Animated Solution for Mathematics - Statistics: For observations $x_i$ given ${\sum }_{i=1}^{10}(x_i-5)=10$ and ${\sum }_{i=1}^{10}(x_i-5{)}^2=40$. If mean and variance of observations $(x_1-3),(x_2-3)\dots (x_{10}-3)$ is $\lambda$ & $\mu$ respectively then ordered pair $(\lambda ,\mu )$ is

Welcome, future engineers! Today, we are diving into a problem that might look like a tedious calculation at first glance, but it is actually a beautiful lesson in the geometry of statistics.

We are given ten observations, $x_i$, and some information about their relationship with the number five. Specifically, we know:

Our goal is to find the mean and variance of a new set of observations: $(x_i-3)$. Many students immediately panic, trying to find the individual values of $x_1,x_2,\dots ,x_{10}$. Stop! That is a trap. In statistics, we rarely care about the individual trees; we care about the shape of the forest.

The Shift of Origin

Let us simplify our life by defining a new variable, $y_i=x_i-5$. Visually, imagine your original data points as a distribution on a number line. By subtracting five, we are simply shifting the entire distribution to the left by five units.

Now, let us look at what we know about $y_i$. The mean of $y_i$ is $\bar{y}=\frac{\sum y_i}{n}$.

Since we know $\sum y_i=10$ and $n=10$, the mean is:

This tells us the center of our new distribution is at one. Next, the variance of $y_i$ is given by ${\sigma }_y^2=\frac{\sum y_i^2}{n}-(\bar{y}{)}^2$.

Plugging in our values, we get:

This value, three, is the spread of our distribution.

The Transformation

Now, we look at our target variable, $z_i=x_i-3$. We want the mean and variance of this new set.

Notice the elegant relationship between our target $z_i$ and our known $y_i$:

This is the core of the problem! To get from our distribution $y_i$ to our target distribution $z_i$, we simply shift everything to the right by two units.

The Invariance Principle

Here is the masterclass concept: how do mean and variance react to this shift? The mean is sensitive to shifts.

If you add two to every observation, the average must also increase by two. So, the new mean $\lambda$ is:

But variance? Variance is the measure of spread. If you slide a rigid object across a table, does its width change? Of course not!

Variance is independent of the change of origin. Therefore, the variance of $z_i$ is identical to the variance of $y_i$. Thus, the variance $\mu$ is:

We have arrived at our destination. The mean $\lambda$ is three, and the variance $\mu$ is three. The ordered pair $(\lambda ,\mu )$ is $(3,3)$.

This problem teaches us that in the world of JEE mathematics, the most powerful tool is not brute force calculation, but the ability to see the underlying symmetry. Shifting data changes the mean, but the variance remains perfectly invariant. Keep this in your toolkit, and you will conquer any statistics problem that comes your way!