The mean and variance of a data of 10 observations are 10 and 2, respectively. If an observations α

Q: Wait, why do we need to calculate the sum of squares? Can't we just use the variance formula directly?

That is a great question! While you *can* write the variance formula directly, you will find that the variables $\alpha$ and $\beta$ are trapped inside the summation. By isolating the sum of squares, you turn a complex statistical problem into a clean algebraic one. It is all about simplifying the landscape before you start the climb.

Q: Why does the variance formula change to $\frac{\sum x_i^2}{n} - (\bar{x})^2$? Is this always true?

Yes, this is the computational formula for variance, derived from $\frac{\sum (x_i - \bar{x})^2}{n}$. It is a lifesaver in JEE problems because it allows us to handle the sum of squares as a single entity, rather than dealing with individual deviations. Trust the formula; it is your best friend in statistics.

Q: Is it possible to solve this without finding $\alpha$ and $\beta$ individually?

Absolutely! In fact, that is the elegant path. Notice how the problem asks for $\alpha + \beta$. By using the identity $\beta^2 - \alpha^2 = (\beta - \alpha)(\beta + \alpha)$, we bypass the need to solve for the individual values. In JEE Advanced, looking for these shortcuts is exactly what separates the top rankers from the rest.

Question

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

Options:

(A) 15

(B) 5

(C) 10

(D) 20

Sigma Percentile · Accepted Answer

### Analyzing the Initial State

To understand the change, we must first master the initial state. We know the number of observations $n = 10$. The mean $\bar{x}_{old} = 10$ is the anchor.

By the definition $\bar{x} = \frac{\sum x_i}{n}$, we immediately see that the sum of our initial observations is:

$$\sum x_i = 10 	imes 10 = 100$$

The variance $\sigma^2_{old} = 2$ tells us about the spread. Using the powerful computational formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$, we can extract the hidden sum of squares.

Substituting our values, we get:

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

With a quick rearrangement, $\frac{\sum x_i^2}{10} = 2 + 100 = 102$, leading us to the initial sum of squares:

$$\sum x_i^2 = 1020$$

This is our baseline.

### The Transformation

Now, the swap happens. We remove $\alpha$ and insert $\beta$. The new mean is $\bar{x}_{new} = 10.1$.

The new sum of observations must be:

$$\sum x_{new} = 10.1 	imes 10 = 101$$

Mathematically, this transformation is expressed as $\sum x_{old} - \alpha + \beta = 101$. Substituting our initial sum, we get $100 - \alpha + \beta = 101$, which simplifies beautifully to:

$$\beta - \alpha = 1$$

This is our first key equation. Now, let us look at the variance. The new variance is $1.99$.

Using the same variance formula, we have:

$$1.99 = \frac{\sum x_{new}^2}{10} - (10.1)^2$$

Calculating $(10.1)^2 = 102.01$, we find $\frac{\sum x_{new}^2}{10} = 1.99 + 102.01 = 104$. Thus, the new sum of squares is:

$$\sum x_{new}^2 = 1040$$

### The Algebraic Climax

We know that the sum of squares changes by the removal of $\alpha^2$ and the addition of $\beta^2$. So, $\sum x_{old}^2 - \alpha^2 + \beta^2 = \sum x_{new}^2$.

Plugging in our values:

$$1020 - \alpha^2 + \beta^2 = 1040$$

This gives us:

$$\beta^2 - \alpha^2 = 20$$

Here is where the magic happens. We have $\beta - \alpha = 1$ and $\beta^2 - \alpha^2 = 20$. We use the difference of squares identity:

$$\beta^2 - \alpha^2 = (\beta - \alpha)(\beta + \alpha)$$

Substituting our known values, $20 = (1)(\beta + \alpha)$. Therefore, the final result is:

$$\alpha + \beta = 20$$

The elegance of this result lies in how the complex statistical shifts collapsed into a simple, beautiful algebraic identity. You have mastered the system.

Animated Solution for Mathematics - Statistics: 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

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Visualized Solution

Initial Data Setup

Initial Sum of Observations

Initial Sum of Squares

Effect of Replacement on Mean

First Equation: $β - α$

Effect of Replacement on Variance

Calculating New Sum of Squares

Relating the Sum of Squares

Second Equation: $β^{2} - α^{2}$

Final Calculation using Algebraic Identity

The Sigma Insight: Variance and Standard Deviation

Analyzing the Initial State

The Transformation

The Algebraic Climax

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