/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 94 Lori just bought a new set of fo... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 94

Lori just bought a new set of four tires for her car. The life of each tire is normally distributed with a mean of 45,000 miles and a standard deviation of 2000 miles. Find the probability that all four tires will last for at least 46,000 miles. Assume that the life of each of these tires is independent of the lives of other tires.

Short Answer

Expert verified
The probability that all four tires will last for at least 46000 miles is approximately 0.2286.

Step by step solution

01

Standardizing the Normal Variable

First, it is needed to convert the normal random variable (X ~ N(45000, 2000^2)) to a standard normal random variable (Z ~ N(0,1)). This is done using the following formula: Z = (X - 渭) / 蟽, where X is a variable value, 渭 is the mean, and 蟽 is the standard deviation. In this case, X = 46000, 渭 = 45000, and 蟽 = 2000. So the Z-score corresponding to a tire life of 46000 miles would be Z = (46000 - 45000) / 2000.
02

Calculating Probability Using Z-Score

Next, the above Z-score is used to find a probability that one tire will last for at least 46000 miles. We are asked 'greater than', so that will be a right-tailed test. Lookup the Z value in the standard normal (Z) table or use a calculator to find that the cumulative probability against this Z value is 0.3085. Since it is a right-tailed test, the probability will be 1 - 0.3085 or 0.6915.
03

Probability for All Four Tires

In the end, use the independence property of the tire life events, it means that the probability that all four tires last for at least 46000 miles is the product of the individual probabilities. This is calculated as 0.6915 * 0.6915 * 0.6915 * 0.6915 = 0.2286.

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