$$\begin{array}{l}
\begin{array} { | l | l | l | }
\hline \text { Grade } & \text { Class Boundaries } & \text { Frequency } \
\hline \mathrm { A } & 89.5 - 99.5 & 4 \
\hline \mathrm { B } & 79.5 - 89.5 & 7 \
\hline \mathrm { C } & 69.5 - 79.5 & 11 \
\hline \mathrm { D } & 59.5 - 69.5 & 3 \
\hline \mathrm { F } & 49.5 - 59.5 & 3 \
\hline
\end{array}\
\text { For the table below, calculate the variance }
\end{array}$$
A) 141.4
B) 145.6
C) 133.4
D) 136.0

Question

$$\begin{array}{l}
\begin{array} { | l | l | l | } 
\hline \text { Grade } & \text { Class Boundaries } & \text { Frequency } \
\hline \mathrm { A } & 89.5 - 99.5 & 4 \
\hline \mathrm { B } & 79.5 - 89.5 & 7 \
\hline \mathrm { C } & 69.5 - 79.5 & 11 \
\hline \mathrm { D } & 59.5 - 69.5 & 3 \
\hline \mathrm { F } & 49.5 - 59.5 & 3 \
\hline
\end{array}\
\text { For the table below, calculate the variance }
\end{array}$$
A) 141.4
B) 145.6
C) 133.4
D) 136.0

Quizplus · Accepted Answer

First, we need to find the mean, denoted by $\bar{x}$, which can be calculated using the formula:
$$\bar{x} = \frac{\sum f_ix_i}{n}$$
where $f_i$ is the frequency of the $i$th class, $x_i$ is the midpoint of the $i$th class, and $n$ is the total frequency. The midpoint is simply the average of the upper and lower class boundaries.

Using the table, we can compute:
\begin{align*}
\bar{x} &= \frac{(4 	imes 94.5) + (7 	imes 84.5) + (11 	imes 74.5) + (3 	imes 64.5) + (3 	imes 54.5)}{28}\
&= 77.8214
\end{align*}

Next, we need to find the variance, denoted by $s^2$, which can be calculated using the formula:
$$s^2 = \frac{\sum f_i(x_i - \bar{x})^2}{n-1}$$
where $n$ is the total frequency.

Using the table, we can compute:
\begin{align*}
s^2 &= \frac{(4 	imes (94.5 - 77.8214)^2) + (7 	imes (84.5 - 77.8214)^2)\
&\quad + (11 	imes (74.5 - 77.8214)^2) + (3 	imes (64.5 - 77.8214)^2)\
&\quad + (3 	imes (54.5 - 77.8214)^2)}{28-1}\
&\approx 136.0
\end{align*}
Therefore, the best choice is D, with a variance of approximately 136.0.

A) 1414 B) 145

A) 1414
B) 145