The following table presents the heights (in inches) of a sample of college basketball play.
$\begin{array} { c c }
\hline \text { Height (in.) } & \text { Frequency } \
\hline 68 - 71 & 18 \
72 - 75 & 49 \
76 - 79 & 66 \
80 - 83 & 48 \
84 - 87 & 10 \
\hline
\end{array}$
Considering the data to be a population, approximate the standard deviation of the heights.
A)77.6
B)4
C)17.4
D)4.2

Question

The following table presents the heights (in inches) of a sample of college basketball play.
$\begin{array} { c c } 
\hline \text { Height (in.) } & \text { Frequency } \
\hline 68 - 71 & 18 \
72 - 75 & 49 \
76 - 79 & 66 \
80 - 83 & 48 \
84 - 87 & 10 \
\hline
\end{array}$
Considering the data to be a population, approximate the standard deviation of the heights.
A)77.6
B)4
C)17.4
D)4.2

Quizplus · Accepted Answer

To approximate the standard deviation of the heights, we first find the midpoint of each class (which represents the height value for that class), multiply each midpoint by the class frequency to find the sum of all heights, and then use the formula for standard deviation. The midpoint for each class is found by averaging the upper and lower bounds of each class. Then, using these midpoints and frequencies, we calculate the mean height and use the formula for standard deviation, which involves subtracting the mean from each midpoint, squaring the result, multiplying by the frequency, summing these values, dividing by the total number of observations, and finally taking the square root. The correct approximation is 4.2 inches, indicating variability in the heights of the college basketball players.

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