Mean of Grouped Data

Only the counts are left

A gardener measures 10 seedlings. She does not keep each height; she records only the class it falls in. Writing h for a height in centimeters, 4 seedlings are in the class $0\le h<10$0 ≤ h < 10 and 6 are in the class $10\le h<20$10 ≤ h < 20.

The mean is the total of the values divided by how many values there are. But the values are gone. Her record says that 4 seedlings are shorter than 10 cm, and nothing about how much shorter, so there are no heights to add.

The histogram shows how many seedlings are in each class, 4 and 6, and not where each height lies inside its class.

One table, many possible data sets

Here are two different sets of heights that both give this table. In the first, every seedling is near the bottom of its class: 1, 1, 2 and 2 cm, then 10, 11, 11, 12, 12 and 13 cm. Their total is 6 + 69 = 75 cm, and their mean is 75 ÷ 10 = 7.5 cm.

In the second, every seedling is near the top of its class: 8, 9, 9 and 9 cm, then 17, 18, 18, 19, 19 and 19 cm. Their total is 35 + 110 = 145 cm, and their mean is 145 ÷ 10 = 14.5 cm.

Both sets have 4 heights in the first class and 6 in the second, so the table cannot tell them apart, yet their means differ by 7 cm. From grouped data alone the mean cannot be found exactly. It can only be estimated.

The first set: 4 dots below 10 and 6 dots from 10 up to 20, all low in their classes. The mean is 7.5 cm.

The second set has the same counts, 4 and 6, with every dot high in its class. The mean is 14.5 cm.

Midpoints stand in for the values

The usual value to stand in for every value in a class is its midpoint, the number halfway between its two ends. The midpoint of $0\le h<10$0 ≤ h < 10 is $\frac{\left(0+10\right)}{2}=5$(0 + 10)/2 = 5, and the midpoint of $10\le h<20$10 ≤ h < 20 is $\frac{\left(10+20\right)}{2}=15$(10 + 20)/2 = 15.

So treat every seedling in the first class as 5 cm tall and every seedling in the second class as 15 cm tall. When the real heights are spread through each class, some lie above the midpoint and some below, and those errors partly cancel when the heights are added. Then average as usual.

The midpoints, 5 and 15, sit halfway along the classes 0 to 10 and 10 to 20.

The estimated mean

With every value replaced by its midpoint, the data is 5 four times and 15 six times. Multiply each midpoint by its frequency and add: 5 × 4 + 15 × 6 = 20 + 90 = 110 cm. Divide by the number of seedlings: 110 ÷ 10 = 11 cm.

So the estimated mean height is 11 cm. Always say that it is an estimate. The two sets of heights with this same table have means of 7.5 cm and 14.5 cm, and the estimate, 11 cm, lies between them.

The estimate, 11, is closer to 15 than to 5, because the second class holds more seedlings. Each midpoint is weighted by its frequency. Taking the mean of the two midpoints, $\frac{\left(5+15\right)}{2}=10$(5 + 15)/2 = 10, treats the two classes as if they held equally many seedlings, and they do not.

Four seedlings at 5 cm and six at 15 cm balance at 11, pulled toward the class with more seedlings in it.

Classes with unequal ends

Classes do not always start at round numbers or have the same width. The midpoint is always found the same way: add the two ends of the class and halve. For times t in minutes, the class $20\le t<35$20 ≤ t < 35 has midpoint $\frac{\left(20+35\right)}{2}=27.5$(20 + 35)/2 = 27.5, and the class $35\le t<60$35 ≤ t < 60 has midpoint $\frac{\left(35+60\right)}{2}=47.5$(35 + 60)/2 = 47.5.

Do not take 5 more than the start of each class out of habit: 25 is the midpoint of a class from 20 to 30, not of one from 20 to 35. A class 15 wide has its midpoint 7.5 above its start.

The modal class

With grouped data there is no single most common value, because the values are gone. What the table does show is the class with the largest frequency, called the modal class. For the seedlings it is $10\le h<20$10 ≤ h < 20, with 6 seedlings. The modal class is an interval of heights; the 6 is its frequency, not the answer.