Statistical inference — Question 18

QCAA 2024, Paper 1 · 6 marks

Q18 · 2024 · Technology-freeComplex unfamiliar6 marks

QUESTION 18 (6 marks)

A random variable

X

has a probability density function given by

f(x)=\begin{cases}k\sin^{-1}(x),&0\le x\le1,\\0,&\text{otherwise},\end{cases}

where

k

is a positive constant. Determine the value of

k

WORKED SOLUTION

6 marks

ANSWER

k=\frac{2}{\pi-2}

Worked solution

Using the pdf property,

\int_0^1k\sin^{-1}(x)dx=1.

Integration by parts gives

\int k\sin^{-1}(x)dx=kx\sin^{-1}(x)+k\sqrt{1-x^2}+c.

Hence

k\left[x\sin^{-1}x+\sqrt{1-x^2}\right]_0^1=1,

k\left(\frac{\pi}{2}-1\right)=1,

and therefore

k=\frac{2}{\pi-2}.

correctly uses a suitable pdf property

[1 mark]

correctly determines both results required to progress integration by parts

[1 mark]

uses integration by parts

[1 mark]

uses a suitable substitution method to progress the developed integral within the use of integration by parts

[1 mark]

determines a general result for the required integral

[1 mark]

determines the value of

k

[1 mark]

One QCAA sample method typeset for web; criterion wording is adapted from the official marking guide.

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