Question

One tire manufacturer claims that his tires last an average of 42,000 miles with a standard deviation of 7800 miles. A random sample of 100 of his tires is taken. What is the probability that the average of these 100 tires will last greater than 41,000 miles?

Answer 1

Answer:

89.97% probability that the average of these 100 tires will last greater than 41,000 miles.

Step-by-step explanation:

The solve this problem, it is important to know 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 $\sigma$, a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 42000, \sigma = 7800, n = 100, s = \frac{7800}{\sqrt{100}} = 780$

What is the probability that the average of these 100 tires will last greater than 41,000 miles?

This is 1 subtracted by the pvalue of Z when X = 41000.

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

By the Central Limit Theorem, we use s instead of $\sigma$.

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

$Z = \frac{41000 - 42000}{780}$

$Z = -1.28$

$Z = -1.28$ has a pvalue of 0.1003.

So there is a 1-0.1003 = 0.8997 = 89.97% probability that the average of these 100 tires will last greater than 41,000 miles.