For a statistical data x_1, x_2, ..., x_{10} of 10 values, a student obtained the mean as 5.5 and ∑_

Q: Wait, why do we subtract the squares of the wrong values and add the squares of the correct values? Shouldn't we just correct the mean first?

That is a great question! You are absolutely right that the mean must be corrected first, but the sum of squares is an independent entity. The variance formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ relies on two distinct pieces of information: the mean and the sum of squares. Both are 'infected' by the incorrect data points. We correct the sum of squares by removing the contribution of the wrong values ($4^2$ and $5^2$) and adding the contribution of the correct ones ($6^2$ and $8^2$). Think of it as cleaning two separate data streams.

Q: Is there a shortcut? Can I just find the change in variance directly?

While you could theoretically derive a formula for the change in variance, it is often more complex and prone to algebraic errors than simply recalculating the sums. In JEE Advanced, the most robust and reliable method is to treat the sum of observations and the sum of squares as fundamental building blocks. Stick to the 'correction of sums' method; it is elegant, logical, and minimizes the risk of silly mistakes.

Q: What if the number of observations $n$ had changed? What would happen then?

If the number of observations $n$ had changed—for example, if the student had accidentally added an extra value instead of just misrecording one—the entire structure of the variance formula would shift. The denominator $n$ would change, and the mean would be calculated over a different count. Always verify if the count of observations remains constant. In this problem, $n=10$ throughout, which keeps our life simple!

Question

For a statistical data $x_{1},x_{2},\dots,x_{10}$ of 10 values, a student obtained the mean as 5.5 and $\sum_{i=1}^{10}x_{i}^{2}=371.$ He later found that he had noted two values in the data incorrectly as 4 and 5, instead of the correct values 6 and 8, respectively. The variance of the corrected data is

Options:

(A) 9

(B) 5

(C) 7

(D) 4

Sigma Percentile · Accepted Answer

### The Data Detective: Uncovering the Truth in Statistics

Welcome, future engineer. Today, we are stepping into the shoes of a data detective. In the world of JEE Advanced, statistics isn't just about plugging numbers into formulas; it is about understanding the integrity of data.

We are presented with a dataset of $10$ values where a student has made a classic recording error. They calculated a mean of $5.5$ and a sum of squares of $371$, but they mistakenly recorded $4$ and $5$ instead of the true values, $6$ and $8$. Our mission is to restore the truth and calculate the correct variance.

### Phase 1: The Anatomy of the Error

Before we dive into the math, let us visualize what happened. The student's calculations are based on a 'corrupted' dataset with $n = 10$ observations.

The incorrect mean is $\bar{x}_{	ext{incorrect}} = 5.5$, and the incorrect sum of squares is $\sum x_{i, 	ext{incorrect}}^2 = 371$. The error is localized: two values are wrong.

To fix this, we must first understand the 'incorrect' sum of all observations. We know that the mean is defined as:

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

Rearranging this, we find that the total sum of the incorrect data is:

$$ \sum x_{i, 	ext{incorrect}} = n 	imes \bar{x}_{	ext{incorrect}} = 10 	imes 5.5 = 55 $$

This $55$ is the sum of the ten values as they were incorrectly recorded.

### Phase 2: The Correction Ritual

Now, we perform the correction. We have a sum of $55$ that includes the wrong values $4$ and $5$. To get the true sum, we must perform a simple, logical adjustment: subtract the wrong values and add the correct ones.

The formula is:

$$ \sum x_{i, 	ext{correct}} = \sum x_{i, 	ext{incorrect}} - (4 + 5) + (6 + 8) $$

Substituting our values, we get:

$$ \sum x_{i, 	ext{correct}} = 55 - 9 + 14 = 60 $$

With this corrected sum of $60$, finding the correct mean is trivial:

$$ \bar{x}_{	ext{correct}} = \frac{60}{10} = 6 $$

### Phase 3: The Sum of Squares

Variance is a measure of spread, and it is highly sensitive to the squares of the values. The student's sum of squares, $371$, is also corrupted.

We apply the same correction logic here, but we must square the values before adding or subtracting them. The logic is:

$$ \sum x_{i, 	ext{correct}}^2 = \sum x_{i, 	ext{incorrect}}^2 - (4^2 + 5^2) + (6^2 + 8^2) $$

Let's calculate this carefully: $4^2 = 16$, $5^2 = 25$, $6^2 = 36$, and $8^2 = 64$. The correction becomes:

$$ 371 - (16 + 25) + (36 + 64) = 371 - 41 + 100 = 430 $$

This $430$ is the true sum of squares for our corrected dataset.

### Phase 4: The Final Synthesis

We are now at the finish line. We have the correct sum of squares ($430$) and the correct mean ($6$). The variance formula is our final tool:

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

Plugging in our corrected values, we get:

$$ \sigma^2 = \frac{430}{10} - (6)^2 $$

This simplifies to $43 - 36$, which equals $7$. The variance of the corrected data is **$7$**.

Select Answer:

Visualized Solution

Identify Given Parameters

Calculate Incorrect $\sum x_{i}$

Logic for Correcting $\sum x_{i}$

Execute Sum Correction

Calculate Correct Mean $\overset{x}{ˉ}$

Logic for Correcting $\sum x_{i}^{2}$

Substitute Values for $\sum x_{i}^{2}$

Execute Sum of Squares Correction

Apply Variance Formula $σ^{2}$

Substitute Correct Values

Final Calculation

The Sigma Insight: Variance and Standard Deviation

The Data Detective

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

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

The mean and standard deviation of $15$ observations are found to be $8$ and $3$ respectively. On rechecking it was found that, in the observations, $20$ was misread as $5$ . Then, the correct variance is equal to _______.

The mean and standard deviation of $50$ observations are $15$ and $2$ respectively. It was found that one incorrect observation was taken such that the sum of correct and incorrect observations is $70$ . If the correct mean is $16$ , then the correct variance is equal to :

The mean and standard deviation of 40 observations are 30 and 5 respectively. It was noticed that two of these observations 12 and 10 were wrongly recorded. If $σ$ is the standard deviation of the data after omitting the two wrong observations from the data, then $38 σ^{2}$ is equal to

The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation by mistake was taken 8 instead of 12. The correct standard deviation is

Let the mean and variance of 12 observations be $\frac{9}{2}$ and 4 respectively. Later on, it was observed that two observations were considered as 9 and 10 instead of 7 and 14 respectively. If the correct variance is $\frac{m}{n}$ , where $m$ and $n$ are coprime, then $m + n$ is equal to

Let the Mean and Variance of five observations $x_{1} = 1, x_{2} = 3, x_{3} = a, x_{4} = 7$ and $x_{5} = b, a > b$ , be 5 and 10 respectively. Then the Variance of the observations $n + x_{n}, n = 1, 2, \dots, 5$ is

The mean and variance of a data of 10 observations are 10 and 2, respectively. If an observations $α$ in this data is replaced by $β$ , then the mean and variance become 10.1 and 1.99, respectively. Then $α + β$ equals