Question 14 Marks
Suppose the error involved in making a certain measurement is a continuous r.v. X with p.d.f.
$f(x) = k(4 – x^2), -2 \leq x \leq 2$ and $= 0$ otherwise.
Compute
$P(X < -0.5$ or $X > 0.5)$
$f(x) = k(4 – x^2), -2 \leq x \leq 2$ and $= 0$ otherwise.
Compute
$P(X < -0.5$ or $X > 0.5)$
Answer
View full question & answer→(iii) $P ( X <-0.5$ or $X >0.5)$
Since, $f$ is the p.d.f. of $X$,
$\int_{-\infty}^{\infty} f(x) d x=1$
$\therefore \int_{-\infty}^{-2} f(x) d x+\int_{-2}^2 f(x) d x+\int_2^{\infty} f(x) d x=1$
$\therefore 0+\int_{-2}^2 k\left(4-x^2\right) d x=1$
$\therefore k \int_{-2}^2\left(4-x^2\right) d x=1$
$\therefore k\left[4 x-\frac{x^3}{3}\right]_{-2}^2=1$
$\therefore k\left[\left(8-\frac{8}{3}\right)-\left(-8+\frac{8}{3}\right)\right]=1$
$\therefore k\left(\frac{16}{3}+\frac{16}{3}\right)=1$
$\therefore k\left(\frac{32}{3}\right)=1$
$\therefore k=\frac{3}{32}$
$P(-0.5<x \text { or } x>0.5)$
$=P(x<-0.5)+P(x>-0.5)$
$=\int_{-\infty}^{-0.5} f(x) d x+\int_{0.5}^{\infty} f(x) d x$
$=\int_{-\infty}^{-2} f(x) d x+\int_{-2}^{-0.5} f(x) d x+\int_{0.5}^2 f(x) d x+\int_2^{\infty} f(x) d x$
$=0+\frac{\int_{-2}^{-1}}{2} k\left(4-x^2\right) d x+\int_{\frac{1}{2}}^2 k\left(4-x^2\right) d x+0$
$=k \frac{\int_{-2}^2}{2}\left(4-x^2\right) d x+\int_{\frac{1}{2}}^2\left(4-x^2\right) d x$
$=\frac{3}{32}\left[4 x-\frac{x^3}{3}\right]_{-2}^{-\frac{1}{2}}+\frac{3}{32}\left[4 x-\frac{x^3}{3}\right]_{\frac{1}{2}}^2 . \ldots . .\left[\because k=\frac{3}{32}\right]$
$=\frac{3}{32}\left[\left(-2+\frac{1}{24}\right)-\left(-8+\frac{8}{3}\right)\right]+\frac{3}{32}\left[\left(8-\frac{8}{3}\right)-\left(2-\frac{1}{24}\right)\right]$
$=\frac{3}{32}\left(\frac{-47}{24}+\frac{16}{3}\right)+\frac{3}{32}\left(\frac{16}{3}-\frac{47}{24}\right)$
$=\frac{3}{32}\left(\frac{-47}{24}+\frac{16}{3}+\frac{16}{3}-\frac{47}{24}\right)$
$=\frac{3}{32}\left(\frac{-47+128+128-47}{24}\right)$
$=\frac{3}{32}\left(\frac{162}{24}\right)=\frac{81}{128}$
$=0.6328$
Since, $f$ is the p.d.f. of $X$,
$\int_{-\infty}^{\infty} f(x) d x=1$
$\therefore \int_{-\infty}^{-2} f(x) d x+\int_{-2}^2 f(x) d x+\int_2^{\infty} f(x) d x=1$
$\therefore 0+\int_{-2}^2 k\left(4-x^2\right) d x=1$
$\therefore k \int_{-2}^2\left(4-x^2\right) d x=1$
$\therefore k\left[4 x-\frac{x^3}{3}\right]_{-2}^2=1$
$\therefore k\left[\left(8-\frac{8}{3}\right)-\left(-8+\frac{8}{3}\right)\right]=1$
$\therefore k\left(\frac{16}{3}+\frac{16}{3}\right)=1$
$\therefore k\left(\frac{32}{3}\right)=1$
$\therefore k=\frac{3}{32}$
$P(-0.5<x \text { or } x>0.5)$
$=P(x<-0.5)+P(x>-0.5)$
$=\int_{-\infty}^{-0.5} f(x) d x+\int_{0.5}^{\infty} f(x) d x$
$=\int_{-\infty}^{-2} f(x) d x+\int_{-2}^{-0.5} f(x) d x+\int_{0.5}^2 f(x) d x+\int_2^{\infty} f(x) d x$
$=0+\frac{\int_{-2}^{-1}}{2} k\left(4-x^2\right) d x+\int_{\frac{1}{2}}^2 k\left(4-x^2\right) d x+0$
$=k \frac{\int_{-2}^2}{2}\left(4-x^2\right) d x+\int_{\frac{1}{2}}^2\left(4-x^2\right) d x$
$=\frac{3}{32}\left[4 x-\frac{x^3}{3}\right]_{-2}^{-\frac{1}{2}}+\frac{3}{32}\left[4 x-\frac{x^3}{3}\right]_{\frac{1}{2}}^2 . \ldots . .\left[\because k=\frac{3}{32}\right]$
$=\frac{3}{32}\left[\left(-2+\frac{1}{24}\right)-\left(-8+\frac{8}{3}\right)\right]+\frac{3}{32}\left[\left(8-\frac{8}{3}\right)-\left(2-\frac{1}{24}\right)\right]$
$=\frac{3}{32}\left(\frac{-47}{24}+\frac{16}{3}\right)+\frac{3}{32}\left(\frac{16}{3}-\frac{47}{24}\right)$
$=\frac{3}{32}\left(\frac{-47}{24}+\frac{16}{3}+\frac{16}{3}-\frac{47}{24}\right)$
$=\frac{3}{32}\left(\frac{-47+128+128-47}{24}\right)$
$=\frac{3}{32}\left(\frac{162}{24}\right)=\frac{81}{128}$
$=0.6328$
