Questions: The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 62 ounces and a standard deviation of 7 ounces. Use the Empirical Rule. Suggestion: sketch the distribution in order to answer these questions. a) 68% of the widget weights lie between and b) What percentage of the widget weights lie between 41 and 69 ounces? % c) What percentage of the widget weights lie below 76 ?

The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 62 ounces and a standard deviation of 7 ounces. Use the Empirical Rule. Suggestion: sketch the distribution in order to answer these questions. a) 68% of the widget weights lie between and b) What percentage of the widget weights lie between 41 and 69 ounces? % c) What percentage of the widget weights lie below 76 ?
Transcript text: The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 62 ounces and a standard deviation of 7 ounces. Use the Empirical Rule. Suggestion: sketch the distribution in order to answer these questions. a) $68 \%$ of the widget weights lie between $\square$ and $\square$ b) What percentage of the widget weights lie between 41 and 69 ounces? $\square$ \% c) What percentage of the widget weights lie below 76 ? $\square$
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Solution

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Solution Steps

Step 1: Understanding the Empirical Rule

The Empirical Rule states that for a bell-shaped (normal) distribution:

  • Approximately \( 68\% \) of the data lies within \( 1 \) standard deviation of the mean.
  • Approximately \( 95\% \) of the data lies within \( 2 \) standard deviations of the mean.
  • Approximately \( 99.7\% \) of the data lies within \( 3 \) standard deviations of the mean.

Given:

  • Mean (\( \mu \)) = \( 62 \) ounces
  • Standard deviation (\( \sigma \)) = \( 7 \) ounces
Step 2: Solving Part (a)

To find the range where \( 68\% \) of the widget weights lie:

  • Calculate \( 1 \) standard deviation below the mean: \( \mu - \sigma = 62 - 7 = 55 \) ounces.
  • Calculate \( 1 \) standard deviation above the mean: \( \mu + \sigma = 62 + 7 = 69 \) ounces.

Thus, \( 68\% \) of the widget weights lie between \( 55 \) and \( 69 \) ounces.

Step 3: Solving Part (b)

To find the percentage of widget weights between \( 41 \) and \( 69 \) ounces:

  • First, determine how many standard deviations \( 41 \) and \( 69 \) are from the mean.
    • \( 41 \) is \( \frac{41 - 62}{7} = -3 \) standard deviations from the mean.
    • \( 69 \) is \( \frac{69 - 62}{7} = 1 \) standard deviation from the mean.
  • According to the Empirical Rule:
    • The range from \( \mu - 3\sigma \) to \( \mu + 3\sigma \) covers \( 99.7\% \) of the data.
    • The range from \( \mu - \sigma \) to \( \mu + \sigma \) covers \( 68\% \) of the data.
  • Therefore, the percentage between \( 41 \) and \( 69 \) ounces is \( 99.7\% - (100\% - 68\%) = 68\% + 16\% = 84\% \).
Step 4: Solving Part (c)

To find the percentage of widget weights below \( 76 \) ounces:

  • Determine how many standard deviations \( 76 \) is from the mean:
    • \( \frac{76 - 62}{7} = 2 \) standard deviations above the mean.
  • According to the Empirical Rule:
    • The range from \( \mu - 2\sigma \) to \( \mu + 2\sigma \) covers \( 95\% \) of the data.
    • Since the distribution is symmetric, \( 2.5\% \) of the data lies above \( \mu + 2\sigma \).
  • Therefore, the percentage below \( 76 \) ounces is \( 100\% - 2.5\% = 97.5\% \).

The remaining parts of the question are not addressed as per the guidelines.

Final Answer

a) \( \boxed{55 \text{ and } 69} \)
b) \( \boxed{84\%} \)
c) \( \boxed{97.5\%} \)

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