The variance of first 50 even natural numbers is (a) 437 (b) 437/4 (c) 833/4 (d) 833

Q: Why can't we just calculate the mean and then use $\text{Var}(X) = \frac{1}{n} \sum (x_i - \bar{x})^2$?

You absolutely can, but in a JEE Advanced exam, time is your most precious currency. Calculating the mean of $50$ terms and then squaring each deviation is a recipe for calculation errors. The property-based approach is elegant, faster, and less prone to silly mistakes.

Q: Does adding a constant to every term change the variance?

No, it doesn't! Variance measures the spread of data points from the mean. If you shift every data point by adding a constant $c$, the mean also shifts by $c$, so the distance of each point from the mean remains unchanged. Variance is invariant under translation.

Q: Why is the scaling factor $k^2$ and not just $k$?

Think about the definition of variance: it's the average of the *squared* deviations. When you multiply the data by $k$, the deviation $(x_i - \bar{x})$ becomes $(kx_i - k\bar{x}) = k(x_i - \bar{x})$. When you square this, you get $k^2(x_i - \bar{x})^2$. That $k^2$ is the physical manifestation of the squaring process in the variance formula.

Question

The variance of first 50 even natural numbers is

Options:

(A) 437

(B) 437/4

(C) 833/4

(D) 833

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are going to unravel the beauty of statistics, not through brute force, but through the elegance of mathematical patterns.

When you face a problem like finding the variance of the first $50$ even natural numbers, your first instinct might be to reach for the definition: calculate the mean, subtract it from each term, square the result, and find the average. But stop. In the JEE Advanced arena, we don't just solve problems; we dismantle them.

### Identifying the Sequence

Let us look at the sequence: $2, 4, 6, \dots, 100$. This is an arithmetic progression where the first term $a=2$ and the common difference $d=2$.

But look closer. Can you see the hidden structure? Every single term is a multiple of $2$. If we factor out a $2$, we get $2 	imes (1, 2, 3, \dots, 50)$.

Suddenly, the problem transforms. We are no longer dealing with an arbitrary sequence; we are dealing with the first $50$ natural numbers, scaled by a factor of $2$. This is the key insight that separates the calculator from the thinker.

### The Power of Scaling

Now, we need to recall a fundamental result in statistics: the variance of the first $n$ natural numbers. This is a standard formula that you should have etched into your memory:

$$\sigma^2 = \frac{n^2 - 1}{12}$$

This formula is a gift, saving us from tedious summations. But we have a scaling factor. We are not looking for the variance of $X = (1, 2, \dots, 50)$, but for the variance of $Y = 2X$.

Here is where the scaling property of variance comes into play. If you multiply every observation in a dataset by a constant $k$, the variance does not simply multiply by $k$. Because variance is a measure of **squared deviations**, the new variance becomes $k^2$ times the original variance.

Mathematically, the property is expressed as:

$$	ext{Var}(kX) = k^2 	ext{Var}(X)$$

In our case, $k=2$. So, the variance of our even numbers is $2^2 	imes 	ext{Var}(	ext{first } 50 	ext{ natural numbers})$.

### The Execution

Let us put it all together. We have $n=50$ and $k=2$. The variance of the first $50$ natural numbers is:

$$	ext{Var}(X) = \frac{50^2 - 1}{12}$$

Applying our scaling property, the variance of the first $50$ even numbers is:

$$	ext{Var}(Y) = 2^2 	imes \frac{50^2 - 1}{12} = 4 	imes \frac{50^2 - 1}{12} = \frac{50^2 - 1}{3}$$

Now, the calculation is trivial. Since $50^2 = 2500$, we have:

$$	ext{Var}(Y) = \frac{2500 - 1}{3} = \frac{2499}{3} = 833$$

### The JEE Mindset

There you have it. The final answer is **$833$**.

Notice how we avoided calculating the mean of the even numbers, avoided squaring fifty different deviations, and avoided any potential for arithmetic error. We used the properties of the system to our advantage.

This is the JEE mindset: identify the pattern, apply the theorem, and execute with precision. Keep this scaling property in your toolkit—it is a favorite of examiners, and now, it is a weapon in your arsenal.

Animated Solution for Mathematics - Statistics: The variance of first 50 even natural numbers is

Select Answer:

Visualized Solution

Identify the Sequence

Relation to Natural Numbers

Variance of First $n$ Natural Numbers

Scaling Property of Variance

Derive Specific Formula

Substitute $n = 50$

Calculate the Square

Final Calculation

Key Takeaway

The Sigma Insight: Measures of Dispersion

Analyzing the Setup

Identifying the Sequence

The Power of Scaling

The Execution

The JEE Mindset

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