Q.Find Mean, Variance and Standard Deviation of the following data : | Class | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 | 90-100 |
| Frequency | 3 | 7 | 12 | 15 | 8 | 3 | 2 |

Corrected Mean and Standard Deviation

Imagine a teacher who has computed the average marks of a class and the standard deviation, only to discover afterwards that one mark was entered wrongly, or that a student must be added or removed. Recomputing everything from the raw list of 100 marks would be tedious. Corrected mean and standard deviation is the technique for updating these two summary measures when the data changes by a small amount — without going back to the full dataset.

The Two Quantities Everything Rests On

Both the mean and the standard deviation can be rebuilt from just two running totals:

• the sum of the observations, $\sum x_i$
• the sum of the squares of the observations, $\sum x_i^2$

Using the population formulas, a very useful rearrangement is:

so that from a known mean and standard deviation we can recover both totals:

Updating for a Change in the Data

Once you hold $\sum x_i$ and $\sum x_i^2$, every kind of correction is just simple bookkeeping.

• Remove an observation $a$: $\sum x_i \to \sum x_i - a$, $\sum x_i^2 \to \sum x_i^2 - a^2$, and $n \to n-1$.
• Add an observation $b$: $\sum x_i \to \sum x_i + b$, $\sum x_i^2 \to \sum x_i^2 + b^2$, and $n \to n+1$.
• Replace a wrong value $a$ by the correct value $b$: do both at once — $\sum x_i \to \sum x_i - a + b$ and $\sum x_i^2 \to \sum x_i^2 - a^2 + b^2$, with $n$ unchanged.

Then recompute:

Worked Example (a mis-recorded value)

Problem. The mean and standard deviation of $100$ observations were found to be $40$ and $5.1$. Later it was found that one observation was wrongly read as $50$ instead of its correct value $40$. Find the correct mean and standard deviation.

Step 1 — recover the totals.

From $\sigma^2 = \dfrac{1}{n}\sum x_i^2 - \bar{x}^2$ with $\sigma = 5.1$:

Step 2 — correct the totals (replace $50$ by $40$; $n$ stays $100$):

Step 3 — corrected mean and standard deviation.

$\bar{x}_{\text{new}} = \frac{3990}{100} = 39.9$ …