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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 8, Problem 8.9.42a

Lifetime of a tire Assume the random variable L in Example 2f is normally distributed with mean μ = 22,000 miles and σ = 4,000 miles.
a. In a batch of 4000 tires, how many can be expected to last for at least 18,000 miles?

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1
Identify the given information: the lifetime L of tires is normally distributed with mean \( \mu = 22000 \) miles and standard deviation \( \sigma = 4000 \) miles. The batch size is 4000 tires.
We want to find the number of tires expected to last at least 18,000 miles. This means we need to find \( P(L \geq 18000) \), the probability that a tire lasts 18,000 miles or more.
Convert the lifetime value 18,000 miles to a standard normal variable \( Z \) using the formula: \[ Z = \frac{X - \mu}{\sigma} = \frac{18000 - 22000}{4000} \]. This standardizes the value to the standard normal distribution.
Use the standard normal distribution table or a calculator to find \( P(Z \geq z) \), where \( z \) is the value calculated in the previous step. This gives the probability that a tire lasts at least 18,000 miles.
Multiply the probability \( P(L \geq 18000) \) by the total number of tires (4000) to find the expected number of tires lasting at least 18,000 miles: \[ \text{Expected number} = 4000 \times P(L \geq 18000) \].

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Normal Distribution

The normal distribution is a continuous probability distribution characterized by its bell-shaped curve, defined by the mean (μ) and standard deviation (σ). It models many natural phenomena, including tire lifetimes, and allows calculation of probabilities for values within certain ranges.
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Standardization (Z-Score)

Standardization converts a normal random variable to a standard normal variable (Z) with mean 0 and standard deviation 1. This is done using the formula Z = (X - μ) / σ, enabling the use of standard normal tables to find probabilities associated with specific values.
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Expected Value in Probability

The expected value represents the average or mean outcome in a probabilistic scenario. When dealing with large batches, multiplying the probability of an event by the total number of trials estimates how many items meet a certain criterion, such as tires lasting at least 18,000 miles.
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