CIE A level Math -Probability & Statistics 2: 6.3 Continuous random variables: Exam Style Questions Paper 6

Solution: –

(a) Finding

The probability density function

represents a quarter-circle with radius

, meaning it satisfies the condition:

$$\int_0^a f(x) \,dx = 1$$

Since the area under the quarter-circle is given by:

$$\frac{1}{4} \pi a^2$$

Setting this equal to 1:

$$\frac{1}{4} \pi a^2 = 1$$

Solving for

:

$$a^2 = \frac{4}{\pi}$$ $$a = \frac{2}{\sqrt{\pi}}$$

(b) Finding

Since the graph is part of a quarter-circle centered at

with radius

, the equation of the full circle is:

$$x^2 + y^2 = a^2$$

Solving for

:

$$f(x) = \sqrt{a^2 – x^2}$$

Substituting

:

$$f(x) = \sqrt{\frac{4}{\pi} – x^2}$$

(c) Finding

The expected value of

is:

$$E(X) = \int_0^a x f(x) \, dx$$

Substituting

:

$$E(X) = \int_0^a x \sqrt{\frac{4}{\pi} – x^2} \, dx$$

Using the standard result for this type of integral:

$$\int_0^a x \sqrt{a^2 – x^2} \,dx = \frac{a^3}{3}$$

Substituting

:

$$E(X) = \frac{1}{3} \left(\frac{2}{\sqrt{\pi}}\right)^3$$ $$E(X) = \frac{1}{3} \times \frac{8}{\pi^{3/2}}$$ $$E(X) = \frac{8}{3\sqrt{\pi^3}}$$

Final Answers:

• (a):
  $a = \frac{2}{\sqrt{\pi}}$
• (b):
  $f(x) = \sqrt{\frac{4}{\pi} – x^2}$
• (c):
  $E(X) = \frac{8}{3\sqrt{\pi^3}}$

—-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) $\int_{0}^{\frac{2}{\sqrt{\pi}}}x\sqrt{\frac{4}{\pi}-x^{2}}dx$

$=-\frac{1}{3}\left[\left(\frac{4}{\pi}-x^{2}\right)^{\frac{3}{2}}\right]\begin{matrix}\frac{2}{\sqrt{\pi}}\\ 0\end{matrix}$

$\frac{8}{3\sqrt{\pi^{3}}}$