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Question Number 6604 by WAI LIN last updated on 05/Jul/16
$${If}\:{the}\:{density}\:{function}\:{of}\:{X}\:{equals} \\ $$ $${f}\left({x}\right)\:=\:\begin{cases}{{c}\:{e}^{−\mathrm{2}{x}} \:,\:\mathrm{0}<{x}<\infty}\\{\mathrm{0}\:,\:\:\:\:\:\:\:\:\:\:{x}<\mathrm{0}}\end{cases} \\ $$ $${find}\:{c}.\:{What}\:{is}\:{P}\left\{{X}>\mathrm{2}\right\}? \\ $$
Commented byprakash jain last updated on 05/Jul/16
$$\int_{\mathrm{0}} ^{\:\infty} {ce}^{−\mathrm{2}} {dx}\neq\mathrm{1} \\ $$ $$\mathrm{So}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{valid}\:\mathrm{probability}\:\mathrm{density} \\ $$ $$\mathrm{function}.\: \\ $$
Commented byWAI LIN last updated on 05/Jul/16
$${Now}\:{I}\:{have}\:{answer}. \\ $$
Commented byWAI LIN last updated on 05/Jul/16
$${Sorry}!\:{Befroe}\:{my}\:{question}\:{is}\:{wrong}. \\ $$
Answered by WAI LIN last updated on 05/Jul/16
![∫_0 ^∞ f(x)dx =1 ∫_0 ^∞ c e^(−2x) dx = 1 −(c/2) [e^(−2) ]_0 ^∞ =1 (c/2) =1 ∴c = 2 Next,P(X>2)=∫_2 ^∞ 2 e^(−2x) dx = −[e^(−2x) ]_2 ^∞ = e^(−4)](Q6623.png)
$$\int_{\mathrm{0}} ^{\infty} \:{f}\left({x}\right){dx}\:=\mathrm{1} \\ $$ $$\int_{\mathrm{0}} ^{\infty} \:{c}\:{e}^{−\mathrm{2}{x}} {dx}\:=\:\mathrm{1} \\ $$ $$−\frac{{c}}{\mathrm{2}}\:\left[{e}^{−\mathrm{2}} \right]_{\mathrm{0}} ^{\infty} \:=\mathrm{1} \\ $$ $$\frac{{c}}{\mathrm{2}}\:=\mathrm{1} \\ $$ $$\therefore{c}\:=\:\mathrm{2} \\ $$ $${Next},{P}\left({X}>\mathrm{2}\right)=\int_{\mathrm{2}} ^{\infty} \:\mathrm{2}\:{e}^{−\mathrm{2}{x}} \:{dx}\:=\:−\left[{e}^{−\mathrm{2}{x}} \right]_{\mathrm{2}} ^{\infty} \:=\:{e}^{−\mathrm{4}} \\ $$