Question

the heights of 8th graders are modeled using the normal distribution shown below. the mean of the distribution is 62.7 in and the standard deviation is 1.1 in. 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 u v w height (in inches)

Explanation:

Step1: Find the value of \( V \)

In a normal distribution, the highest point of the curve is at the mean. Given the mean \(\mu = 62.7\), so \( V=\mu = 62.7\)

Step2: Calculate the distance from \( V \) to \( U \) and \( V \) to \( W \)

Since \(U\) and \(W\) are equidistant from \(V\) and using the empirical rule (68 - 95 - 99.7 rule). Looking at the scale, the distance from \(V\) to \(U\) (or \(V\) to \(W\)) is \( 3\) standard deviations. Given \(\sigma=1.1\)
The distance \(d = 3\sigma\)
\(d=3\times1.1 = 3.3\)

Step3: Calculate \( U \) and \( W \)

\(U=V - 3\sigma\)
\(U=62.7-3\times1.1=62.7 - 3.3=59.4\)
\(W=V + 3\sigma\)
\(W=62.7+3\times1.1=62.7 + 3.3=66\)

Step4: Determine the percentage of the shaded area

By the empirical rule, approximately \(99.7\%\) of the data lies within \(3\) standard deviations of the mean in a normal distribution.

Answer:

Percentage of total area shaded: \(99.7\%\)
\(U = 59.4\), \(V=62.7\), \(W = 66\)