Let X = {x in mathbb{N} : 1 le x le 19} and f | vedclass

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(D) The mean of $X$ is $\bar{x} = \frac{1+19}{2} = 10$.The variance of $X$ is $\sigma_x^2 = \frac{n^2-1}{12} = \frac{19^2-1}{12} = \frac{360}{12} = 30

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$60$

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(D) The mean of $X$ is $\bar{x} = \frac{1+19}{2} = 10$.The variance of $X$ is $\sigma_x^2 = \frac{n^2-1}{12} = \frac{19^2-1}{12} = \frac{360}{12} = 30$.Given $Y = aX + b$, the mean of $Y$ is $\bar{y} = a\bar{x} + b = 10a + b = 30$.The variance of $Y$ is $\sigma_y^2 = a^2 \sigma_x^2 = a^2(30) = 750$.Thus, $a^2 = 25$, which means $a = 5$ or $a = -5$.If $a = 5$, then $10(5) + b = 30 \Rightarrow b = -20$.If $a = -5$, then $10(-5) + b = 30 \Rightarrow b = 80$.The sum of all possible values of $b$ is $-20 + 80 = 60$.

$x_i$	$4$	$8$	$11$	$17$	$20$	$24$	$32$
$f_i$	$3$	$5$	$9$	$5$	$4$	$3$	$1$

Let $X = \{x \in \mathbb{N} : 1 \le x \le 19\}$ 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

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