To solve this problem, we'll use the properties of the normal distribution.

Given:

• Mean ($\mu$) = 2735 hours
• Standard deviation ($\sigma$) = 113 hours
• We want to find $P(2540 < X < 2743)$.

Step 1: Convert the raw scores to Z-scores

The Z-score formula is:

$Z = \frac{X - \mu}{\sigma}$

First, calculate the Z-scores for 2540 and 2743.

For $X = 2540$:

$Z_{2540} = \frac{2540 - 2735}{113} = \frac{-195}{113} \approx -1.7257$

For $X = 2743$:

$Z_{2743} = \frac{2743 - 2735}{113} = \frac{8}{113} \approx 0.0708$

Step 2: Find the probabilities corresponding to these Z-scores

Using the standard normal distribution table or a calculator:

• The probability corresponding to $Z = -1.7257$ is approximately 0.0423.
• The probability corresponding to $Z = 0.0708$ is approximately 0.5282.

Step 3: Calculate the probability $P(2540 < X < 2743)$

$P(2540 < X < 2743) = P(Z_{2743}) - P(Z_{2540}) \approx 0.5282 - 0.0423 = 0.4859$

So, the probability is:

$\boxed{0.4859}$

Would you like further details or have any questions?

---

1. What is the probability that a bulb lasts more than 3000 hours?
2. How does changing the mean affect the probability distribution?
3. How is the standard deviation interpreted in the context of lightbulb lifespan?
4. What does the Z-score represent in a normal distribution?
5. How would you find the probability for a different interval of hours?

Tip: The Z-score helps standardize data points within a normal distribution, making it easier to calculate probabilities.