JEE MAIN - Mathematics (2025 - 3rd April Evening Shift - No. 12)
Explanation
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, \quad 3+2=5, \quad 10+3=13, \quad 7+4=11, \quad 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.
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