Question
The diagram shows the graph of the probability density function, $f$, of a random variable $X$. The graph is a quarter circle entirely in the first quadrant with centre $(0,0)$ and radius $a$, where $a$ is a positive constant. Elsewhere $f(x) = 0$.
(a) Show that $a = \frac{2}{\sqrt{\pi}}$.
(b) Show that $f(x) = \sqrt{\frac{4}{\pi} – x^2}$.
(c) Show that $E(X) = \frac{8}{3\sqrt{\pi^3}}$.
▶️Answer/Explanation
Solution: –
(a) Finding
The probability density function
represents a quarter-circle with radius
, meaning it satisfies the condition:
Since the area under the quarter-circle is given by:
Setting this equal to 1:
Solving for
:
(b) Finding
Since the graph is part of a quarter-circle centered at
with radius
, the equation of the full circle is:
Solving for
:
Substituting
:
(c) Finding
The expected value of
is:
Substituting
:
Using the standard result for this type of integral:
Substituting
:
Final Answers:
- (a):
- (b):
- (c):
—-Markscheme—————
6(a) $\frac{1}{4}\pi a^{2}=1$
$a=\frac{2}{\sqrt{\pi}}$
6(b) $x^{2}+y^{2}=(\frac{2}{\sqrt{\pi}})^{2}$
$[y^{2}=\frac{4}{\pi}-x^{2}]$
$f(x)=\sqrt{\frac{4}{\pi}-x^{2}}$
6(c) $\int_{0}^{\frac{2}{\sqrt{\pi}}}x\sqrt{\frac{4}{\pi}-x^{2}}dx$
$=-\frac{1}{3}\left[\left(\frac{4}{\pi}-x^{2}\right)^{\frac{3}{2}}\right]\begin{matrix}\frac{2}{\sqrt{\pi}}\\ 0\end{matrix}$
$\frac{8}{3\sqrt{\pi^{3}}}$