Question

the acme company manufactures widgets. the distribution of widget weights is bell - shaped with a mean of 48 ounces and a standard deviation of 7 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 27 and 55 ounces? %
c) what percentage of the widget weights lie below 62? %
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Explanation:

Step1: Recall the Empirical Rule for part (a)

The Empirical Rule states that for a bell - shaped (normal) distribution, about 68% of the data lies within 1 standard deviation of the mean.
If the mean \(\mu = 48\) and the standard deviation \(\sigma=7\), then the lower bound is \(\mu-\sigma\) and the upper bound is \(\mu + \sigma\).

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Step2: Calculate the number of standard deviations for part (b)

For \(x = 27\), \(\frac{\mu - x}{\sigma}=\frac{48 - 27}{7}=\frac{21}{7}=3\)
For \(x = 55\), \(\frac{x-\mu}{\sigma}=\frac{55 - 48}{7}=1\)
The percentage of data within \(k = 3\) standard deviations is 99.7% and within \(k = 1\) standard deviation is 68%.
The percentage of data from \(-3\sigma\) to \(- 1\sigma\) is \(\frac{99.7\%-68\%}{2}=15.85\%\)
The percentage of data from \(-1\sigma\) to \(+\infty\) is \(50\%+\frac{68\%}{2}=84\%\)
The percentage of data from \(27\) (i.e., \(-3\sigma\)) to \(55\) (i.e., \(+\sigma\)) is \(84\% - 15.85\%=83.85\%\)

Step3: Calculate the number of standard deviations for part (c)

For \(x = 62\), \(\frac{x-\mu}{\sigma}=\frac{62 - 48}{7}=\frac{14}{7}=2\)
The percentage of data within \(k = 2\) standard deviations is 95%. The percentage of data above \(k = 2\) standard deviations is \(\frac{100\%-95\%}{2}=2.5\%\)
The percentage of data below \(x = 62\) (i.e., \(+2\sigma\)) is \(100\%-2.5\% = 97.5\%\)

Answer:

a) \(41\) and \(55\)
b) \(83.85\)
c) \(97.5\)