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Let's go through the questions step by step using the **Empirical Rule** (also known as the **68-95-99.7 Rule**) and the provided information:

- **Mean (μ)** = 50 ounces
- **Standard deviation (σ)** = 10 ounces
- The distribution is **bell-shaped** (normal distribution).

### a) 99.7% of the widget weights lie between ______ and ______.

According to the Empirical Rule:
- **99.7%** of the data falls within **three standard deviations (±3σ)** from the mean.

So, the range would be:
- **Lower bound** = \( \mu - 3\sigma = 50 - 3 \times 10 = 50 - 30 = 20 \) ounces.
- **Upper bound** = \( \mu + 3\sigma = 50 + 3 \times 10 = 50 + 30 = 80 \) ounces.

Thus, **99.7% of the widget weights lie between 20 and 80 ounces**.

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### b) What percentage of the widget weights lie between 40 and 80 ounces?

- The range from **40 to 80** encompasses **40** (which is one standard deviation below the mean, μ - 1σ) to **80** (which is three standard deviations above the mean, μ + 3σ).

Using the Empirical Rule:
- **68%** of the data lies within **1 standard deviation** (between 40 and 60).
- **95%** of the data lies within **2 standard deviations** (between 30 and 70).
- **99.7%** of the data lies within **3 standard deviations** (between 20 and 80).

Since we're looking for the percentage between 40 and 80:
- From **40 to 50** (mean), it covers half of 68% (so, 34%).
- From **50 to 80**, this covers the remaining part of 99.7%, which is **49.85%** (half of 99.7%).

Thus, the total percentage from 40 to 80 is **34% + 49.85% = 83.85%** (approximately **84%**).

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### c) What percentage of the widget weights lie below 70?

- **70** is **2 standard deviations** above the mean (μ + 2σ).

Using the Empirical Rule:
- **95%** of the data lies within **two standard deviations** (between 30 and 70).
- Since the distribution is symmetrical, half of this data lies **below 70**, which is **95%** of the total.

Thus, **97.5%** of the widget weights lie below 70 ounces.

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Would you like more details or have any questions?

### 5 Related Questions:
1. What percentage of widget weights lie between 30 and 70 ounces?
2. What percentage of the weights lie above 60 ounces?
3. How many standard deviations from the mean do 95% of the weights lie within?
4. How would the range change if the standard deviation was 5 ounces instead of 10?
5. What percentage of widget weights lie below 30 ounces?

### Tip:
The Empirical Rule is very useful for approximating probabilities in normal distributions without needing detailed calculations!

The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 50 ounces and a standard deviation of 10 ounces. Use the Standard Deviation Rule (Empirical Rule). 99.7% of the widget weights lie between ______ and ______. What percentage of the widget weights lie between 40 and 80 ounces? What percentage of the widget weights lie below 70?

Learn how to apply the Empirical Rule (68-95-99.7 Rule) to solve a problem involving the distribution of widget weights, which have a mean of 50 ounces and a standard deviation of 10 ounces. This example covers calculating percentages of weights within a normal distribution, suitable for high school students.

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