Animated Solution for Mathematics - Statistics: Let the Mean and Variance of five observations $x_1=1,x_2=3,x_3=a,x_4=7$ and $x_5=b,a>b$, be 5 and 10 respectively. Then the Variance of the observations $n+x_n,n=1,2,\dots ,5$ is

Welcome, future engineer. Today, we are not just solving a statistics problem; we are performing detective work. In the world of JEE Advanced, statistics is often dismissed as 'easy' or 'formula-based,' but that is a grave mistake.

Statistics is the language of uncertainty, and mastering it requires a blend of algebraic precision and logical intuition. Let us dive into this problem, where we have five observations, two of which are shrouded in mystery: $a$ and $b$.

Imagine you are standing before a set of data points: $1,3,a,7,b$. We are given the mean $\bar{x}=5$ and the variance ${\sigma }^2=10$. The mean is the 'center of gravity' of your data, defined by the formula:

When we apply this to our set, we get:

Do not rush the arithmetic. When you multiply the $5$ from the denominator to the other side, you get $25$. Summing the known integers $1+3+7$ gives us $11$.

Thus, our first equation is born: $11+a+b=25$, or simply:

This is our first anchor point. We know the sum of our unknowns is $14$.

Now, we need the second piece of the puzzle. We have the variance, which measures the 'spread' or the 'volatility' of our data. The formula ${\sigma }^2=\frac{\sum x_i^2}{n}-(\bar{x}{)}^2$ is your best friend here.

Substituting our knowns, we have:

The square of the mean, $5^2=25$, moves to the left to join the $10$, giving us $35$. Multiplying by $5$ gives us $175$. On the right, we have $1+9+49+a^2+b^2$, which simplifies to $59+a^2+b^2$.

Solving for $a^2+b^2$, we find $175-59=116$. Now we have two beautiful equations: $a+b=14$ and $a^2+b^2=116$.

How do we extract $a$ and $b$ from these? We use the algebraic identity $(a+b{)}^2=a^2+b^2+2ab$. This is the bridge that connects the sum to the product.

Substituting our values:

This leads to $196=116+2ab$, which simplifies to $2ab=80$, or $ab=40$. Now, we need two numbers that add to $14$ and multiply to $40$.

You could solve the quadratic $t^2-14t+40=0$, but your intuition might already be screaming the answer: $10$ and $4$. Since the problem explicitly states $a>b$, we assign $a=10$ and $b=4$. The mystery is solved; our data set is $1,3,10,7,4$.

But wait, the problem isn't over! We are asked to find the variance of a new set of observations, $y_n=n+x_n$. This is a transformation where we add the index $n$ to each observation.

Let us calculate these new values: - $y_1=1+1=2$ - $y_2=2+3=5$ - $y_3=3+10=13$ - $y_4=4+7=11$ - $y_5=5+4=9$

Our new set is $2,5,13,11,9$. To find the variance, we need the new mean $\bar{y}$. The sum is $2+5+13+11+9=40$. Dividing by $5$, we get $\bar{y}=8$.

Finally, we calculate the sum of squares for the new set:

Applying the variance formula one last time:

There it is. The final answer is $16$. You navigated the algebra, respected the constraints, and performed the transformation with precision.