https://file.aiquickdraw.com/mathbot/file/17269781063994lp2fq4u.jpg

Let's analyze this problem.

### Given information:
- The mean weight of the widgets is \( \mu = 61 \) ounces.
- The standard deviation is \( \sigma = 5 \) ounces.
- The distribution of widget weights is bell-shaped, meaning it's normally distributed.

#### Part (a):
We are asked to determine the range for which **99.7%** of the widget weights lie.

By the **Empirical Rule** (68-95-99.7 rule):
- **99.7% of the data lies within 3 standard deviations** from the mean in a normal distribution.

So, the range will be:
\[
\text{Lower bound} = \mu - 3\sigma = 61 - 3(5) = 61 - 15 = 46 \, \text{ounces}
\]
\[
\text{Upper bound} = \mu + 3\sigma = 61 + 3(5) = 61 + 15 = 76 \, \text{ounces}
\]

Thus, **99.7% of the weights lie between 46 and 76 ounces**.

#### Part (b):
We are asked what percentage of widget weights lie between **51** and **76** ounces.

To solve this, let's use the standard normal distribution:
1. Convert the weights to **z-scores** using the formula:
\[
z = \frac{x - \mu}{\sigma}
\]
For \( x = 51 \):
\[
z = \frac{51 - 61}{5} = \frac{-10}{5} = -2
\]
For \( x = 76 \):
\[
z = \frac{76 - 61}{5} = \frac{15}{5} = 3
\]

2. Now, using a standard normal distribution table or calculator:
- The area under the curve for \( z = -2 \) is approximately **0.0228** (2.28% below 51).
- The area under the curve for \( z = 3 \) is approximately **0.9987** (99.87% below 76).

3. The percentage of widget weights between 51 and 76 ounces is:
\[
0.9987 - 0.0228 = 0.9759 \, \text{or} \, 97.59\%
\]

Thus, **97.59%** of the widget weights lie between 51 and 76 ounces.

---

Would you like further details or explanations on any part of this?

Here are some related questions:
1. How does the Empirical Rule work for 68% and 95% intervals?
2. What is the formula for finding a z-score?
3. How can I calculate percentages for values outside the 3-standard deviation range?
4. How does standard deviation affect the spread of data in a normal distribution?
5. Can the normal distribution be used for skewed data sets?

**Tip:** When using the Empirical Rule, remember that it applies specifically to normal distributions, where the data is symmetrically distributed around the mean.

The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 61 ounces and a standard deviation of 5 ounces.

Use the Empirical Rule and a sketch of the normal distribution in order to answer these questions:

a) 99.7% of the widget weights lie between ___ and ___.
b) What percentage of the widget weights lie between 51 and 76 ounces?

Solve this Empirical Rule problem involving the normal distribution of widget weights with a mean of 61 ounces and standard deviation of 5 ounces. Learn how to calculate percentages and apply the 68-95-99.7 rule for normally distributed data.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation