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Date November 2011 Marks available 5 Reference code 11N.1.hl.TZ0.10
Level HL only Paper 1 Time zone TZ0
Command term Find Question number 10 Adapted from N/A

Question

A continuous random variable X has the probability density function

\[f(x) = \left\{ {\begin{array}{*{20}{c}}
  {k\sin x,}&{0 \leqslant x \leqslant \frac{\pi }{2}} \\
  {0,}&{{\text{otherwise}}{\text{.}}}
\end{array}} \right.\]

Find the value of k.

[2]
a.

Find \({\text{E}}(X)\).

[5]
b.

Find the median of X.

[3]
c.

Markscheme

\(k\int_0^{\frac{\pi }{2}} {\sin x{\text{d}}x = 1} \)     M1

\(k[ - \cos x]_0^{\frac{\pi }{2}} = 1\)

k = 1     A1

[2 marks]

a.

\({\text{E}}(X) = \int_0^{\frac{\pi }{2}} {x\sin x{\text{d}}x} \)     M1

integration by parts     M1

\([ - x\cos x]_0^{\frac{\pi }{2}} + \int_0^{\frac{\pi }{2}} {\cos x{\text{d}}x} \)     A1A1

= 1     A1

[5 marks]

b.

\(\int_0^M {\sin x{\text{d}}x}  = \frac{1}{2}\)     M1

\([ - \cos x]_0^M = \frac{1}{2}\)     A1

\(\cos M = \frac{1}{2}\)

\(M = \frac{\pi }{3}\)     A1

Note: accept \(\arccos \frac{1}{2}\)

 

[3 marks]

c.

Examiners report

 

Most candidates scored maximum marks on this question. A few candidates found k = –1.

 

a.

 

Most candidates scored maximum marks on this question. A few candidates found k = –1.

 

b.

 

Most candidates scored maximum marks on this question. A few candidates found k = –1.

 

c.

Syllabus sections

Topic 5 - Core: Statistics and probability » 5.5 » Expected value (mean), mode, median, variance and standard deviation.
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