Math Problem Statement

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, also known as the Empirical Rule. Suggestion: sketch the 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 30 and 80 ounces? % c) What percentage of the widget weights lie below 60 ?

Solution

The Empirical Rule, also known as the 68-95-99.7 Rule, applies to normally distributed data. It gives us approximate percentages of data that fall within 1, 2, and 3 standard deviations from the mean.

We know that the widget weights are normally distributed with:

  • Mean (μ\mu) = 50 ounces
  • Standard deviation (σ\sigma) = 10 ounces

(a) 99.7% of the widget weights lie between:

  • 99.7% of the data lies within 3 standard deviations from the mean.

    The range is calculated as: Lower bound=μ3σ=503(10)=5030=20 ounces\text{Lower bound} = \mu - 3\sigma = 50 - 3(10) = 50 - 30 = 20 \text{ ounces} Upper bound=μ+3σ=50+3(10)=50+30=80 ounces\text{Upper bound} = \mu + 3\sigma = 50 + 3(10) = 50 + 30 = 80 \text{ ounces}

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

(b) What percentage of the widget weights lie between 30 and 80 ounces?

  • 30 ounces is 2 standard deviations below the mean: Number of deviations from mean=305010=2\text{Number of deviations from mean} = \frac{30 - 50}{10} = -2
  • 80 ounces is 3 standard deviations above the mean: Number of deviations from mean=805010=+3\text{Number of deviations from mean} = \frac{80 - 50}{10} = +3

According to the Empirical Rule:

  • From 2σ-2\sigma to +3σ+3\sigma covers from 30 to 80 ounces.
  • The percentage between 2σ-2\sigma and +3σ+3\sigma can be found by combining:
    • 95% of data is between 2σ-2\sigma and +2σ+2\sigma (i.e., 30 to 70 ounces)
    • From +2σ+2\sigma to +3σ+3\sigma covers an additional 2.35% of data.

Thus, the percentage of widget weights between 30 and 80 ounces is 95% + 2.35% = 97.35%.

(c) What percentage of the widget weights lie below 60 ounces?

  • 60 ounces is 1 standard deviation above the mean: 605010=+1\frac{60 - 50}{10} = +1

According to the Empirical Rule, 68% of the data lies within 1 standard deviation of the mean, so:

  • From 5010=4050 - 10 = 40 to 50+10=6050 + 10 = 60, 68% of data lies between 40 and 60 ounces.
  • Since the distribution is symmetric, 34% of the data lies between the mean and 60 ounces, and half of the data (50%) is below the mean.

Thus, the total percentage below 60 ounces is: 50%+34%=84%50\% + 34\% = 84\%

Summary:

  • (a) 99.7% of the widget weights lie between 20 and 80 ounces.
  • (b) 97.35% of the widget weights lie between 30 and 80 ounces.
  • (c) 84% of the widget weights lie below 60 ounces.

Would you like further details or explanations?


Here are five related questions you can explore:

  1. What percentage of widget weights lie between 40 and 60 ounces?
  2. What percentage of widget weights lie above 70 ounces?
  3. If the mean widget weight increased to 55 ounces, how would this affect the distribution?
  4. What is the probability that a randomly chosen widget weighs more than 80 ounces?
  5. How would the percentage of widget weights change if the standard deviation was reduced to 5 ounces?

Tip: The Empirical Rule is a quick approximation for normal distributions, but more precise calculations can be done using a Z-score table.

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Math Problem Analysis

Mathematical Concepts

Statistics
Normal Distribution
Empirical Rule

Formulas

Empirical Rule: 68-95-99.7
Standard deviation range: Mean ± k(Standard deviation)

Theorems

Empirical Rule (68-95-99.7 Rule)

Suitable Grade Level

Grades 9-12