Animated Solution for Mathematics - Statistics: If ${\sum }_{i=1}^9(x_i-5)=9$ and ${\sum }_{i=1}^9(x_i-5{)}^2=45$, then the standard deviation of the 9 items $x_1,x_2,\dots ,x_9$ is :

When you encounter the expression ${\sum }_{i=1}^9(x_i-5)=9$, your first instinct might be to panic. You might think, "How can I possibly find nine different values of $x$ with only two equations?"

But here is the secret: you don't have to. In the world of statistics, we often look for the "soul" of the data rather than its individual components.

Imagine you have a collection of nine points scattered on a number line. The term $(x_i-5)$ is simply a way of saying, "Let's shift our perspective."

By defining a new variable $y_i=x_i-5$, we are essentially sliding our entire coordinate system so that the center of our data is easier to manage. This is what we call a "change of origin."

Why do we do this? Because the standard deviation, denoted by $\sigma$, is a measure of how spread out the points are.

If you take a group of people and move them all five steps to the left, the distance between them doesn't change. Therefore, the standard deviation of our original set $x$ is identical to the standard deviation of our new set $y$.

Mathematically, we state this as ${\sigma }_x={\sigma }_y$. We have just simplified our problem significantly!

Now that we have our new variable $y_i$, our given information looks much cleaner:

We are looking for the standard deviation, which is the square root of the variance. The formula for variance is the "mean of the squares minus the square of the mean."

It is a beautiful, compact expression that captures the essence of dispersion:

Let's breathe and look at this. We have $n=9$ items.

The sum of our new variables is $9$, so the mean $\bar{y}$ is simply $\bar{y}=\frac{9}{9}=1$.

The sum of the squares is $45$, so the mean of the squares is $\frac{45}{9}=5$.

Now, we substitute these values into our variance engine:

And there it is. The result is $2$.

Because we established that the standard deviation is invariant under a change of origin, the standard deviation of our original set $x$ is also $2$.

Think about what we just achieved. We didn't need to know a single value of $x_i$. We used the properties of the data to bypass the complexity.

This is the hallmark of a JEE topper—not brute force, but the elegant application of mathematical principles. Keep this "change of origin" tool in your kit; it will serve you well when the algebra gets heavy and the time is ticking.