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Let's break down this problem step by step using the information provided. The lifetime of a battery is normally distributed with:

- Mean \( \mu = 400 \) hours
- Standard deviation \( \sigma = 45 \) hours

We'll use the Z-score formula to solve the sub-questions. The Z-score is calculated by:

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

Where:
- \( X \) is the value of interest,
- \( \mu \) is the mean,
- \( \sigma \) is the standard deviation.

### a. Proportion of batteries with a lifetime of at least 470 hours

We need to find \( P(X \geq 470) \). First, calculate the Z-score for 470 hours:

\[
Z = \frac{470 - 400}{45} = \frac{70}{45} \approx 1.56
\]

Using the standard normal distribution table, a Z-score of 1.56 corresponds to a cumulative probability of about 0.9406. Since we want the proportion greater than 470 hours, we calculate:

\[
P(X \geq 470) = 1 - 0.9406 = 0.0594
\]

So, approximately 5.94% of the batteries have a lifetime of at least 470 hours.

### b. Proportion of batteries with a lifetime between 385 and 415 hours

We need to find \( P(385 \leq X \leq 415) \). First, calculate the Z-scores for 385 and 415:

For 385 hours:
\[
Z = \frac{385 - 400}{45} = \frac{-15}{45} = -0.33
\]
For 415 hours:
\[
Z = \frac{415 - 400}{45} = \frac{15}{45} = 0.33
\]

Using the Z-table:
- The cumulative probability for \( Z = -0.33 \) is approximately 0.3707,
- The cumulative probability for \( Z = 0.33 \) is approximately 0.6293.

Thus, the probability of a battery having a lifetime between 385 and 415 hours is:

\[
P(385 \leq X \leq 415) = 0.6293 - 0.3707 = 0.2586
\]

So, approximately 25.86% of the batteries have a lifetime between 385 and 415 hours.

### c. Minimum life of the best 5% of the batteries

For the best 5% of the batteries, we need to find the lifetime corresponding to the 95th percentile. The Z-score corresponding to the 95th percentile is about 1.645.

Now, convert the Z-score back to the corresponding battery lifetime using the Z-score formula:

\[
X = \mu + Z \cdot \sigma = 400 + 1.645 \cdot 45 \approx 400 + 74.025 = 474.025
\]

Thus, the minimum lifetime of the best 5% of the batteries is approximately 474.03 hours.

---

Would you like more details or explanations on any part of this? 

Here are 5 related questions for further exploration:
1. How would the answers change if the standard deviation was larger?
2. What is the probability that a battery lasts less than 350 hours?
3. Can you explain why the Z-score approach is appropriate for normal distributions?
4. How would you calculate the median battery lifetime if the distribution was skewed?
5. What is the relationship between standard deviation and the spread of battery lifetimes?

**Tip**: When working with normal distributions, always remember that about 68% of data falls within one standard deviation of the mean, 95% within two, and 99.7% within three.

The lifetime of a certain kind of battery is found to have an average equal to 400 hours and standard deviation equal to 45 hours. Assuming the distribution of lifetime to be normal, find: a) The proportion of batteries with a lifetime of at least 470 hours, b) The proportion of batteries with a lifetime between 385 and 415 hours, c) The minimum life of the best 5% of the batteries.

Solve a statistics problem involving the lifetime of batteries. Given a normal distribution with a mean of 400 hours and a standard deviation of 45 hours, find the proportion of batteries that last at least 470 hours, the proportion lasting between 385 and 415 hours, and the minimum lifetime of the top 5%. Learn how to use the Z-score and normal distribution in this Grade 11-12 level problem.

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