The mean and variance of 10 observations were calculated as 15 and 15 respectively by a student who

Q: Wait, why do we use the incorrect mean to find the incorrect sum of squares? Shouldn't we use the correct mean?

That is a classic trap! The incorrect variance of 15 was calculated using the incorrect mean of 15. The formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ is a snapshot of the calculation at that specific moment. To reverse-engineer the 'sum of squares' that the student actually used, you must use the values they had at that time.

Q: Can't I just adjust the variance directly by subtracting the difference in values?

Unfortunately, no. Variance is a non-linear measure. It depends on the sum of squares, not just the sum of the observations. You must always go back to the 'sum of squares' ($\sum x_i^2$) to perform any corrections.

Q: Why does the standard deviation change so drastically?

Standard deviation measures the spread. When you replace a value like 25 with 15, you are pulling that data point closer to the mean. This reduces the overall 'spread' or 'dispersion' of the data, which is why the variance (and consequently the standard deviation) decreases.

Question

The mean and variance of 10 observations were calculated as 15 and 15 respectively by a student who took by mistake 25 instead of 15 for one observation. Then, the correct standard deviation is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The foundation of our analysis is the mean, which acts as the "center of gravity" of your data. If you change even one value, the entire balance shifts.

The formula for the mean is defined as:

$$\bar{x} = \frac{\sum x_i}{n}$$

We are given that the incorrect mean is $15$ and the number of observations is $n=10$. This implies the incorrect sum of observations was:

$$\sum x_{wrong} = 15 	imes 10 = 150$$

### Phase 1: The Mean Correction

To find the correct sum, we perform a simple adjustment by subtracting the incorrect value and adding the correct one:

$$\sum x_{correct} = \sum x_{wrong} - x_{wrong} + x_{correct}$$

Substituting our values, we get:

$$\sum x_{correct} = 150 - 25 + 15 = 140$$

Now, the new mean is calculated as:

$$\bar{x}_{new} = \frac{140}{10} = 14$$

Notice how the mean dropped; this is because we replaced a larger value ($25$) with a smaller one ($15$).

### Phase 2: The Variance Correction

The variance formula is $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$. To fix the variance, we first need to find the "sum of squares" that the student originally calculated.

Using the incorrect variance of $15$ and the incorrect mean of $15$, we set up the equation:

$$15 = \frac{\sum x_{wrong}^2}{10} - (15)^2$$

Solving this, we find:

$$\frac{\sum x_{wrong}^2}{10} = 15 + 225 = 240 \Rightarrow \sum x_{wrong}^2 = 2400$$

Now, we perform the same surgery on the sum of squares:

$$(\sum x_i^2)_{actual} = \sum x_{wrong}^2 - (x_{wrong})^2 + (x_{correct})^2$$

Plugging in the numbers:

$$(\sum x_i^2)_{actual} = 2400 - (25)^2 + (15)^2 = 2400 - 625 + 225 = 2000$$

This value represents the "true" sum of squares.

### Phase 3: The Final Calculation

We are now ready to calculate the actual variance using our corrected values:

$$\sigma_{actual}^2 = \frac{(\sum x_i^2)_{actual}}{n} - (\bar{x}_{new})^2$$

Substituting the corrected values:

$$\sigma_{actual}^2 = \frac{2000}{10} - (14)^2 = 200 - 196 = 4$$

Finally, the question asks for the standard deviation, which is the square root of the variance:

$$(S.D.)_{actual} = \sqrt{4} = 2$$

By methodically correcting the sum and the sum of squares, we have navigated the trap. The **actual standard deviation is 2**.

Animated Solution for Mathematics - Statistics: The mean and variance of 10 observations were calculated as 15 and 15 respectively by a student who took by mistake 25 instead of 15 for one observation. Then, the correct standard deviation is

Enter Numerical Value:

Visualized Solution

Given Data

Incorrect Sum of Observations

Correcting the Sum

Correct Mean

Incorrect Sum of Squares Setup

Calculate Incorrect Sum of Squares

Correcting Sum of Squares

Calculate Correct Variance

Final Standard Deviation

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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