2000}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668

0.9332 - 0.0668 = 0.8664

86.64% probability that the mean tire life of these four tires is between 57,000 and 63,000 miles