The assumed mean method

What you will be able to do

• find the deviation of each class mark from an assumed mean;
• find the mean by the assumed mean method;
• explain why the answer does not depend on the number you assume.

Big numbers, big mistakes

An electricity office records how many units (kilowatt-hours, kWh) each of $50$ homes used in a month.

The direct method works: $\bar{x} = \dfrac{7940}{50} = 158.8$ kWh. But the products are large, and one slip in the multiplying spoils the answer. Can we use smaller numbers?

The idea

All the class marks are near $150$. So measure each one from $150$. The distance of a class mark $x$ from $150$ is the deviation $d = x - 150$. The number we measure from is the assumed mean, and we write it $a$. Here $a = 150$, and the deviations are $-40, -20, 0, 20, 40, 60$.

A deviation is negative when the class mark is below $a$, and positive when it is above. Now write each class mark as $x = a + d$. The total of all the values is

Divide by $N$:

In words, the mean is the assumed mean plus the average deviation. The sum $\sum fd$ is the correction. It is negative when the true mean lies below $a$.

The method, step by step

1. Find the class mark $x$ of every class.
2. Choose an assumed mean $a$. Pick a class mark near the middle, often the one with the largest frequency.
3. Find the deviation $d = x - a$ for every class.
4. Multiply: $fd$. Add: $\sum fd$.
5. Use $\bar{x} = a + \dfrac{\sum fd}{N}$.

This agrees with the direct method. The numbers in the $fd$ column are much smaller than in the $fx$ column.

Does the choice of $a$ matter? Let us try another one. Take $a = 170$. The deviations are $-60, -40, -20, 0, 20, 40$, and $\sum fd = -560$. Then $\bar{x} = 170 + \dfrac{-560}{50} = 170 - 11.2 = 158.8$ kWh. The same mean. Use the simulation to try every class mark.

The mean does not depend on $a$. If $a$ rises by $20$, every deviation falls by $20$, so $\sum fd$ falls by $20N$. Then $\dfrac{\sum fd}{N}$ falls by $20$ while $a$ rises by $20$, and the mean stays the same. Choose the $a$ that makes the arithmetic easy: near the middle, so that the deviations are small and the positive and negative products partly cancel.

What other books call it

A-Level books call this method coding: they code the data as $y = x - a$, find the mean of $y$, and add $a$ back. IGCSE and GCSE questions usually expect the direct method with midpoints, with a calculator. The assumed mean method is the one boards ask for when they say "by the assumed mean method".

Practise with new numbers

Try it in Python

This program uses every class mark in turn as the assumed mean. For each one it finds $\sum fd$ and the mean. Watch the mean.

Common mistakes

• Subtracting the wrong way round. The deviation is $x - a$, the class mark minus the assumed mean. Then a class mark below $a$ gets a negative deviation.
• Forgetting to add $a$ back. The fraction $\dfrac{\sum fd}{N}$ is only the correction. The mean is $a + \dfrac{\sum fd}{N}$.
• Dividing by the number of classes. Divide by $N = \sum f$.
• Losing a minus sign. When $\sum fd$ is negative, the mean is below $a$. Check that the answer lies between the lowest and the highest class mark.

Remember these

• The deviation is $d = x - a$, where $a$ is the assumed mean.
• $\bar{x} = a + \dfrac{\sum fd}{\sum f}$.
• Any $a$ gives the same mean. A class mark near the middle gives the smallest numbers.
• A negative $\sum fd$ means that the mean is below $a$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. The class marks are $20, 30, 40, 50, 60$ and the assumed mean is $a = 40$. List the deviations.

2. For $N = 20$, $a = 50$ and $\sum fd = 30$, find the mean.

3. The sum $\sum fd$ is negative. Is the mean above or below $a$?

4. A student chose $a = 150$ and another chose $a = 170$ for the same table. Will they get the same mean? Will they get the same $\sum fd$?

5. Why do we usually choose $a$ near the middle of the class marks?

References

• NCERT, Mathematics for Class 10, the chapter Statistics: the assumed mean method.
• OpenStax, Introductory Statistics, Chapter 2 Descriptive Statistics: the mean of grouped data, for the direct form.