Animated Solution for Mathematics - Statistics: The variance ${\sigma }^2$ of the data $$\begin{array}{cccccccc}{x}_i& 0& 1& 5& 6& 10& 12& 17\\ f_i& 3& 2& 3& 2& 6& 3& 3\end{array}$$ is ________.

Mastering Variance

Welcome, future engineers! Today, we are going to demystify the concept of variance. It is not just a dry statistical formula; it is a measure of how 'spread out' a set of data is.

Imagine you are looking at a distribution of particles; the variance tells you how far, on average, these particles are from their center of gravity, which we call the mean. We are given a table of data points $x_i$ and their frequencies $f_i$, and our goal is to find the variance, denoted as ${\sigma }^2$.

The Foundation

Before we calculate anything, we must find the total number of observations, $N$. Think of $N$ as the total mass of our system.

We find it by summing all the frequencies:

With $N=22$ in our pocket, we are ready to find the center of our data.

Finding the Balance Point

The mean, $\bar{x}$, is the balance point of our distribution. The formula is defined as:

We calculate the numerator by multiplying each data point by its frequency:

This gives us $0+2+15+12+60+36+51=176$. Dividing this by $N=22$, we get:

Our balance point is exactly $8$.

The Measure of Spread

Now, we use the computational formula for variance:

This formula is elegant because it separates the 'mean of the squares' from the 'square of the mean'. First, we need $\sum f_i{x}_i^2$.

We square each $x_i$ and multiply by its frequency:

This results in $0+2+75+72+600+432+867=2048$.

The Final Calculation

Now, we bring it all together. We have $\sum f_i{x}_i^2=2048$, $N=22$, and $\bar{x}=8$.

Substituting these into our variance formula:

Simplifying the fraction, $\frac{2048}{22}=\frac{1024}{11}\approx 93.09$. Subtracting the square of the mean, $64$, we get:

In the context of JEE problems, we often round to the nearest integer, which gives us 29. You have just navigated the statistical landscape of variance with precision.