https://file.aiquickdraw.com/mathbot/file/1726583268590aruob3c4.jpg

This question refers to the **Empirical Rule** (also known as the 68-95-99.7 rule), which applies to a normal (bell-shaped) distribution. The rule states the following:

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

### Given:
- Mean (\(\mu\)) = 44 ounces
- Standard deviation (\(\sigma\)) = 7 ounces

Let's solve the questions.

### a) 95% of the widget weights lie between:
Using the Empirical Rule, 95% of the data lies within 2 standard deviations from the mean. So:

\[
\mu - 2\sigma = 44 - 2(7) = 44 - 14 = 30
\]
\[
\mu + 2\sigma = 44 + 2(7) = 44 + 14 = 58
\]

Thus, 95% of the widget weights lie between **30 and 58** ounces.

### b) What percentage of the widget weights lie between 23 and 58 ounces?
We calculate how many standard deviations 23 and 58 ounces are from the mean:

- \(23\) ounces: 
\[
\frac{23 - 44}{7} = \frac{-21}{7} = -3 \quad (\text{3 standard deviations below the mean})
\]

- \(58\) ounces:
\[
\frac{58 - 44}{7} = \frac{14}{7} = 2 \quad (\text{2 standard deviations above the mean})
\]

From the Empirical Rule:
- Between \(-3\sigma\) and \(+2\sigma\), this includes the range from \(-3\sigma\) to \(+\3\sigma\) (99.7%) minus the tail from \(+2\sigma\) to \(+3\sigma\) (2.5%).

Thus, approximately **97.5%** of the weights lie between 23 and 58 ounces.

### c) What percentage of the widget weights lie below 51 ounces?
For 51 ounces:
\[
\frac{51 - 44}{7} = \frac{7}{7} = 1 \quad (\text{1 standard deviation above the mean})
\]

From the Empirical Rule, 1 standard deviation above the mean accounts for 68% of the data. Half of this is below the mean, so 34% is above the mean. Thus, the percentage of weights below 51 ounces is:

\[
50\% + 34\% = 84\%
\]

So, **84%** of the weights lie below 51 ounces.

---

Let me know if you'd like further details on any part of the solution. Here are five related questions to deepen understanding:

1. What percentage of widget weights lie between 37 and 51 ounces?
2. How would the Empirical Rule apply if the standard deviation was 10 ounces instead of 7?
3. What percentage of widget weights lie above 58 ounces?
4. How does the Empirical Rule change if the data is not normally distributed?
5. What range would contain 99.7% of widget weights based on this distribution?

**Tip**: The Empirical Rule is useful for approximating probabilities in a normal distribution when you don't have access to exact z-scores!

The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped with a mean of 44 ounces and a standard deviation of 7 ounces. Using the Empirical Rule, answer the following questions: a) 95% of the widget weights lie between __ and __. b) What percentage of the widget weights lie between 23 and 58 ounces? c) What percentage of the widget weights lie below 51?

Solve a normal distribution problem involving widget weights with a mean of 44 ounces and a standard deviation of 7 ounces using the Empirical Rule. This problem helps students understand how to find percentages and ranges of data based on the standard deviation in a bell-shaped curve.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation