The following table presents the heights (in inches) of a sample of college basketball players.

$\begin{array} { c c }
\hline \text { Height (in.) } & \text { Frequency } \
\hline 68 - 71 & 2 \
72 - 75 & 2 \
76 - 79 & 4 \
80 - 83 & 2 \
84 - 87 & 3 \
\hline
\end{array}$

Considering the data to be a population, approximate the variance of the heights.
A) 5.4
B) 5.6
C) 29.2
D) 31.6

Question

The following table presents the heights (in inches) of a sample of college basketball players.

$\begin{array} { c c } 
\hline \text { Height (in.) } & \text { Frequency } \
\hline 68 - 71 & 2 \
72 - 75 & 2 \
76 - 79 & 4 \
80 - 83 & 2 \
84 - 87 & 3 \
\hline
\end{array}$
 
Considering the data to be a population, approximate the variance of the heights.
A) 5.4
B) 5.6
C) 29.2
D) 31.6

Quizplus · Accepted Answer

To approximate the variance of a population, we use the formula $	ext{Var} \approx \dfrac{\sum f(x-\overline{x})^2}{n}$ where $f$ is the frequency of each class, $x$ is the midpoint of each class, $\overline{x}$ is the mean of the data, and $n$ is the total number of data points.

Using the given data, we can find the mean height $\overline{x}$ by taking the weighted average of the class midpoints:
$$\overline{x} = \dfrac{(69.5)(2) + (73.5)(2) + (77.5)(4) + (81.5)(2) + (85.5)(3)}{13} = 78$$

Then, we can plug in the values for each class to find the numerator of the variance formula:
\begin{align*}
\sum f(x-\overline{x})^2 &= (2)(69.5-78)^2 + (2)(73.5-78)^2 + (4)(77.5-78)^2 \
&+ (2)(81.5-78)^2 + (3)(85.5-78)^2 \
&= 390.5
\end{align*}

Finally, we can calculate the variance approximation:
$$	ext{Var} \approx \dfrac{\sum f(x-\overline{x})^2}{n} = \dfrac{390.5}{13} \approx 30.04$$

Therefore, the best choice is (C) 29.2, which is the closest approximation to our calculated value of 30.04.

The Following Table Presents the Heights (In Inches) of a Sample