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

Let's solve the problem using basic concepts of statistics. Given:

• \(X = \{x \in \mathbb{N} : 1 \leq x \leq 19\}\)
• Set \(Y = \{ax + b : x \in X\}\)
• The mean of \(Y\) is 30
• The variance of \(Y\) is 750

First, calculate the mean and variance of \(X\):

• Mean of \(X\) is given by:

\(\text{Mean of } X = \frac{\sum_{x=1}^{19} x}{19} = \frac{\frac{19 \times (19+1)}{2}}{19} = 10\)

• The variance of the set \({1, 2, ..., 19}\) is:

\(\text{Variance of } X = \frac{\sum_{x=1}^{19} (x - 10)^2}{19} = \frac{\sum_{x=1}^{19} x^2}{19} - 10^2\)

\(\sum_{x=1}^{19} x^2 = \frac{19 \times (19+1)(2 \times 19+1)}{6} = 2470\)

\(\text{Variance of } X = \frac{2470}{19} - 100 = 30\)

Now, convert these values to \(Y\):

• The mean of \(Y\) can be represented using the expected value:

\(\text{Mean of } Y = \text{Mean of } (ax + b) = a \times \text{Mean of } X + b\)

\(30 = 10a + b\)

• The variance relationship:

\(\text{Variance of } (ax+b) = a^2 \times \text{Variance of } X\)

\(750 = a^2 \times 30\)

\(a^2 = 25 \Rightarrow a = \pm 5\)

Substitute values of \(a\) into the mean equation:

• For \(a = 5\):

\(30 = 10 \times 5 + b \quad \Rightarrow \quad b = -20\)

• For \(a = -5\):

\(30 = 10 \times (-5) + b \quad \Rightarrow \quad b = 80\)

Finally, sum all possible values of \(b\):

\(b = -20 + 80 = 60\)

Therefore, the sum of all possible values of \(b\) is 60.