Question Number 63296 by Rio Michael last updated on 02/Jul/19
$${A}\:{random}\:{Variable}\:{Y}\:{has}\:{probability}\:{function}\:{P},\:{defined}\:{by} \\ $$$$\:{P}\left({y}\right)\:=\:\begin{cases}{\frac{{y}^{\mathrm{2}} }{{k}}\:,\:{y}=\:\mathrm{1},\mathrm{2},\mathrm{3}}\\{\frac{\left({y}−\mathrm{7}\right)^{\mathrm{2}} }{{k}}\:,\:{y}=\:\mathrm{4},\mathrm{5},\mathrm{6}}\\{\mathrm{0}\:\:\:\:,\:{otherwise}.}\end{cases} \\ $$$${Find}\: \\ $$$$\left({i}\right)\:{The}\:{value}\:{of}\:{the}\:{constant}\:{k}. \\ $$$$\left({ii}\right)\:{the}\:{mean}\:{and}\:{varriance}\:{of}\:{Y}. \\ $$$$\left({iii}\right)\:{The}\:{variance}\:{R},\:{where}\:{R}=\:\mathrm{2}{Y}\:−\mathrm{3}. \\ $$
Commented by peter frank last updated on 02/Jul/19
$${mean}=\mathrm{225}.\mathrm{75} \\ $$$${var}\left({y}\right)=−\mathrm{38656}.\mathrm{8} \\ $$$${var}\left(\mathrm{2}{R}−\mathrm{3}\right)=\mathrm{4}{var}\left(\mathrm{2}{R}−\mathrm{3}\right) \\ $$$${please}\:{check} \\ $$
Answered by peter frank last updated on 02/Jul/19
![∫^3 _1 (y^2 /k)=1 (1/k)[(y^3 /3)]_1 ^3 =1 (1/k)[((27)/3)−(1/3)]=1 3−(1/3)=k (8/3)=k please check](Q63316.png)
$$\underset{\mathrm{1}} {\int}^{\mathrm{3}} \frac{{y}^{\mathrm{2}} }{{k}}=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{{k}}\left[\frac{{y}^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{1}} ^{\mathrm{3}} =\mathrm{1} \\ $$$$\frac{\mathrm{1}}{{k}}\left[\frac{\mathrm{27}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\right]=\mathrm{1} \\ $$$$\mathrm{3}−\frac{\mathrm{1}}{\mathrm{3}}={k} \\ $$$$\frac{\mathrm{8}}{\mathrm{3}}={k} \\ $$$${please}\:{check} \\ $$
Commented by Rio Michael last updated on 02/Jul/19
$${yeah}..\:{i}\:{think}\:{from}\:{my}\:{solution} \\ $$$$\mathrm{8}\:=\:\mathrm{3}{k}. \\ $$
Commented by Rio Michael last updated on 02/Jul/19
$${if}\:{i}\:{use}\:{table}\:{of}\:{values}\:{for}\:{y}=\:\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7}. \\ $$$${then}\:{compute}\:{for}\:{P}\left({y}\right)\: \\ $$$${and}\:{use}\:\:\Sigma{P}\left({x}\right)\:=\mathrm{1}? \\ $$