The 68-95-99.7 Rule (also called the Empirical Rule) states that for a normal distribution:

1. About 68% of the data falls within 1 standard deviation of the mean.
2. About 95% of the data falls within 2 standard deviations of the mean.
3. About 99.7% of the data falls within 3 standard deviations of the mean.

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Given Information:

• Mean ($\mu$) = 56
• Standard Deviation ($\sigma$) = 4
• Interval of interest: $52 \leq x \leq 56$

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Step-by-Step Solution:

1. Determine how many standard deviations the interval endpoints are from the mean:

   • For $x = 52$: $z = \frac{x - \mu}{\sigma} = \frac{52 - 56}{4} = -1$
   • For $x = 56$: $z = \frac{x - \mu}{\sigma} = \frac{56 - 56}{4} = 0$

   Thus, the interval $52 \leq x \leq 56$ corresponds to $-1 \leq z \leq 0$.

2. Use the Empirical Rule:

   • The 68% of the data lies within 1 standard deviation of the mean ($-1 \leq z \leq 1$).
   • The interval $-1 \leq z \leq 1$ is symmetric, so half of this range falls between $-1 \leq z \leq 0$.

   Therefore: $\text{Percentage of data in } -1 \leq z \leq 0 = \frac{68\%}{2} = 34\%$

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Final Answer:

The approximate percentage of lightbulb replacement requests numbering between 52 and 56 is 34%.

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Would you like me to explain further or calculate another interval?

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Related Questions:

1. What percentage of requests fall between 52 and 60?
2. What percentage of requests are below 52?
3. What percentage of requests are above 60?
4. How many requests would you expect in a day between 48 and 64?
5. Can you calculate the probability for a non-standard interval, such as 54 to 58?

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Tip:

For precise percentages, use the z-score formula combined with a standard normal distribution table or software.