Math Problem Statement

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?

Solution

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 (μ\mu) = 44 ounces
  • Standard deviation (σ\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:

μ2σ=442(7)=4414=30\mu - 2\sigma = 44 - 2(7) = 44 - 14 = 30 μ+2σ=44+2(7)=44+14=58\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:

  • 2323 ounces: 23447=217=3(3 standard deviations below the mean)\frac{23 - 44}{7} = \frac{-21}{7} = -3 \quad (\text{3 standard deviations below the mean})

  • 5858 ounces: 58447=147=2(2 standard deviations above the mean)\frac{58 - 44}{7} = \frac{14}{7} = 2 \quad (\text{2 standard deviations above the mean})

From the Empirical Rule:

  • Between 3σ-3\sigma and +2σ+2\sigma, this includes the range from 3σ-3\sigma to +\3σ+\3\sigma (99.7%) minus the tail from +2σ+2\sigma to +3σ+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: 51447=77=1(1 standard deviation above the mean)\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%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!

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Math Problem Analysis

Mathematical Concepts

Statistics
Empirical Rule
Normal Distribution
Standard Deviation

Formulas

Mean ± k * Standard Deviation
z = (X - μ) / σ

Theorems

Empirical Rule (68-95-99.7 Rule)

Suitable Grade Level

Grades 10-12