Combined mean and combined variance

What you will be able to do

• find the mean of two or more groups taken together;
• find the combined variance and standard deviation from the group means and standard deviations;
• correct a data set when one reading was wrong, without going back to every value.

The head teacher's question

A head teacher has the results of two classes. Class A has $30$ students, with mean $60$ and standard deviation $8$. Class B has $20$ students, with mean $70$ and standard deviation $6$. She wants the mean and the standard deviation of all $50$ students. The marks are not in front of her. Only the three numbers for each class are. Can she still find the answer?

The combined mean

The total marks of a group is its size times its mean. So the total for both groups is $n_1\bar{x}_1 + n_2\bar{x}_2$, and there are $n_1 + n_2$ students. The combined mean is

It is a weighted mean: each group mean counts as many times as its group has members. Here $\bar{x} = \dfrac{30 \times 60 + 20 \times 70}{50} = \dfrac{3200}{50} = 64$. Notice that $64$ is not the middle of $60$ and $70$, because the larger group pulls it nearer to $60$.

You cannot average the standard deviations

It is tempting to find the combined standard deviation as $\dfrac{8 + 6}{2} = 7$, or as a weighted mean of $8$ and $6$. Both are wrong. Think of two groups where every value is the same, so each has standard deviation $0$. If the two values differ, say all the values in one group are $10$ and in the other $20$, the combined data are certainly spread out. So the spread of the whole depends on how far apart the group means are, as well as on the spread inside each group.

The combined variance

In the last course we met the identity: for any centre $a$, the mean square deviation about $a$ equals $\sigma^2 + (\bar{x} - a)^2$. Apply it to group $i$ with the centre $a = \bar{x}$, the combined mean. The sum of the squared distances of group $i$ from $\bar{x}$ is then $n_i\left(\sigma_i^2 + d_i^2\right)$, where $d_i = \bar{x}_i - \bar{x}$. Add the two groups and divide by $n_1 + n_2$:

Read it in words: the combined variance is the weighted mean of the spread inside each group ($\sigma_i^2$) and the spread of the group means ($d_i^2$).

For the two classes, $d_1 = 60 - 64 = -4$ and $d_2 = 70 - 64 = 6$. Then

The combined standard deviation is a little larger than both $8$ and $6$, because the means $60$ and $70$ are apart.

For samples, A-Level and IB courses combine groups in the same spirit, but with the sums of squares and the divisor $n - 1$. The formula above is for data that are the whole group, which is how the Indian syllabus states it.

Correcting a reading

A reading was copied wrongly. We keep the two totals, $\sum x$ and $\sum x^2$, and then replace the bad value by the good one.

For $10$ values with mean $20$ and standard deviation $4$, a reading of $12$ should have been $22$.

1. The old totals. $\sum x = 10 \times 20 = 200$. $\sum x^2 = 10\,(16 + 400) = 4160$.
2. The corrected totals. $\sum x = 200 - 12 + 22 = 210$. $\sum x^2 = 4160 - 144 + 484 = 4500$.
3. The new mean. $\dfrac{210}{10} = 21$.
4. The new variance. $\dfrac{4500}{10} - 21^2 = 450 - 441 = 9$, so the standard deviation is $3$.

The same steps handle a value that is added or removed.

Practise with new numbers

Try it in Python

The program combines the two classes of the lesson. It finds the combined mean, the two distances $d_1$ and $d_2$, and the combined variance.

Common mistakes

• Averaging the means without weights. Weight each mean by the size of its group.
• Averaging the standard deviations. Use the combined variance formula.
• Forgetting $d_i^2$. The distance of each group mean from the combined mean matters.
• Dividing by $n_1$ or $n_2$. Divide by $n_1 + n_2$.
• Correcting only the mean. A wrong reading changes $\sum x^2$ too.

Remember these

• $\bar{x} = \dfrac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2}$.
• $\sigma^2 = \dfrac{n_1\left(\sigma_1^2 + d_1^2\right) + n_2\left(\sigma_2^2 + d_2^2\right)}{n_1 + n_2}$, with $d_i = \bar{x}_i - \bar{x}$.
• When the two means are equal, $d_1 = d_2 = 0$ and the combined variance is the weighted mean of the two variances.
• To correct a reading, change $\sum x$ and $\sum x^2$, then find the mean and the variance again.

Check yourself

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

1. A group of $15$ boys has mean height $150\ \text{cm}$ and a group of $25$ girls has mean height $142\ \text{cm}$. Find the mean height of all $40$.

2. Two groups both have mean $50$ and standard deviation $6$. What are the mean and the standard deviation of the two groups together?

3. In one group of $10$ values every value is $10$. In another group of $10$ values every value is $20$. Find the mean and the standard deviation of all $20$ values.

4. The mean of $8$ readings is $15$. One reading, written as $12$, should have been $20$. Find the correct mean.

5. Group A has $40$ values with variance $25$. Group B has $60$ values with variance $16$. Both groups have mean $50$. Find the combined variance.

References

• OpenStax, Introductory Statistics, Chapter 2, section on the mean and the spread.
• NCERT, Mathematics for Class XI, Chapter 13 (Statistics), exercises on the combined mean and the corrected standard deviation.