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This page sets out the method for averages from frequency tables in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the Averages from frequency tables topic page.
When each row is a single value with its frequency.
When rows are classes like — you cannot know the exact values.
When you need the middle value of data in a table.
For the mean, add a column of value × frequency.
The total of that column, divided by the total frequency, is the mean.
Five 2s and one 10 are six numbers, not two rows.
The denominator is the total frequency.
The median is the middle of the ordered list.
For 21 items it is the 11th, even if that sits in a later row of the table.
The mode is the value with the highest frequency, not the highest value.
Look at the frequency column.
For grouped data, use the midpoint of each class as the value.
The mean is then an estimate, because the raw values are hidden.
The range uses the largest and smallest values, not the frequencies.
In grouped data the range is estimated from the class ends.
The mean from a table uses frequency times value, then divides by the total frequency. Five homes with 2 pets contribute 2 × 5 = 10, not one row. Dividing by the number of rows ignores how often each value appears.
For values 2, 6 and 10 with frequencies 5, 3 and 2, the products are 10, 18 and 20. The total is 48 and the frequencies add to 10, so the mean is 4.8. Averaging 2, 6 and 10 to get 6 uses the wrong count.
Method: 1. Multiply every value by how often it appears. 2. Add those products. 3. Add the frequency column to get the count. 4. Divide the product total by the frequency total.
The mode is the value with the highest frequency, not the greatest value. Frequency 5 on the value 2 beats frequency 2 on the value 10, so the mode is 2. The greatest value belongs in a range, not in the mode.
If two values share the top frequency, state both modes. In grouped data, name the modal class. Do not call its midpoint the mode. The midpoint is only a stand-in for the estimated mean.
Method: 1. Read down the frequency column. 2. Find the greatest frequency. 3. Take the value sitting in the row you marked. 4. With grouped data, give the class that has the highest frequency.
Grouped data hide the raw values, so a mean from classes is an estimate. Use the midpoint, halfway from the start of the class to the end. For 0 ≤ t < 10 the midpoint is 5. Multiply each midpoint by the class frequency, add, and divide by the total frequency.
Say estimate, because every item in a class was treated as the midpoint. Using the class end, such as 10 instead of 5, is a different calculation. The class width is not the divisor.
Method: 1. Find the midpoint of each class. 2. Multiply the midpoint by the class frequency. 3. Add the products and add the frequencies. 4. Divide, and call the result an estimate.
For a mean from a frequency table, add a third column headed f × x and a totals row. Fill the column by multiplying each value by its frequency. At the bottom write the total frequency and the total of the f × x column. The mean is the second total divided by the first. A table like this keeps the working tidy and lets a marker award the method mark even if one product is wrong.
Check the total frequency against the number of people in the question. If the question says 32 people took the survey, your frequencies must add to 32. A wrong total frequency means a wrong mean. Do not divide by the number of rows. A table with five rows may hold thirty values. Give the answer as a decimal if the question asks, and check that it lies between the smallest and the largest value.
Method: 1. Add a column for f × x and a row for totals. 2. Multiply each value by its frequency and write the product. 3. Add the frequencies and add the products. 4. Divide the total of the products by the total frequency.
Tip: The mean must be between the smallest and largest values. If it is not, a product or a total is wrong.
A frequency table does not list every value, so find the position of the median first. With n values in total, the median is in position (n + 1)/2. For 45 values that is the 23rd value. For 31 values it is the 16th. If n is even, the median is half way between the two middle positions.
Then build a running total of the frequencies down the table. The value whose running total first reaches or passes the position is the median. With frequencies 6, 11, 13, 9 and 6 for the scores 1 to 5, the running totals are 6, 17, 30, 39 and 45. The 23rd value is in the third group, so the median is 3. The median is the score, not the frequency.
Method: 1. Add the frequencies to find n. 2. Work out the position (n + 1) ÷ 2. 3. Write a running total of the frequencies. 4. Find the first running total that reaches the position and read the value.
The mode is the value with the highest frequency. The question asks for the value, not for the frequency. In a table where the score 3 has frequency 13 and is the highest, the mode is 3, not 13. For grouped data you give a modal class, such as 160 < h ≤ 170. The modal class is the class with the highest frequency, which is not always the class that contains the median.
This is why tables need all three averages. The mean uses every value and is pulled by extreme values. The median is the middle value and ignores them. The mode is the most frequent. A question may ask which average you would use, or ask you to compare two tables. Quote the numbers and say what they mean in the context.
Method: 1. Find the largest frequency in the table. 2. Read off the value or the class that goes with it. 3. For grouped data write the class using the inequality. 4. Do not confuse the mode with the frequency, and do not confuse the modal class with the median class.
For grouped data the exact values are not known, so use the midpoint of each class as a stand-in. Add the two ends and halve. A class 20 < t ≤ 30 has midpoint 25. Multiply each midpoint by its frequency, add, and divide by the total frequency. The answer is an estimate, because the true values in each class could be anywhere between the ends.
Harder questions give the mean and leave one frequency as a letter. Write the mean as a fraction with f in both the top and the bottom, set it equal to the given mean, then multiply out and solve. If the estimated mean of three classes with midpoints 5, 15, 25 is 16, then (5 × 4 + 15f + 25 × 6)/(4 + f + 6) = 16 gives f = 10. Check by recalculating the mean.
Method: 1. Find the midpoint of every class. 2. Multiply each midpoint by its frequency and add the products. 3. Divide by the total frequency for the estimated mean. 4. If a frequency is unknown, write an equation, solve it, and check the mean.
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