Q.For a random variable X, $E(X) = 3$ and $E(X^2) = 11$. The variance of X is : (A) 8 (B) 5 (C) 2 (D) 1

Variance: Measuring Spread

You already know about averages. If I tell you the average height in a class is 165 cm, you have a central value. But that single number hides a lot. Are all students close to 165 cm, or are some very short and some very tall? Variance answers that question.

The Intuition: How "Spread Out" Are the Numbers?

Imagine two small datasets:

• Dataset A: 160, 162, 165, 168, 170
• Dataset B: 140, 150, 165, 180, 190

Both have the same mean (165). But Dataset B is much more "spread out." Variance is a single number that captures this spread. The bigger the variance, the more the numbers are scattered away from the mean.

Building the Formula Step by Step

We need a way to measure, on average, how far each data point is from the mean. Let's call the mean $\bar{x}$.

Step 1: Find the deviation of each point.

For a data point $x_i$, the deviation is $x_i - \bar{x}$. This tells us how far that point is from the mean, with a sign (positive if above, negative if below).

Step 2: Square each deviation.

Why square? Two reasons:

• Squaring removes the sign. A deviation of +5 and -5 both become 25. We care about distance, not direction.
• Squaring gives more weight to points that are far from the mean. A point 10 units away contributes 100, while a point 2 units away contributes only 4. This is usually what we want — outliers matter more.

Step 3: Average the squared deviations.

For a population (all data), we divide by $N$ (the number of data points). For a sample (a subset), we divide by $n-1$ (this is called Bessel's correction — it gives a better estimate of the population variance). For a first introduction, focus on the population formula.

A Worked Example

Take Dataset A: 160, 162, 165, 168, 170. Mean $\bar{x} = 165$.

Sum of squared deviations = $25 + 9 + 0 + 9 + 25 = 68$.

Variance $\sigma^2 = \frac{68}{5} = 13.6$.

Now Dataset B: 140, 150, 165, 180, 190. Mean $\bar{x} = 165$.