JEE Main 2025 — Statistics Question with Solution

First, find the values of $a$ and $b$ using the given conditions:

Mean Condition:

$5=\frac{1+3+a+7+b}{5}$

This implies:

$a+b=14$

Variance Condition:

$\frac{1+9+a^2+49+b^2}{5}-(5{)}^2=10$

Simplifying, we get:

$a^2+b^2=116$

Using $a+b=14$, we have:

$b=14-a$

Substitute $b$ into the equation for $a^2+b^2$:

$a^2+(14-a{)}^2=116$

Expand and simplify:

$a^2+196-28a+a^2=116$

$2a^2-28a+80=0$

Simplify further:

$a^2-14a+40=0$

Solve the quadratic equation:

$(a-10)(a-4)=0$

Thus, $a=10$ and $b=4$ (since $a>b$).

New Observations:

The transformed observations are:

$1+1=2,3+2=5,10+3=13,7+4=11,4+5=9$

Calculate the Variance of the New Observations:

First, find the mean of the new data:

$\text{Mean}=\frac{2+5+13+11+9}{5}=8$

Calculate the variance:

$\text{Variance}=\frac{(2-8{)}^2+(5-8{)}^2+(13-8{)}^2+(11-8{)}^2+(9-8{)}^2}{5}$

$=\frac{36+9+25+9+1}{5}$

$=\frac{80}{5}=16$

Therefore, the variance of the transformed observations is 16.