The mean of useful life of car batteries is 54 months. They have a standard deviation of 2. Assume the useful life of batteries is normally distributed. Calculate the percent of batteries with a useful life of less than 52 months. Calculate the percent of batteries that will last longer than 60 months

To solve this problem, we need to use the properties of the normal distribution. The question provides the following information:

- The mean (\(\mu\)) of the useful life of car batteries is 54 months.
- The standard deviation (\(\sigma\)) is 2 months.

We'll calculate two things:
1. The percentage of batteries with a useful life of **less than 52 months**.
2. The percentage of batteries with a useful life of **more than 60 months**.

### 1. Percent of batteries with a useful life of less than 52 months

We need to find the **Z-score** for 52 months. The Z-score formula is:

\[
Z = \frac{X - \mu}{\sigma}
\]

Where:
- \(X\) is the value we're interested in (52 months),
- \(\mu\) is the mean (54 months),
- \(\sigma\) is the standard deviation (2 months).

Substituting the values:

\[
Z = \frac{52 - 54}{2} = \frac{-2}{2} = -1
\]

Using a Z-table (or a calculator for the cumulative distribution function), we find the probability associated with a Z-score of -1. 

For \(Z = -1\), the cumulative probability is approximately 0.1587. This means that about **15.87%** of batteries will have a useful life of less than 52 months.

### 2. Percent of batteries with a useful life of more than 60 months

Now, we calculate the Z-score for 60 months:

\[
Z = \frac{60 - 54}{2} = \frac{6}{2} = 3
\]

From the Z-table, the cumulative probability for \(Z = 3\) is approximately 0.9987. This means that about 99.87% of batteries will have a useful life of **less than 60 months**.

To find the percentage of batteries that will last **more than** 60 months, we subtract this value from 1:

\[
1 - 0.9987 = 0.0013
\]

This corresponds to **0.13%** of batteries lasting longer than 60 months.

### Final Answers:
- **15.87%** of batteries have a useful life of less than 52 months.
- **0.13%** of batteries have a useful life of more than 60 months.

Would you like further explanation or more details on this? 

Here are 5 related questions you might find helpful:
1. How do we interpret the Z-score in a normal distribution?
2. Can we use the same method for any other value of battery life?
3. How does the standard deviation affect the distribution of battery life?
4. What if the distribution was not normal? How would we approach it?
5. How can we calculate the probability between two given values?

**Tip:** For normally distributed data, approximately 68% of the data lies within one standard deviation from the mean.

The mean of useful life of car batteries is 54 months. They have a standard deviation of 2. Assume the useful life of batteries is normally distributed. Calculate the percent of batteries with a useful life of less than 52 months. Calculate the percent of batteries that will last longer than 60 months.

Calculate the percent of car batteries with a useful life of less than 52 months and those lasting more than 60 months, using the normal distribution. With a mean of 54 months and a standard deviation of 2 months, this problem explores Z-scores and probability, ideal for Grades 10-12.

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