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To solve this problem, we need to find the probability that the lifetime of a 2-volt non-rechargeable battery exceeds 520 hours. This involves using the properties of the normal distribution.

### Step-by-Step Solution:

1. **Identify the given values:**
   - Mean (\(\mu\)) = 516 hours
   - Standard deviation (\(\sigma\)) = 20 hours
   - We need to find the probability \( P(X > 520) \).

2. **Standardize the value to a Z-score:**
   The Z-score formula is given by:
   \[
   Z = \frac{X - \mu}{\sigma}
   \]
   Substituting the values:
   \[
   Z = \frac{520 - 516}{20} = \frac{4}{20} = 0.2
   \]

3. **Find the probability corresponding to the Z-score:**
   We need the probability of \( Z > 0.2 \).

   Using standard normal distribution tables or a calculator, the probability of \( Z < 0.2 \) is approximately 0.5793.

4. **Calculate the required probability:**
   Since we need \( P(Z > 0.2) \):
   \[
   P(Z > 0.2) = 1 - P(Z < 0.2) = 1 - 0.5793 = 0.4207
   \]

### Final Answer:
The proportion of batteries with lifetimes exceeding 520 hours is approximately **0.4207**.

Would you like more details on how to read values from the Z-table, or have any other questions?

### Related Questions:
1. How do you find the probability for a value below a certain point in a normal distribution?
2. What are the steps to standardize a random variable using the Z-score?
3. How can you use a Z-table to find probabilities associated with the normal distribution?
4. What are the implications of changing the mean or standard deviation on the probability calculations?
5. How can technology (like calculators or software) assist in finding probabilities for normal distributions?

**Tip:** Always check whether the question asks for probability above or below a certain point, as this changes the interpretation of the Z-score value.

The lifetime of a 2-volt non-rechargeable battery in constant use has a Normal distribution, with a mean of 516 hours and a standard deviation of 20 hours. The proportion of batteries with lifetimes exceeding 520 hours is approximately?

Learn how to calculate the proportion of batteries with lifetimes exceeding 520 hours in a normal distribution, given a mean of 516 hours and a standard deviation of 20 hours. This solution uses the Z-score formula and standard normal distribution tables.

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