The mean and variance of the marks obtained by the students in a test are 10 and 4 respectively. Lat

Q: Wait, why can't I just add the difference in mean to the variance? Isn't variance just a measure of spread?

That is a common trap! Variance is not a linear operator. It depends on the sum of squares, which changes non-linearly when you change a single data point. You must recalculate the sum of squares from scratch.

Q: Why do we need to find $n$ first? Can't we solve it without knowing the number of students?

Great question! The variance formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ explicitly contains $n$ in the denominator. Without knowing the sample size, we cannot determine how much weight that single change of $4$ marks has on the overall distribution.

Q: Why do we subtract $8^2$ and add $12^2$ in the sum of squares? Shouldn't we just use the difference?

Remember, the variance formula uses the sum of the *squares* of the marks, not the square of the sum. When you change a value from $8$ to $12$, you are removing the contribution of $8^2$ and adding the contribution of $12^2$ to the total sum of squares. It is the individual values that are squared, not the total sum.

Question

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 :

Options:

(A) 4.04

(B) 4.08

(C) 3.96

(D) 3.92

Sigma Percentile · Accepted Answer

### The Symphony of Statistics

Welcome, student. Today, we are not just solving a problem; we are conducting a symphony of data. Statistics is the art of understanding how a collective behaves, and when one individual changes, the entire collective shifts.

Let us unravel this mystery together.

### Phase 1: The Detective Work

Imagine you are looking at a class of $n$ students. We have clues: the initial mean $\bar{x}$ is $10$, and the variance $\sigma^2$ is $4$.

The mean is the 'center of gravity' of the marks. If the mean is $10$, then the sum of all marks must be $\sum x_i = n \cdot \bar{x} = 10n$.

A student's mark, previously recorded as $8$, is updated to $12$. This is a net gain of $4$ marks for the class, so our new sum becomes $\sum x_i' = 10n + 4$.

Given the new mean is $10.2$, we use the definition $\bar{x}' = \frac{\sum x_i'}{n}$ to set up our equation:

$$\frac{10n + 4}{n} = 10.2$$

Solving this algebraic dance: $10n + 4 = 10.2n$, which leads to $0.2n = 4$. Thus, the class size is **$n = 20$**.

### Phase 2: The Sum of Squares

Variance is not just about the mean; it is about the spread. The formula for variance is $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.

We know the initial variance is $4$ and the mean is $10$. Let's isolate the sum of squares, $\sum x_i^2$:

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

$$4 = \frac{\sum x_i^2}{20} - 100$$

$$\frac{\sum x_i^2}{20} = 104$$

$$\sum x_i^2 = 2080$$

This $2080$ represents the 'energy' of the original distribution.

### Phase 3: The Update

We must now update this sum of squares by removing the incorrect mark's square and adding the correct one:

$$\sum (x_i')^2 = \sum x_i^2 - 8^2 + 12^2$$

$$\sum (x_i')^2 = 2080 - 64 + 144 = 2160$$

Our new sum of squares is $2160$. We are now ready for the final act.

### Phase 4: The Final Calculation

We plug our new values into the variance formula one last time:

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

$$(\sigma')^2 = \frac{2160}{20} - (10.2)^2$$

$$(\sigma')^2 = 108 - 104.04$$

$$(\sigma')^2 = 3.96$$

The new variance is **$3.96$**. You have successfully navigated the non-linear nature of variance and mastered the logic!

Animated Solution for Mathematics - Statistics: 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 :

Select Answer:

Visualized Solution

Defining the Initial Parameters

Relating Sum and Mean

Updating the Total Sum

Using the New Mean

Solving for $n$

The Variance Formula

Finding Initial Sum of Squares

Updating the Sum of Squares

Calculating New Variance

Final Result

The Sigma Insight: Measures of Dispersion

The Symphony of Statistics

Phase 1

Phase 2

Phase 3

Phase 4

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