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Question Number 86668 by jagoll last updated on 30/Mar/20
$$\mathrm{what}\:\mathrm{is}\:\mathrm{P}\left(\mid\mathrm{x}\mid\:>\:\mathrm{1}\:\right)\:\mathrm{if}\:\mathrm{x}\:\mathrm{has}\:\mathrm{a}\:\mathrm{PDF}\:\mathrm{of} \\ $$ $$\mathrm{f}\left(\mathrm{x}\right)\:=\begin{cases}{\frac{\mathrm{1}}{\mathrm{4}}\:,\:\:−\mathrm{2}<\mathrm{x}<\mathrm{2}}\\{\mathrm{0}\:,\:\mathrm{elsewhere}}\end{cases} \\ $$
Commented byjohn santu last updated on 30/Mar/20
![⇒P(x<−1 ∪ x>1) = ∫_(−2) ^(−1) f(x)dx+∫_1 ^2 f(x)dx = ∫_(−2) ^(−1) (1/4)dx + ∫_1 ^2 (1/4)dx = (1/4)[−1+2] +(1/4)[2−1] = (1/2)](Q86669.png)
$$\Rightarrow\mathrm{P}\left(\mathrm{x}<−\mathrm{1}\:\cup\:\mathrm{x}>\mathrm{1}\right) \\ $$ $$=\:\int_{−\mathrm{2}} ^{−\mathrm{1}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}+\int_{\mathrm{1}} ^{\mathrm{2}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \\ $$ $$=\:\underset{−\mathrm{2}} {\overset{−\mathrm{1}} {\int}}\frac{\mathrm{1}}{\mathrm{4}}\mathrm{dx}\:+\:\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\frac{\mathrm{1}}{\mathrm{4}}\mathrm{dx} \\ $$ $$=\:\frac{\mathrm{1}}{\mathrm{4}}\left[−\mathrm{1}+\mathrm{2}\right]\:+\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{2}−\mathrm{1}\right] \\ $$ $$=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$