Question ID: #897
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
- (1) 20
- (2) 80
- (3) 100
- (4) 60
Solution:
Set $X = \{1, 2, …, 19\}$. Number of terms $n = 19$.
Mean of X ($\bar{x}$) = $\frac{n+1}{2} = \frac{19+1}{2} = 10$.
Variance of X ($\sigma_x^2$) = $\frac{n^2-1}{12} = \frac{19^2-1}{12} = \frac{360}{12} = 30$.
Given $y_i = ax_i + b$.
Mean of Y ($\bar{y}$) = $a\bar{x} + b$.
$$30 = a(10) + b \Rightarrow 10a + b = 30 \quad \dots(1)$$
Variance of Y ($\sigma_y^2$) = $a^2 \sigma_x^2$.
$$750 = a^2(30) \Rightarrow a^2 = 25 \Rightarrow a = \pm 5$$
Case 1: If $a = 5$
From (1): $10(5) + b = 30 \Rightarrow 50 + b = 30 \Rightarrow b = -20$.
Case 2: If $a = -5$
From (1): $10(-5) + b = 30 \Rightarrow -50 + b = 30 \Rightarrow b = 80$.
Sum of all possible values of b = $-20 + 80 = 60$.
Ans. (4)
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