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.

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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.

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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.

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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.

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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!