Let X = {x ∈ N : 1 ≤ x ≤ 19} and for some a, b ∈ R, Y = {ax + b : x ∈ 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

Let X = {x ∈ N : 1 ≤ x ≤ 19} and Y = {ax + b : x ∈ X}. If the mean and variance of Y are 30 and 750, respectively, then the sum of all possible values of b is

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

(A) 20

(B) 100

(D) 60

Correct Answer: 60

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.

Explanation

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