The mean and variance of 20 observations are found to be 10 and 4, respectively. On rechecking, it w

Q: Wait, why do we need to calculate $\sum x_i^2$? Can't we just adjust the variance directly?

That is a classic trap! Variance is not a linear property. Because the formula involves the square of the mean and the mean of the squares, you cannot simply add or subtract a constant from the variance. You must reconstruct the underlying 'raw' sums first.

Q: Is it possible that the mean stays the same even if the observation changes?

Yes, if the difference between the correct and incorrect observation is zero (i.e., $x_{correct} - x_{wrong} = 0$), the mean remains unchanged. However, the variance would still change because the squared sum $\sum x_i^2$ would change!

Q: Why did we use $10.1^2$ instead of $10^2$ in the final step?

Because the mean itself has shifted! Since we changed an observation from $9$ to $11$, the total sum increased by $2$, making the new mean $\frac{202}{20} = 10.1$. Always remember to update your mean before calculating the final variance.

Q: What if I forget to square the observations when correcting $\sum x_i^2$?

That is the most common error in JEE exams! Remember, the variance formula relies on the 'mean of squares'. If you only correct the linear sum, your variance will be completely wrong. Always treat the linear sum and the squared sum as two separate, sacred entities.

Q: Is there a shortcut for this? This feels like a lot of arithmetic.

While there are algebraic identities for small changes, the 'sum-correction' method is the most robust and least prone to errors. In the heat of the JEE Advanced, reliability beats shortcuts every single time.

Question

The mean and variance of 20 observations are found to be 10 and 4, respectively. On rechecking, it was found that an observation 9 was incorrect and the correct observation was 11. Then the correct variance is :

Options:

(A) $3.99$

(B) $3.98$

(C) $4.01$

(D) $4.02$

Sigma Percentile · Accepted Answer

### The Anatomy of Data: Unlocking the Variance Mystery

Welcome, future engineer! Today, we are going to peel back the layers of a classic statistical problem. Often, students view statistics as a dry collection of formulas, but I want you to see it as a detective story.

We have a dataset of $20$ observations, but there is a 'crime'—an incorrect entry. Our mission is to restore the integrity of the data and find the true variance.

### Phase 1: The Detective's Toolkit

Imagine you have a box of $20$ numbers. You calculated their mean as $10$ and their variance as $4$. But then, you realize one number was recorded as $9$ when it should have been $11$.

To fix this, we cannot just 'tweak' the variance. We must go back to the source. The variance formula is our North Star:

$$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$$

This formula tells us that variance is essentially the difference between the 'average of the squares' and the 'square of the average'. To find the new variance, we need the new $\sum x_i$ and the new $\sum x_i^2$.

### Phase 2: Reconstructing the Past

First, let's find the original sum of our observations. Since the mean $\bar{x}_{old} = 10$ and $n = 20$, the total sum is simply:

$$\sum x_{old} = n 	imes \bar{x}_{old} = 20 	imes 10 = 200$$

Now, for the trickier part: the sum of squares. Using our variance formula, we substitute the known values:

$$4 = \frac{\sum x_{old}^2}{20} - (10)^2$$

Solving this, we get $4 + 100 = \frac{\sum x_{old}^2}{20}$, which leads us to $\sum x_{old}^2 = 104 	imes 20 = 2080$. We have successfully reconstructed the 'DNA' of our original dataset!

### Phase 3: The Correction

Now, we perform the surgery. We remove the 'wrong' value ($9$) and insert the 'correct' value ($11$).

For the linear sum:

$$\sum x_{new} = 200 - 9 + 11 = 202$$

For the squared sum, we must be careful—we square the values before adding or subtracting:

$$\sum x_{new}^2 = 2080 - (9)^2 + (11)^2 = 2080 - 81 + 121 = 2120$$

### Phase 4: The Final Calculation

With our corrected sums, the rest is a beautiful descent to the finish line. First, the new mean:

$$\bar{x}_{new} = \frac{202}{20} = 10.1$$

Finally, we plug everything back into the variance formula:

$$\sigma^2_{new} = \frac{2120}{20} - (10.1)^2$$

$$\sigma^2_{new} = 106 - 102.01 = 3.99$$

### The Takeaway

Look at that result: **$3.99$**. It is so close to the original $4$, yet distinct.

This problem teaches us that in data science and physics, precision is everything. A tiny error in a single observation ripples through the entire calculation.

By mastering the art of correcting the sums, you have gained a powerful tool that will serve you well in experimental physics, where error analysis is the difference between a successful experiment and a failed one. Keep questioning, keep calculating, and never lose that curiosity!

Animated Solution for Mathematics - Statistics: The mean and variance of 20 observations are found to be 10 and 4, respectively. On rechecking, it was found that an observation 9 was incorrect and the correct observation was 11. Then the correct variance is :

Select Answer:

Visualized Solution

Identify Given Parameters

Calculate Incorrect Sum $\sum x_{i}$

Apply Variance Formula

Calculate Incorrect $\sum x_{i}^{2}$

Correct the Linear Sum $\sum x_{i}$

Correct the Squared Sum $\sum x_{i}^{2}$

Calculate Correct Mean $\overset{x}{ˉ}_{n e w}$

Calculate Correct Variance $σ_{n e w}^{2}$

Conclusion & Key Takeaway

The Sigma Insight: Measures of Dispersion

The Anatomy of Data

Phase 1

Phase 2

Phase 3

Phase 4

The Takeaway

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