In calculating the mean and variance of 10 readings, a student wrongly used the figure 52 for the co

Q: Wait, why can't I just subtract the difference (52-25=27) from the mean? The mean is just an average, right?

That is a very common trap! The mean is a linear operator, but it depends on the total sum. Changing one value changes the sum, which changes the mean by $\frac{\Delta}{n}$. It is not just subtracting the difference from the mean itself. You must adjust the total sum first, then divide by $n$ to get the new mean.

Q: Why do we have to use $\sum x^2$? Can't we just adjust the variance directly?

Variance is not linear. It involves squares. If you change a data point, the variance changes in a non-linear way because of the $(\bar{x})^2$ term and the $\sum x^2$ term. You must reconstruct the sum of squares from the old variance, correct it, and then calculate the new variance. There is no simple 'add or subtract' shortcut for variance.

Q: What if there were multiple errors? Does the method change?

The beauty of this method is that it is robust! If there were multiple errors, you would simply subtract all the incorrect values and add all the correct values to the initial sums. The fundamental principle—correcting the raw sums $\sum x$ and $\sum x^2$—remains exactly the same.

Question

In calculating the mean and variance of 10 readings, a student wrongly used the figure 52 for the correct figure of 25. He obtained the mean and variance as 45.0 and 16.0 respectively. Determine the correct mean and variance.

Sigma Percentile · Accepted Answer

### The Detective's Guide to Statistical Errors

Imagine you are in a high-stakes laboratory, meticulously recording ten different data points. You are the lead scientist, and your results are crucial. But then, a transcription error occurs.

A single digit is misread, a single value is recorded incorrectly. Suddenly, your mean and variance are skewed. This is the classic 'transcription trap' that often appears in JEE Advanced problems.

Today, we are going to play detective. We will not just solve the problem; we will learn how to reverse-engineer the truth from contaminated data.

### The Anatomy of the Error

When you are given an incorrect mean $\bar{x}_{old} = 45.0$ and an incorrect variance $\sigma^2_{old} = 16.0$ for $n = 10$ readings, you are looking at the final output of a process. The error is not in the formula; it is in the input.

The student recorded $52$ instead of $25$. To fix this, we must go back to the fundamental building blocks of statistics: the sum of observations $\sum x$ and the sum of squares $\sum x^2$.

### Phase 1: Purifying the Mean

The mean is defined as $\bar{x} = \frac{\sum x}{n}$. This means the total sum is simply $\sum x = n \cdot \bar{x}$.

Our incorrect sum is:
$$\sum x_{old} = 10 \cdot 45.0 = 450$$

Now, we perform the 'purification'. We remove the intruder ($52$) and welcome the correct guest ($25$). The new sum is:
$$\sum x_{new} = 450 - 52 + 25 = 423$$

With this pure sum, the corrected mean is simply:
$$\bar{x}_{new} = \frac{423}{10} = 42.3$$

### Phase 2: The Non-Linearity of Variance

The variance formula is $\sigma^2 = \frac{\sum x^2}{n} - (\bar{x})^2$. This is the most important equation in this problem.

Notice that it depends on the sum of squares $\sum x^2$. To correct the variance, we must first find the incorrect sum of squares. Rearranging the formula, we get:
$$\sum x^2 = n(\sigma^2 + (\bar{x})^2)$$

Plugging in our incorrect values:
$$\sum x^2_{old} = 10 \cdot (16.0 + (45.0)^2) = 10 \cdot (16 + 2025) = 10 \cdot 2041 = 20410$$

Now, we correct this sum of squares. Just as we did with the mean, we subtract the square of the wrong value and add the square of the correct value:
$$\sum x^2_{new} = 20410 - (52)^2 + (25)^2 = 20410 - 2704 + 625 = 18331$$

### Phase 3: The Final Synthesis

Now that we have the corrected sum of squares $\sum x^2_{new} = 18331$ and the corrected mean $\bar{x}_{new} = 42.3$, we can find the true variance. The formula is $\sigma^2_{new} = \frac{\sum x^2_{new}}{n} - (\bar{x}_{new})^2$.

Substituting our values:
$$\sigma^2_{new} = \frac{18331}{10} - (42.3)^2 = 1833.1 - 1789.29 = 43.81$$

### The Takeaway

We have successfully navigated the trap. The key takeaway for your JEE preparation is this: never try to adjust the mean or variance directly.

They are derived values. Always drill down to the fundamental sums—$\sum x$ and $\sum x^2$—correct those first, and then build your final answers from there.

This approach is not just a technique; it is a mindset. It is the difference between guessing and knowing. Keep practicing this, and you will find that even the most complex statistical problems become a simple matter of detective work.

Animated Solution for Mathematics - Statistics: In calculating the mean and variance of 10 readings, a student wrongly used the figure 52 for the correct figure of 25. He obtained the mean and variance as 45.0 and 16.0 respectively. Determine the correct mean and variance.

Visualized Solution

Given Information

Calculating Incorrect Sum $\sum x_{o l d}$

Correcting the Sum $\sum x_{n e w}$

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

Formula for Variance

Finding Incorrect Sum of Squares $\sum x_{o l d}^{2}$

Correcting the Sum of Squares $\sum x_{n e w}^{2}$

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

Summary and Takeaway

The Sigma Insight: Measures of Dispersion

The Detective's Guide to Statistical Errors

The Anatomy of the Error

Phase 1

Phase 2

Phase 3

The Takeaway

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