Consider the following grouped data and calculate the mean, the modal group and the median group.
Mass (\(\text{kg}\)) | Count |
\(40 < m \le 45\) | \(\text{7}\) |
\(45 < m \le 50\) | \(\text{10}\) |
\(50 < m \le 55\) | \(\text{15}\) |
\(55 < m \le 60\) | \(\text{12}\) |
\(60 < m \le 65\) | \(\text{6}\) |
To find the mean we use the middle value for each group. The count then tells us how many times that value occurs in the data set. Therefore the mean is:
\begin{align*}
\text{mean } & = \frac{7(43) + 10(48) + 15(53) + 12(58) + 6(63)}{7 + 10 + 15 + 12 + 6} \\
& = \frac{2650}{50} \\
& = 53
\end{align*}
The modal group is the group with the highest number of data values. This is \(50 <m \le 55\) with 15 data values.
The median group is the central group. There are 5 groups and so the central group is the third one: \(50 <m \le 55\).
Mean: \(\text{52}\); Modal group: \(50 <m \le 55\); Median group: \(50 < m \le 55\).
Find the mean, the modal group and the median group in this data set of how much time people needed to complete a game.
Time (s) | Count |
\(35 < t \le 45\) | \(\text{5}\) |
\(45 < t \le 55\) | \(\text{11}\) |
\(55 < t \le 65\) | \(\text{15}\) |
\(65 < t \le 75\) | \(\text{26}\) |
\(75 < t \le 85\) | \(\text{19}\) |
\(85 < t \le 95\) | \(\text{13}\) |
\(95 < t \le 105\) | \(\text{6}\) |
To find the mean we use the middle value for each group. The count then tells us how many times that value occurs in the data set. Therefore the mean is:
\begin{align*}
\text{mean } & = \frac{5(\text{40,5}) + 11(\text{50,5}) + 15(\text{60,5}) + 26(\text{70,5}) + 19(\text{80,5}) + 13(\text{90,5}) + 6(\text{100,5})}{5 + 11 + 15 + 26 + 19 + 13 + 6} \\
& = \frac{\text{6 807,5}}{95} \\
& = \text{71,66}
\end{align*}
The modal group is the group with the highest number of data values. This is \(65 <m \le 75\) with 26 data values.
The median group is the central group. There are 7 groups and so the central group is the fourth one: \(65 <m \le 75\).
Mean: \(\text{70,66}\); Modal group: \(65 < t \le 75\); Median group: \(65 < t \le 75\).