https://file.aiquickdraw.com/mathbot/file/1730126697329a2sthwuh.jpg

To solve this problem, let's use the properties of a normal distribution.

### Given Information
- **Mean (μ)** = 750 hours
- **Standard Deviation (σ)** = 50 hours
- We want to find the percentage of the distribution between **600 hours** and **900 hours**.

### Step-by-Step Solution
1. **Convert Hours to Z-scores**:
   - The Z-score formula is: 
     \[
     Z = \frac{X - \mu}{\sigma}
     \]
   
   - **For 600 hours**:
     \[
     Z = \frac{600 - 750}{50} = \frac{-150}{50} = -3
     \]
   
   - **For 900 hours**:
     \[
     Z = \frac{900 - 750}{50} = \frac{150}{50} = 3
     \]

2. **Determine the Percentage Between Z-scores -3 and 3**:
   - In a standard normal distribution, approximately **99.74%** of data falls within 3 standard deviations from the mean (from Z = -3 to Z = 3).

### Answer
The correct answer is:
**d. 99.74%**

This means that about 99.74% of the light bulbs will have a lifespan between 600 and 900 hours.

Would you like further details or have any questions?

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### Related Questions
1. What is the probability of a light bulb lasting between 700 and 800 hours?
2. How would the percentage change if the standard deviation was 100 hours instead?
3. What is the Z-score for a lifespan of 850 hours?
4. If the average lifespan was 800 hours, how would the answer change?
5. How can we calculate the percentage of bulbs lasting less than 600 hours?

### Tip
For quick estimations with a normal distribution, remember that roughly 68% of data falls within 1 standard deviation, 95% within 2, and 99.7% within 3 standard deviations of the mean.

A new extended-life light bulb has an average life of 750 hours, with a standard deviation of 50 hours. If the life of these light bulbs approximates a normal distribution, about what percent of the distribution will be between 600 hours and 900 hours?

Find the percentage of light bulbs lasting between 600 and 900 hours, with a mean lifespan of 750 hours and standard deviation of 50 hours. This problem uses concepts in normal distribution and Z-scores, suitable for high school statistics students.

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