Let X = {x in N : 1 leq x leq 17} and Y = {ax | vedclass

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(A) Given $X = \{1, 2, \dots, 17\}$. The mean of $X$ is $\bar{x} = \frac{1+17}{2} = 9$. The variance of $X$ is $\sigma_X^2 = \frac{17^2 - 1}{12} = \fr

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$-7$

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(A) Given $X = \{1, 2, \dots, 17\}$. The mean of $X$ is $\bar{x} = \frac{1+17}{2} = 9$. The variance of $X$ is $\sigma_X^2 = \frac{17^2 - 1}{12} = \frac{288}{12} = 24$.For $Y = aX + b$,the mean is $\bar{Y} = a\bar{x} + b = 9a + b = 17$ (Equation $1$).The variance of $Y$ is $\sigma_Y^2 = a^2 \sigma_X^2 = a^2(24) = 216$.$a^2 = \frac{216}{24} = 9$. Since $a > 0$,we have $a = 3$.Substituting $a = 3$ into Equation $1$: $9(3) + b = 17$ $\Rightarrow 27 + b = 17$ $\Rightarrow b = -10$.Thus,$a + b = 3 + (-10) = -7$.

Let $X = \{x \in N : 1 \leq x \leq 17\}$ and $Y = \{ax + b : x \in X \text{ and } a, b \in R, a > 0\}$. If the mean and variance of the elements of $Y$ are $17$ and $216$ respectively,then $a + b$ is equal to:

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