Problem 8 - Algebra 2

Question

For Exercises \(5-8,\) use the following information. The useful life of a certain car battery is normally distributed with a mean of \(100,000\) miles and a standard deviation of \(10,000\) miles. The company makes \(20,000\) batteries a month. What is the probability that if you buy a car battery at random, it will last between \(80,000\) and \(110,000\) miles?

Step-by-Step Solution

Answer

The probability that a car battery lasts between 80,000 and 110,000 miles is 81.85%.

Step 1: Identify given values

From the problem, we are given the mean \( \mu = 100,000 \) miles and the standard deviation \( \sigma = 10,000 \) miles. We need to find the probability that the battery lasts between \( 80,000 \) and \( 110,000 \) miles.

Step 2: Define the standard normal variable

To find probabilities for a normal distribution, we need to convert our range to the standard normal distribution using the z-score formula: \( z = \frac{x - \mu}{\sigma} \).

Step 3: Calculate z-scores

Calculate the z-scores for both values 80,000 and 110,000 miles: - For \( 80,000 \): \( z_1 = \frac{80,000 - 100,000}{10,000} = -2 \) - For \( 110,000 \): \( z_2 = \frac{110,000 - 100,000}{10,000} = 1 \).

Step 4: Use z-table to find probabilities

Using the z-table, find the probabilities corresponding to \( z_1 = -2 \) and \( z_2 = 1 \): - \( P(Z \leq -2) \approx 0.0228 \) - \( P(Z \leq 1) \approx 0.8413 \).

Step 5: Calculate the probability range

The probability that the battery lasts between 80,000 and 110,000 miles is given by the difference: \[ P(-2 < Z < 1) = P(Z < 1) - P(Z < -2) = 0.8413 - 0.0228 = 0.8185 \].

Step 6: Interpret the result

The final interpretation is that there is an 81.85% chance that a randomly selected car battery will last between 80,000 and 110,000 miles.

Key Concepts

Understanding Z-Score

The z-score is a common way to understand how far away a particular value is from the mean of a set of data, measured in standard deviations. In simpler terms, a z-score tells us where a particular value falls on a standard normal distribution. If you consider a z-score of zero, it means that the value is exactly at the mean. A positive z-score indicates a value above the mean, while a negative z-score signals it is below the average.

To calculate the z-score, we use the formula:

• \( z = \frac{x - \mu}{\sigma} \),

where \( x \) is the value in question, \( \mu \) is the mean of the distribution, and \( \sigma \) is the standard deviation. By converting the original problem's mileages of the car battery (80,000 and 110,000 miles) into z-scores, we can easily interpret the data's position in the normal distribution.

Delving into Mean and Standard Deviation

When discussing normal distributions, two key components come into play: the mean and the standard deviation. The mean, often represented by \( \mu \), is the average value of a dataset. In a normal distribution, it is the peak point where most values congregate—think of it as the center of gravity of the data.

The standard deviation, denoted as \( \sigma \), measures how spread out the data is around the mean. A smaller standard deviation means the data points are tightly clustered around the mean, while a larger standard deviation implies the data is spread over a wider range. Together these two parameters define the bell-shaped curve of a normal distribution.

In our exercise with car batteries, the mean represents an average lifespan of 100,000 miles, while the standard deviation of 10,000 miles shows the typical mileage spread from the mean. Understanding these concepts helps in determining how values vary within a dataset.

Probability Calculation in a Normal Distribution

Probability calculation in a normal distribution is a powerful tool that lets us predict the likelihood of a given outcome—or range of outcomes. By using z-scores and referencing a z-table, we can determine how probable it is for a battery to last a certain number of miles.

Here's a simple illustration of how it works: Once we have our z-scores calculated for 80,000 and 110,000 miles (-2 and 1, respectively), we consult a z-table to find the probability of each score. For \( z = -2 \), the probability is approximately 0.0228, and for \( z = 1 \), it stands at 0.8413. These probabilities reflect the likelihood of a randomly selected battery lasting up to those mileages.

To find the probability of the battery lasting between 80,000 and 110,000 miles, subtract the probability of \( z_1 \) from \( z_2 \):

• \( P(-2 < Z < 1) = 0.8413 - 0.0228 = 0.8185 \).

This calculation indicates that there's an 81.85% chance a battery will fall within those limits, giving a clear picture of how likely it is for the life expectancy of a battery to align with the provided mileage range.