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.