Question

using the empirical rule to identify values and percentages of a normal...
the scores on a standardized test for a certain year are modeled using the normal distribution shown below.
the mean of the distribution is 72.7 points and the standard deviation is 5.4 points.
in the figure, v is a number along the axis and is under the highest part of the curve.
and, u and w are numbers along the axis that are each the same distance away from v.
use the empirical rule to choose the best value for the percentage of the area under the curve that is shaded, and find the values of u, v, and w.
percentage of total area shaded: select

Explanation:

Step1: Identify the value of \( V \)

Since \( V \) is under the highest part of the normal - curve, \( V \) is the mean. Given the mean of the distribution is \( 72.7 \) points, so \( V = 72.7 \).

Step2: Calculate the number of standard deviations from the mean

The distance from \( V\) to \( U\) (or \( V\) to \( W\)): \( 72.7−U=W - 72.7\). Let's find the number of standard deviations \( n\) using the formula \(x=\mu\pm n\sigma\).
If \(x = U\) and \(\mu = 72.7\), \(\sigma=5.4\). Let's assume \(U=\mu - n\sigma\) and \(W=\mu + n\sigma\).
We know that \(72.7 - U=W - 72.7\). If we check \(n = 3\), then \(U=72.7-3\times5.4=72.7 - 16.2=56.5\) (close to \(55\) in the figure, considering possible rounding in the problem - presentation) and \(W=72.7 + 3\times5.4=72.7+16.2 = 88.9\) (close to \(90\) in the figure, considering possible rounding in the problem - presentation).
According to the empirical rule:

• Approximately \(68\%\) of the data lies within \(1\) standard deviation of the mean (\(\mu\pm\sigma\)), approximately \(95\%\) of the data lies within \(2\) standard deviations of the mean (\(\mu\pm2\sigma\)), and approximately \(99.7\%\) of the data lies within \(3\) standard deviations of the mean (\(\mu\pm3\sigma\)).

Answer:

• Percentage of total area shaded: \(99.7\%\)
• \(U = 72.7-3\times5.4=56.5\)
• \(V = 72.7\)
• \(W=72.7 + 3\times5.4=88.9\)