. The lifetime of a certain kind of battery is found to have 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.

Given the problem, we know that the battery lifetime follows a **normal distribution** with a mean (\(\mu\)) of 400 hours and a standard deviation (\(\sigma\)) of 45 hours. We'll solve each part step by step using the properties of the normal distribution.

### a. The proportion of batteries with a lifetime of at least 470 hours.
We need to find the proportion of batteries with a lifetime of at least 470 hours, i.e., \( P(X \geq 470) \).

1. First, we compute the **z-score** for 470 hours:

\[
z = \frac{X - \mu}{\sigma} = \frac{470 - 400}{45} = \frac{70}{45} \approx 1.56
\]

2. Next, we use the standard normal table (or a calculator) to find the proportion of values greater than a z-score of 1.56.

From the standard normal distribution table:
\[
P(Z \leq 1.56) \approx 0.9406
\]

Thus, the proportion of batteries with a lifetime **at least** 470 hours is:
\[
P(X \geq 470) = 1 - P(Z \leq 1.56) = 1 - 0.9406 = 0.0594
\]

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

---

### b. The proportion of batteries with a lifetime between 385 and 415 hours.
We are looking for \( P(385 \leq X \leq 415) \).

1. Compute the z-scores for 385 hours and 415 hours:
   - For \( X = 385 \):
     \[
     z = \frac{385 - 400}{45} = \frac{-15}{45} = -0.33
     \]
   - For \( X = 415 \):
     \[
     z = \frac{415 - 400}{45} = \frac{15}{45} = 0.33
     \]

2. Now, use the standard normal table to find the probabilities corresponding to these z-scores:
   - \( P(Z \leq -0.33) \approx 0.3707 \)
   - \( P(Z \leq 0.33) \approx 0.6293 \)

3. The proportion of batteries with a lifetime between 385 and 415 hours is:
\[
P(385 \leq X \leq 415) = P(Z \leq 0.33) - P(Z \leq -0.33) = 0.6293 - 0.3707 = 0.2586
\]

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

---

### c. The minimum life of the best 5% of the batteries.
We are looking for the minimum lifetime such that only 5% of the batteries exceed this value. This corresponds to finding the z-score for the top 5% of the normal distribution.

1. For the top 5%, the z-score is found using the inverse of the cumulative distribution function (CDF). From the z-table or calculator, the z-score corresponding to the 95th percentile is approximately **1.645**.

2. Now, we convert this z-score back to the original battery lifetime using the formula:
\[
X = \mu + z \cdot \sigma = 400 + 1.645 \cdot 45 = 400 + 74.025 = 474.025
\]

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

---

### Summary:
- **a.** Proportion of batteries with a lifetime of at least 470 hours: **5.94%**.
- **b.** Proportion of batteries with a lifetime between 385 and 415 hours: **25.86%**.
- **c.** Minimum life of the best 5% of the batteries: **474.03 hours**.

---

Do you want more details or have any questions on any part?

Here are 5 related questions:
1. How would the answers change if the standard deviation was 50 hours instead of 45 hours?
2. What is the probability that a randomly selected battery lasts more than 500 hours?
3. What is the z-score for a battery that lasts 350 hours?
4. What percentage of batteries have a lifetime of less than 370 hours?
5. If you sample 10 batteries, what’s the probability that at least 2 of them last over 470 hours?

**Tip:** The z-score is a powerful tool in statistics to standardize values, making comparisons across different distributions easier!

The lifetime of a certain kind of battery is found to have 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.

Learn how to solve a normal distribution problem involving battery lifetimes with an average of 400 hours and a standard deviation of 45 hours. Find the proportion of batteries lasting at least 470 hours, the percentage between 385 and 415 hours, and the minimum life of the best 5%. Suitable for Grades 10-12.

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