Combined mean and SD; effect of transformations

When two datasets are joined into one, we need formulas for the combined mean and combined variance. We also need to know how the mean and SD change when each observation undergoes a linear transformation.

Effect of a linear transformation $y = a + b x$

If $y_i = a + b x_i$ for $i = 1, \dots, n$, then:

• Mean. $\bar{y} = a + b \bar{x}$.
• Variance. $\sigma_y^2 = b^2 \sigma_x^2$.
• Standard deviation. $\sigma_y = |b| \sigma_x$.

In words: the mean shifts by $a$ and scales by $b$; the SD only scales by $|b|$ (it ignores the shift).

Combined mean

Suppose dataset 1 has $n_1$ observations with mean $\bar{x}_1$ and dataset 2 has $n_2$ observations with mean $\bar{x}_2$. The combined mean of all $n_1 + n_2$ observations is the weighted average: $\bar{x}_{12} = \frac{n_1 \bar{x}_1 + n_2 \bar{x}_2}{n_1 + n_2}.$

Combined variance

The combined variance is more involved. With $d_i = \bar{x}_i - \bar{x}_{12}$: $\sigma_{12}^2 = \frac{n_1 (\sigma_1^2 + d_1^2) + n_2 (\sigma_2^2 + d_2^2)}{n_1 + n_2}.$

The terms $d_i^2$ account for the deviation of each subgroup's mean from the combined mean. If both subgroups have the same mean, the formula reduces to a straightforward weighted average of variances.

Why combined variance contains $d_i^2$ terms

In each subgroup, $\sigma_i^2$ measures spread around $\bar{x}_i$ , not around the combined mean. To get spread around the combined mean, we add the "between-group" contribution $d_i^2$.

Worked examples

Example 1. Mean of $20$ items is $50$, SD is $5$. Each is multiplied by $3$ and then $4$ is added (so $y_i = 4 + 3 x_i$). Find new mean and SD.

$\bar{y} = 4 + 3 \cdot 50 = 154$. $\sigma_y = 3 \cdot 5 = 15$.

Example 2. Group A: $50$ items, mean $20$, SD $4$. Group B: $40$ items, mean $30$, SD $5$. Find combined mean and SD.

$\bar{x}_{12} = (50 \cdot 20 + 40 \cdot 30)/90 = (1000 + 1200)/90 = 24.44$.

$d_1 = 20 - 24.44 = -4.44$, $d_2 = 30 - 24.44 = 5.56$.

$\sigma_{12}^2 = (50(16 + 19.75) + 40(25 + 30.86))/90 = (50 \cdot 35.75 + 40 \cdot 55.86)/90 = (1787.5 + 2234.4)/90 = 4021.9/90 \approx 44.69$.

$\sigma_{12} \approx 6.68$.

Example 3. A group of $25$ students has mean weight $40$ kg and SD $3$. Another group of $30$ has mean $45$ and SD $4$. Find the combined SD.

$\bar{x}_{12} = (25 \cdot 40 + 30 \cdot 45)/55 = (1000 + 1350)/55 = 42.73$.

$d_1 = -2.73$, $d_2 = 2.27$.

$\sigma_{12}^2 = (25(9 + 7.45) + 30(16 + 5.15))/55 = (411.25 + 634.55)/55 = 19.01$.

$\sigma_{12} \approx 4.36$.

Example 4. If $y = (x - 50)/4$ and $\bar{y} = 2, \sigma_y = 3$, find $\bar{x}$ and $\sigma_x$.

$x = 50 + 4 y$. $\bar{x} = 50 + 4 \cdot 2 = 58$. $\sigma_x = 4 \cdot 3 = 12$.

Example 5. The mean and SD of $10$ observations are $30$ and $5$. By mistake one observation was recorded as $20$ instead of $40$. Find the corrected mean and SD.

Original $\sum x = 300$. Corrected $\sum x = 300 - 20 + 40 = 320$. Corrected mean $= 32$.

Original $\sigma^2 = 25 \Rightarrow \sum x^2 / 10 - 900 = 25 \Rightarrow \sum x^2 = 9250$.

Corrected $\sum x^2 = 9250 - 400 + 1600 = 10450$.

New $\sigma^2 = 10450/10 - 32^2 = 1045 - 1024 = 21$. $\sigma = \sqrt{21} \approx 4.58$.

Try it yourself

1. Mean of $50$ items is $30$, SD is $4$. Find the new mean and SD if each is multiplied by $2$.
2. Mean of $25$ items is $20$, SD $4$. A new item $80$ is added. Find new mean and SD.
3. Two groups: $n_1 = 100, \bar{x}_1 = 50, \sigma_1 = 5$; $n_2 = 200, \bar{x}_2 = 55, \sigma_2 = 6$. Combined mean and SD.
4. SD of $n$ observations is $k$. What is the SD of these observations plus $7$?
5. If $\bar{x} = 35, \sigma = 8$, and $y = 5(x - 30)$, find $\bar{y}, \sigma_y$.
6. The SD of $5$ numbers $4, 5, 6, 7, 8$ is , compute. If each is multiplied by $3$, find new SD.
7. Combined mean of two sets is $50$. First set: $20$ items mean $45$. If second set has $30$ items, find its mean.
8. The mean of $20$ items is $40$. After deleting one item the new mean of $19$ items becomes $42$. Find the deleted item.
9. Mean and SD of $100$ observations were $50$ and $8$. Two items recorded $30$ and $70$ should be $35$ and $75$. Find corrected mean and SD.
10. For $n_1 = 60, \bar{x}_1 = 25, n_2 = 40, \bar{x}_2 = 30$, find combined mean.
11. If $\sigma_y$ is the SD when $y_i = a + b x_i$, prove $\sigma_y = |b| \sigma_x$.
12. SD of $a, b, c$ is $\sigma$. Find SD of $5a, 5b, 5c$.

Pitfalls / Tricks

• Always use the weighted-average formula for combined means.
• For combined variance, do not just average $\sigma_1^2$ and $\sigma_2^2$ , add the between-group $d^2$ terms.
• Shifts ($+a$) don't affect SD; scales ($\times b$) do, by $|b|$.
• Insight. Variance decomposes into within-group and between-group parts , a foundational idea in statistics that returns in Class XII probability and beyond.