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-mea

Question Number 63296 by Rio Michael last updated on 02/Jul/19

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 ?