A random Variable Y has probability function P, defined by P(y) = { (((y^2 /k) , y= 1,2,3)),(((((y−7)^2 )/k) , y= 4,5,6)),((0 , otherwise.)) :} Find (i) The value of the constant k. (ii) the mean and varriance of Y. (iii) The variance R, where R= 2Y −3.


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Question Number 63296 by Rio Michael last updated on 02/Jul/19
$A random Variable Y has probability function P, defined by P(y) = { (((y^2 /k) , y= 1,2,3)),(((((y−7)^2 )/k) , y= 4,5,6)),((0 , otherwise.)) :} Find (i) The value of the constant k. (ii) the mean and varriance of Y. (iii) The variance R, where R= 2Y −3.$
$A r a n d o m V a r i a b l e Y h a s p r o b a b i l i t y f u n c t i o n P, d e f i n e d b y$ $P (y) = {\begin{cases} \frac{y^{2}}{k}, y = 1, 2, 3 \\ \frac{{(y - 7)}^{2}}{k}, y = 4, 5, 6 \\ 0, o t h e r w i s e . \end{cases}$ $F i n d$ $(i) T h e v a l u e o f t h e c o n s t a n t k .$ $(i i) t h e m e a n a n d v a r r i a n c e o f Y .$ $(i i i) T h e v a r i a n c e R, w h e r e R = 2 Y - 3 .$
Commented by peter frank last updated on 02/Jul/19

$m e a n = 225.75$ $v a r (y) = - 38656.8$ $v a r (2 R - 3) = 4 v a r (2 R - 3)$ $p l e a s e c h e c k$
	Answered by peter frank last updated on 02/Jul/19

	${\int_{1}}^{3} \frac{y^{2}}{k} = 1$ $\frac{1}{k} {[\frac{y^{3}}{3}]}_{1}^{3} = 1$ $\frac{1}{k} [\frac{27}{3} - \frac{1}{3}] = 1$ $3 - \frac{1}{3} = k$ $\frac{8}{3} = k$ $p l e a s e c h e c k$
	Commented by Rio Michael last updated on 02/Jul/19

	$y e a h . . i t h i n k f r o m m y s o l u t i o n$ $8 = 3 k .$
	Commented by Rio Michael last updated on 02/Jul/19

	$i f i u s e t a b l e o f v a l u e s f o r y = 1, 2, 3, 4, 5, 6, 7 .$ $t h e n c o m p u t e f o r P (y)$ $a n d u s e Σ P (x) = 1 ?$
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