Animated Solution for Mathematics - Statistics: Let $X=\{x\in N:1\le x\le 17\}$ and $Y=\{ax+b:x\in X\text{ and }a,b\in R,a>0\}$. If mean and variance of elements of $Y$ are 17 and 216 respectively then $a+b$ is equal to :

Imagine you are standing before a perfectly ordered row of seventeen markers, labeled from $1$ to $17$. This is our set $X$. It is a symmetric arithmetic progression, and in the world of JEE Advanced, we approach such sequences using the elegant tools of statistics.

Analyzing Set $X$

First, let us find the heart of our set $X$. The mean of the first $n$ natural numbers is given by the formula:

With $n=17$, we find ${\mu }_X=\frac{17+1}{2}=9$. This is the balance point of our data.

Next, we look at the spread, or the variance. The variance of the first $n$ natural numbers is defined as:

Substituting $n=17$, we get:

We have now unlocked the DNA of our original set: a mean of $9$ and a variance of $24$.

Scaling and Shifting

Now, we introduce the transformation $Y=aX+b$. We are told that the new mean ${\mu }_Y$ is $17$ and the new variance ${\sigma }_Y^2$ is $216$.

The crucial insight is that variance is a measure of spread. If you shift a set of numbers by adding $b$, the spread does not change; only the scaling factor $a$ affects the variance. The property is:

Plugging in our values, we get:

Dividing both sides, we find $a^2=9$. Since the problem mandates $a>0$, we conclude $a=3$.

Finding the Shift

With $a$ in hand, we turn to the mean. Unlike variance, the mean is sensitive to both scaling and shifting according to the relationship:

We know ${\mu }_Y=17$, $a=3$, and ${\mu }_X=9$. Substituting these, we get:

Solving for $b$, we find $b=17-27=-10$.

The final step is to calculate the sum of the parameters:

Through this journey, we have seen how mean and variance respond to the forces of scaling and shifting. The final result is $-7$.