Suponha que f(x)=0,5 \cos x, para \frac{-\pi}{2} \leqslant x \leqslant \frac{\pi}{2}. Determine a função acumulada de probabilidade e Calcule
(a) P(X<0)
(b) P\left(-\frac{\pi}{4} \leqslant x \leqslant \frac{\pi}{4}\right)
(c) P\left(X<-\frac{\pi}{4}\right)
(d) P\left(X>-\frac{\pi}{4}\right)
(e) Determine x tal que P(X<x)=0,95.
R. F(x)=\left\{\begin{array}{cr}0, & x \leqslant-\frac{\pi}{2}, \\\\ 0,5 \sin (x)+0,5, & \frac{-\pi}{2} \leqslant x \leqslant \frac{\pi}{2}, \\\\ 1, & x \geqslant \frac{\pi}{2} .\end{array}\right.

Passo 1

O enunciado dá uma densidade de probabilidade contínua
f(x)=0,5\cos x,\quad -\tfrac{\pi}{2}\le x\le\tfrac{\pi}{2},
e pede a função acumulada F(x) e várias probabilidades. Vamos usar integração para obter F(x) e então calcular cada caso.

Passo 2

Para -\tfrac{\pi}{2}\le x\le\tfrac{\pi}{2} a função acumulada é
F(x)=\int_{-\frac{\pi}{2}}^{x}0,5\cos t\,dt =0,5\bigl[\sin t\bigr]_{-\frac{\pi}{2}}^{x} =0,5\bigl(\sin x-(-1)\bigr) =0,5\sin x+0,5.
Fora desse intervalo vale
F(x)= \begin{cases} 0,&x\le -\tfrac{\pi}{2},\\\\ 0,5\sin x+0,5,&-\tfrac{\pi}{2}\le x\le\tfrac{\pi}{2},\\\\ 1,&x\ge\tfrac{\pi}{2}. \end{cases}

Passo 3

P(X<0) é simplesmente F(0). Como \sin0=0,
P(X<0)=F(0)=0,5\sin0+0,5=0,5.

Passo 4

Para P\bigl(-\tfrac{\pi}{4}\le X\le\tfrac{\pi}{4}\bigr) usamos
F\bigl(\tfrac{\pi}{4}\bigr)=0,5\sin\tfrac{\pi}{4}+0,5 =0,5\cdot\tfrac{\sqrt2}{2}+0,5 =0,5+\tfrac{\sqrt2}{4},
F\bigl(-\tfrac{\pi}{4}\bigr)=0,5\sin\bigl(-\tfrac{\pi}{4}\bigr)+0,5 =0,5\bigl(-\tfrac{\sqrt2}{2}\bigr)+0,5 =0,5-\tfrac{\sqrt2}{4}.
Logo
P\bigl(-\tfrac{\pi}{4}\le X\le\tfrac{\pi}{4}\bigr) =F\bigl(\tfrac{\pi}{4}\bigr)-F\bigl(-\tfrac{\pi}{4}\bigr) =\Bigl(0,5+\tfrac{\sqrt2}{4}\Bigr) -\Bigl(0,5-\tfrac{\sqrt2}{4}\Bigr) =\tfrac{\sqrt2}{2}.

Passo 5

Para P\bigl(X<-\tfrac{\pi}{4}\bigr) é só
P\bigl(X<-\tfrac{\pi}{4}\bigr)=F\bigl(-\tfrac{\pi}{4}\bigr) =0,5-\tfrac{\sqrt2}{4}.

Passo 6

Para P\bigl(X>-\tfrac{\pi}{4}\bigr) usamos complementar
P\bigl(X>-\tfrac{\pi}{4}\bigr)=1-F\bigl(-\tfrac{\pi}{4}\bigr) =1-\Bigl(0,5-\tfrac{\sqrt2}{4}\Bigr) =0,5+\tfrac{\sqrt2}{4}.

Passo 7

Para achar x tal que P(X<x)=0,95 resolvemos
0,5\sin x+0,5=0,95 \quad\Longrightarrow\quad \sin x=0,9 \quad\Longrightarrow\quad x=\arcsin(0,9)\approx1,1198.

Resposta

(a) 0,5
(b) \tfrac{\sqrt2}{2}
(c) 0,5-\tfrac{\sqrt2}{4}
(d) 0,5+\tfrac{\sqrt2}{4}
(e) x=\arcsin(0,9)\approx1,1198

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