Question

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

• A. 60
• B. 80
• C. 100
• D. 20

Hint:

For linear transformation $Y=ax+b$: Mean changes linearly, variance depends only on $a^2$, not on $b$.

Answer 1

The Correct Option is A

Step 1: Mean and variance of set $X$.
\[ \text{Mean of } X = \frac{1+19}{2}=10 \] Variance of first $n$ natural numbers is \[ \frac{n^2-1}{12} \Rightarrow \text{Var}(X)=\frac{19^2-1}{12}=30 \] Step 2: Use transformation properties.
For $Y=ax+b$: \[ \text{Mean}(Y)=a\,\text{Mean}(X)+b \] \[ \text{Var}(Y)=a^2\text{Var}(X) \] Step 3: Apply variance condition.
\[ a^2(30)=750 \Rightarrow a^2=25 \Rightarrow a=\pm5 \] Step 4: Apply mean condition.
\[ a(10)+b=30 \] For $a=5$: \[ b=30-50=-20 \] For $a=-5$: \[ b=30+50=80 \] Step 5: Sum of all possible values of $b$.
\[ -20+80=60 \] Step 6: Final conclusion.
\[ \boxed{60} \]