The mean and standard deviation of the marks of 10 students were found to be 50 and 12 respectively.

Q: Why can't I just subtract the difference in marks from the variance directly? Variance isn't linear!

That is a brilliant observation! Variance is a measure of spread, not a simple sum. Because it involves squaring the deviations, you cannot simply add or subtract differences. You must reconstruct the 'Sum of Squares' first. Variance is non-linear, so we must work with the raw sums to maintain mathematical integrity.

Q: Why do we need to calculate $\sum x^2$ at all? Can't we just use the definition of variance?

You absolutely can use the definition, but that would require knowing every single individual data point, which we don't have! The formula $\sigma^2 = \frac{\sum x^2}{n} - (\bar{x})^2$ is a powerful tool because it allows us to work with aggregate sums without needing the full list of 10 students' marks. It is the 'shortcut' that makes this problem solvable.

Q: Is there a risk of making a calculation error with the squares? It seems tedious.

It is tedious, but that is the nature of JEE Advanced! Precision is part of the test. Always write down your intermediate sums clearly. If you rush the arithmetic, you lose the marks. Treat the algebra with respect, and the numbers will behave.

Question

The mean and standard deviation of the marks of 10 students were found to be 50 and 12 respectively. Later, it was observed that two marks 20 and 25 were wrongly read as 45 and 50 respectively. Then the correct variance is

Sigma Percentile · Accepted Answer

### The Data Detective's Lab

Welcome, future engineers! Today, we are not just solving a statistics problem; we are acting as Data Detectives. In the world of JEE Advanced, you will often encounter scenarios where the data you are given is 'corrupted' or 'incorrect.'

Your job is not just to calculate, but to reconstruct the truth from the ashes of the error. Let us dive into this problem with the precision of a surgeon.

### Phase 1: The Anatomy of the Error

We start with ten students. We are given an incorrect mean, $\bar{x}_{old} = 50$, and an incorrect standard deviation, $\sigma_{old} = 12$. The problem tells us that two marks were misread: $20$ and $25$ were recorded as $45$ and $50$.

Before we panic, let us remember the definition of the mean: $\bar{x} = \frac{\sum x}{n}$. This tells us that the mean is simply the total sum of all marks divided by the number of students.

If we know the mean and the count, we know the total sum. For our incorrect data, the sum is:

$$\sum x_{old} = n 	imes \bar{x}_{old} = 10 	imes 50 = 500$$

This is our starting point. We have a total of $500$ marks, but it is built on a lie. We need to purge the incorrect values and inject the correct ones.

The correction is straightforward:

$$\sum x_{new} = \sum x_{old} - (45 + 50) + (20 + 25)$$

Calculating this, we get $\sum x_{new} = 500 - 95 + 45 = 450$. Now, we have the correct sum. The new mean is simply $\bar{x}_{new} = \frac{450}{10} = 45$.

Notice how the mean dropped? That makes sense, as we removed higher values and added lower ones.

### Phase 2: The Hidden Engine (Sum of Squares)

Now, here is where most students stumble. They try to adjust the variance directly. But variance is not a linear quantity!

It is a measure of the 'spread' of the data, and it relies on the square of the values. To fix the variance, we must fix the 'Sum of Squares', denoted as $\sum x^2$.

We use the standard variance formula:

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

Let us use our incorrect data to find the incorrect sum of squares. We know $\sigma^2_{old} = 12^2 = 144$. Substituting this into the formula:

$$144 = \frac{\sum x^2_{old}}{10} - (50)^2$$

$$144 = \frac{\sum x^2_{old}}{10} - 2500$$

Solving for $\sum x^2_{old}$, we get:

$$\frac{\sum x^2_{old}}{10} = 2644 \implies \sum x^2_{old} = 26440$$

This number, $26440$, is the 'Sum of Squares' of the incorrect data. It is the engine that drove the incorrect variance.

### Phase 3: The Reconstruction

Now, we perform the same surgery on the Sum of Squares that we did on the Sum of Observations. We subtract the squares of the wrong numbers and add the squares of the correct ones:

$$\sum x^2_{new} = \sum x^2_{old} - (45^2 + 50^2) + (20^2 + 25^2)$$

$$\sum x^2_{new} = 26440 - (2025 + 2500) + (400 + 625)$$

$$\sum x^2_{new} = 26440 - 4525 + 1025 = 22940$$

We are almost there! We have the correct sum of squares, $22940$, and the correct mean, $45$. We just need to plug these into the variance formula one last time to find the truth.

### The Final Victory

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

$$\sigma^2_{new} = \frac{22940}{10} - (45)^2$$

$$\sigma^2_{new} = 2294 - 2025 = 269$$

And there it is: **$269$**. The correct variance.

You see, the math did not change; only our perspective did. By breaking the problem down into the 'Sum of Observations' and the 'Sum of Squares', we turned a complex error into a simple, manageable calculation. Keep this systematic approach in your toolkit, and no statistics problem will ever intimidate you again!

Animated Solution for Mathematics - Statistics: The mean and standard deviation of the marks of 10 students were found to be 50 and 12 respectively. Later, it was observed that two marks 20 and 25 were wrongly read as 45 and 50 respectively. Then the correct variance is

Enter Numerical Value:

Visualized Solution

Identify Given Parameters

Calculate Incorrect Sum of Observations

Correct the Sum of Observations

Calculate Correct Mean

Setup Variance Formula

Calculate Incorrect Sum of Squares

Correct the Sum of Squares

Final Variance Calculation

The Sigma Insight: Measures of Dispersion

The Data Detective's Lab

Phase 1

Phase 2

Phase 3

The Final Victory

Similar Questions

The mean and standard deviation of 10 observations are 20 and 8 respectively. Later on, it was observed that one observation was recorded as 50 instead of 40. Then the correct variance is

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.

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:

The mean and variance of the marks obtained by the students in a test are 10 and 4 respectively. Later, the marks of one of the students is increased from 8 to 12. If the new mean of the marks is 10.2, then their new variance is equal to :

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 :

In an experiment with 15 observations on x, the following results were available: $\sum x^{2} = 2830, \sum x = 170$ . One observation that was 20 was found to be wrong and was replaced by the correct value 30. The corrected variance is

Let the mean of 6 observations 1, 2, 4, 5, $x$ and $y$ be 5 and their variance be 10. Then their mean deviation about the mean is equal to

If the mean and variance of five observations are $\frac{24}{5}$ and $\frac{194}{25}$ respectively and the mean of first four observations is $\frac{7}{2}$ , then the variance of the first four observations in equal to

The mean and variance of 10 observations are 9 and 34.2, respectively. If 8 of these observations are $2, 3, 5, 10, 11, 13, 15, 21$ , then the mean deviation about the median of all the 10 observations is

All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks of 10 to each of the students. Which of the following statistical measures will not change even after the grace marks were given ?