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

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 and standard deviation , the zscore of a measure X is given by:

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 and standard deviation , the sample means with size n of at least 30 can be approximated to a normal distribution with mean and standard deviation

In this problem, we have that:

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

By the Central Limit Theorem

has a pvalue of 0.9332

X = 57000

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

answer: i think it is (1,-2)

need more information

step-by-step explanation: