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Let's break down and solve each part of the question using the Empirical Rule (also known as the 68-95-99.7 rule). This rule applies to a normal (bell-shaped) distribution and gives approximate percentages of data that lie within 1, 2, and 3 standard deviations from the mean.

### Given information:
- Mean (μ) = 47 ounces
- Standard deviation (σ) = 10 ounces

#### a) **68% of the widget weights lie between**:
68% of the data lies within 1 standard deviation of the mean. This means:
- Lower bound = Mean - 1 standard deviation = 47 - 10 = **37 ounces**
- Upper bound = Mean + 1 standard deviation = 47 + 10 = **57 ounces**

Thus, **68% of the widget weights lie between 37 and 57 ounces**.

#### b) **What percentage of the widget weights lie between 27 and 57 ounces?**
- 27 ounces is 2 standard deviations below the mean (47 - 2 * 10 = 27).
- 57 ounces is 1 standard deviation above the mean.

Using the Empirical Rule:
- 95% of the data lies within 2 standard deviations from the mean (between 27 and 67 ounces).
- Since 68% of the data lies between 37 and 57 ounces, the remaining 27% (from 27 to 37 ounces) must be split in half to get the portion from 27 to 37.
- Therefore, the percentage between 27 and 57 is **68% + 13.5% = 81.5%**.

#### c) **What percentage of the widget weights lie below 77 ounces?**
- 77 ounces is 3 standard deviations above the mean (47 + 3 * 10 = 77).
- According to the Empirical Rule, 99.7% of the data lies within 3 standard deviations, so half of the remaining 0.3% (since the question asks for "below" 77) must be subtracted from 100%.
- Thus, the percentage of weights below 77 ounces is approximately **99.85%**.

Would you like any further details or clarification on these calculations?

Here are 5 related questions:
1. How would the answers change if the standard deviation was 5 ounces instead of 10?
2. What percentage of the weights lie above 77 ounces?
3. How does the Empirical Rule differ from using z-scores for precise calculations?
4. If the mean widget weight was increased to 50 ounces, what would be the new range for 68% of the weights?
5. How could you estimate the percentage of widget weights lying between 37 and 67 ounces?

**Tip:** The Empirical Rule is a quick approximation tool, but for exact probabilities, especially for non-standard intervals, you should use the z-score formula and a normal distribution table.

The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped with a mean of 47 ounces and a standard deviation of 10 ounces. Using the Empirical Rule, answer the following questions: a) 68% of the widget weights lie between ___ and ___ ounces. b) What percentage of the widget weights lie between 27 and 57 ounces? c) What percentage of the widget weights lie below 77 ounces?

Solve questions about widget weight distribution using the Empirical Rule. Understand how 68% of data lies within one standard deviation, calculate percentages between specific weight ranges, and determine the percentage of weights below 77 ounces.

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