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

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

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

aa

The probability density function

f(x)f(x)

represents a quarter-circle with radius

aa

, meaning it satisfies the condition:

0af(x)dx=1\int_0^a f(x) \,dx = 1

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

14πa2\frac{1}{4} \pi a^2

Setting this equal to 1:

14πa2=1\frac{1}{4} \pi a^2 = 1

Solving for

aa

:

a2=4πa^2 = \frac{4}{\pi} a=2πa = \frac{2}{\sqrt{\pi}}


(b) Finding

f(x)f(x)

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

(0,0)(0,0)

with radius

aa

, the equation of the full circle is:

x2+y2=a2x^2 + y^2 = a^2

Solving for

f(x)=yf(x) = y

:

f(x)=a2x2f(x) = \sqrt{a^2 – x^2}

Substituting

a2=4πa^2 = \frac{4}{\pi}

:

f(x)=4πx2f(x) = \sqrt{\frac{4}{\pi} – x^2}


(c) Finding

E(X)E(X)

The expected value of

XX

is:

E(X)=0axf(x)dxE(X) = \int_0^a x f(x) \, dx

Substituting

f(x)=4πx2f(x) = \sqrt{\frac{4}{\pi} – x^2}

:

E(X)=0ax4πx2dxE(X) = \int_0^a x \sqrt{\frac{4}{\pi} – x^2} \, dx

Using the standard result for this type of integral:

0axa2x2dx=a33\int_0^a x \sqrt{a^2 – x^2} \,dx = \frac{a^3}{3}

Substituting

a=2πa = \frac{2}{\sqrt{\pi}}

:

E(X)=13(2π)3E(X) = \frac{1}{3} \left(\frac{2}{\sqrt{\pi}}\right)^3 E(X)=13×8π3/2E(X) = \frac{1}{3} \times \frac{8}{\pi^{3/2}} E(X)=83π3E(X) = \frac{8}{3\sqrt{\pi^3}}


Final Answers:

  • (a):
    a=2πa = \frac{2}{\sqrt{\pi}}
  • (b):
    f(x)=4πx2f(x) = \sqrt{\frac{4}{\pi} – x^2}
  • (c):
    E(X)=83π3E(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) $\int_{0}^{\frac{2}{\sqrt{\pi}}}x\sqrt{\frac{4}{\pi}-x^{2}}dx$

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

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

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