The mean and median can be estimated from tables of grouped data.
The class interval which contains the most values is known as the modal class.
Worked Examples
The table below gives data on the heights, in cm, of 51 children.
Estimate the mean height.
To estimate the mean, the mid-point of each interval should be used.
| Mean | = |
| = 161 (to the nearest cm) |
Find the median class.
The median is the 26th value. In this case it lies in the 160 ≤ h < 170 class interval. Note that you can estimate the median height.
The 4th value in the interval is needed. It is estimated as
160 + × 10 = 162 (to the nearest cm)
Find the modal class.
The modal class is 160 ≤ h < 170 as it contains the most values.
The ages of students in a small primary school were recorded in the table below.
Estimate the mean.
To estimate the mean, we must use the mid-point of each interval; so, for example for '5 – 6', which really means
5 ≤ age < 7
the mid-point is taken as 6.
| Mean | = |
| = 8.2 (to 1 decimal place) |
Estimate the median.
The median is given by the 54th value, which we have to estimate. There are 29 values in the first interval, so we need to estimate the 25th value in the second interval. As there are 40 values in the second interval, the median is estimated as being
of the way along the second interval. This has width 9 − 7 = 2 years, so the median is estimated by
× 2 = 1.25
from the start of the interval. Therefore the median is estimated as
7 + 1.25 = 8.25 years
Find the modal class.
The modal class is the 7 – 8 age group.
Worked Example 1 uses what are called continuous data, since height can be of any value. (Other examples of continuous data are weight, temperature, area, volume and time.)
The next example uses discrete data, that is, data which can take only a particular value, such as the integers 1, 2, 3, 4, . . . in this case.
The calculations for mean and mode are not affected but estimation of the median requires replacing the discrete grouped data with an approximate continuous interval.
The number of days that students were missing from school due to sickness in one year was recorded.
Estimate the mean.
The estimate is made by assuming that all the values in a class interval are equal to the midpoint of the class interval.
| Mean | = |
| = 9.875 days |
Find the median class.
As there are 40 students, we need to consider the mean of the 20th and 21st values. These both lie in the 6–10 class interval, which is really the 5.5–10.5 class interval, so this interval contains the median.
[ You could also estimate the median as follows.
As there are 12 values in the first class interval, the median is found by considering the 8th and 9th values of the second interval.
As there are 11 values in the second interval, the median is estimated as being of the way along the second interval.
But the length of the second interval is 10.5 − 5.5 = 5, so the median is estimated by
× 5 ≈ 3.86
from the start of this interval. Therefore the median is estimated as
5.5 + 3.86 = 9.36 ]
Find the modal class.
The modal class is 1–5, as this class contains the most entries.
The table shows the distribution of scores or 40 students on a Mathematics test.
Estimate the mean score obtained on the test.
| Mean | = (11 × 4 + 14 × 6 + 17 × 13 + 20 × 9 + 23 × 8) ÷ 40 |
| = (44 + 84 + 221 +180 + 184) ÷ 40 | |
| = | |
| = 17.825 |
Estimate the probability that a student selected at random would score at most 15 marks on the test.
probability of score at most 15 = = = 0.625
Using the scale of 1 cm to represent 1 unit on the frequency axis and 2 cm to represent 5 units on the scores axis, use graph paper to draw a frequency polygon to represent the distribution of scores shown in the table.
Exercises
