Q. Let \(X=\{x\in\mathbb{N}:1\le x\le19\}\) and for some \(a,b\in\mathbb{R}\),
\(Y=\{ax+b:x\in X\}\). If the mean and variance of the elements of \(Y\)
are \(30\) and \(750\), respectively, then the sum of all possible values of \(b\) is
Explanation
The set \(X=\{1,2,\dots,19\}\).
\[
\text{Mean of }X = \frac{1+19}{2} = 10
\]
\[
\text{Variance of }X = \frac{(19^2-1)}{12} = 30
\]
For the linear transformation \(Y=ax+b\),
\[
\text{Mean}(Y)=a\,\text{Mean}(X)+b
\]
\[
\text{Var}(Y)=a^2\,\text{Var}(X)
\]
Given \(\text{Var}(Y)=750\),
\[
a^2 \cdot 30 = 750 \Rightarrow a^2=25 \Rightarrow a=\pm5
\]
Given \(\text{Mean}(Y)=30\),
\[
10a+b=30
\]
Case 1: \(a=5\)
\[
50+b=30 \Rightarrow b=-20
\]
Case 2: \(a=-5\)
\[
-50+b=30 \Rightarrow b=80
\]
Sum of all possible values of \(b\):
\[
-20+80=60
\]
Therefore, the correct answer is 60.