Questions: Homework: Section 6.1 Score: 6.5/7 Answered: 6 / 7 Question 7 The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped with a mean of 64 ounces and a standard deviation of 11 ounces. Using the Empirical Rule, answer the following questions. Suggestion: Sketch the distribution. a) 68% of the widget weights lie between and b) What percentage of the widget weights lie between 31 and 75 ounces? % c) What percentage of the widget weights lie below 86 ? % Question Help: Video Submit Question

Solution Steps

According to the Empirical Rule, \(68\%\) of the data lies within \(1\) standard deviation of the mean. Given the mean \(\mu = 64\) ounces and standard deviation \(\sigma = 11\) ounces, we can calculate the range as follows:

\[ \text{Lower Bound} = \mu - \sigma = 64 - 11 = 53 \] \[ \text{Upper Bound} = \mu + \sigma = 64 + 11 = 75 \]

Thus, \(68\%\) of the widget weights lie between \(53\) and \(75\) ounces.

To find the percentage of widget weights between \(31\) and \(75\) ounces, we calculate the probabilities using the Z-scores:

\[ Z_{start} = \frac{31 - 64}{11} = -3.0 \] \[ Z_{end} = \frac{75 - 64}{11} = 1.0 \]

Using the cumulative distribution function \(\Phi\):

\[ P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(1.0) - \Phi(-3.0) = 0.84 \]

Thus, the percentage of widget weights between \(31\) and \(75\) ounces is:

\[ 84.0\% \]

Next, we calculate the percentage of widget weights below \(86\) ounces. The Z-score is calculated as follows:

\[ Z_{end} = \frac{86 - 64}{11} = 2.0 \]

Using the cumulative distribution function \(\Phi\):

\[ P = \Phi(Z_{end}) - \Phi(-\infty) = \Phi(2.0) - 0 = 0.9772 \]

Thus, the percentage of widget weights below \(86\) ounces is:

\[ 97.72\% \]

Final Answer