Math Problem Statement

The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 35 ounces and a standard deviation of 8 ounces. Use the Empirical Rule to answer the following questions: a) 95% of the widget weights lie between _____ and _____. b) What percentage of the widget weights lie between 27 and 51 ounces? c) What percentage of the widget weights lie above 11 ounces?

Solution

We can solve these problems using the Empirical Rule (also known as the 68-95-99.7 rule), which applies to normal distributions.

Given:

  • Mean μ=35\mu = 35 ounces
  • Standard deviation σ=8\sigma = 8 ounces
  • The distribution is bell-shaped (normal distribution).

The Empirical Rule states:

  1. 68% of the data falls within 1 standard deviation (μ±1σ\mu \pm 1\sigma).
  2. 95% of the data falls within 2 standard deviations (μ±2σ\mu \pm 2\sigma).
  3. 99.7% of the data falls within 3 standard deviations (μ±3σ\mu \pm 3\sigma).

Let's solve each part:

a) 95% of the widget weights lie between:

For 95% of the data, the range is within 2 standard deviations of the mean.

μ2σ=352(8)=3516=19 ounces\mu - 2\sigma = 35 - 2(8) = 35 - 16 = 19 \text{ ounces} μ+2σ=35+2(8)=35+16=51 ounces\mu + 2\sigma = 35 + 2(8) = 35 + 16 = 51 \text{ ounces}

Thus, 95% of the widget weights lie between 19 and 51 ounces.

b) What percentage of the widget weights lie between 27 and 51 ounces?

Now, we calculate the number of standard deviations each value is from the mean.

  • 27 ounces=μ1σ=35827 \text{ ounces} = \mu - 1\sigma = 35 - 8
  • 51 ounces=μ+2σ=35+1651 \text{ ounces} = \mu + 2\sigma = 35 + 16

From the Empirical Rule:

  • Between μ1σ\mu - 1\sigma and μ+2σ\mu + 2\sigma is 34% (for μ1σ\mu - 1\sigma to μ\mu) and 47.5% (for μ\mu to μ+2σ\mu + 2\sigma).

Adding them up: 34%+47.5%=81.5%34\% + 47.5\% = 81.5\%

Thus, 81.5% of the widget weights lie between 27 and 51 ounces.

c) What percentage of the widget weights lie above 11 ounces?

We calculate how many standard deviations 11 ounces is from the mean.

11=μ3σ=353(8)=352411 = \mu - 3\sigma = 35 - 3(8) = 35 - 24

From the Empirical Rule, 99.7% of the data lies within 3 standard deviations, which leaves:

100%99.7%=0.3%100\% - 99.7\% = 0.3\%

Half of this 0.3% is below 11 ounces (since the distribution is symmetric), so the percentage above 11 ounces is:

100%0.15%=99.85%100\% - 0.15\% = 99.85\%

Thus, 99.85% of the widget weights lie above 11 ounces.


Would you like more details on any part of this solution?

Here are some follow-up questions for practice:

  1. What percentage of the weights lie between 35 and 43 ounces?
  2. What percentage of the weights lie below 19 ounces?
  3. If the mean increased to 40 ounces but the standard deviation remained the same, how would the ranges change?
  4. What weight corresponds to 68% of the distribution above it?
  5. What is the probability that a randomly selected widget weighs more than 51 ounces?

Tip: When using the Empirical Rule, always compare your values to the mean and standard deviation to understand their relative positions.

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Math Problem Analysis

Mathematical Concepts

Normal Distribution
Empirical Rule
Standard Deviation

Formulas

Empirical Rule: 68% within 1 standard deviation, 95% within 2 standard deviations, 99.7% within 3 standard deviations
Standard deviation calculation: x = mean ± n * standard deviation

Theorems

Empirical Rule

Suitable Grade Level

Grades 10-12