Date | May 2017 | Marks available | 4 | Reference code | 17M.1.hl.TZ1.10 |
Level | HL only | Paper | 1 | Time zone | TZ1 |
Command term | Find | Question number | 10 | Adapted from | N/A |
Question
The continuous random variable X has a probability density function given by
f(x)={ksin(πx6),0⩽.
Find the value of k.
By considering the graph of f write down the mean of X;
By considering the graph of f write down the median of X;
By considering the graph of f write down the mode of X.
Show that P(0 \leqslant X \leqslant 2) = \frac{1}{4}.
Hence state the interquartile range of X.
Calculate P(X \leqslant 4|X \geqslant 3).
Markscheme
attempt to equate integral to 1 (may appear later) M1
k\int\limits_0^6 {\sin \left( {\frac{{\pi x}}{6}} \right){\text{d}}x = 1}
correct integral A1
k\left[ { - \frac{6}{\pi }\cos \left( {\frac{{\pi x}}{6}} \right)} \right]_0^6 = 1
substituting limits M1
- \frac{6}{\pi }( - 1 - 1) = \frac{1}{k}
k = \frac{\pi }{{12}} A1
[4 marks]
mean = 3 A1
Note: Award A1A0A0 for three equal answers in (0,{\text{ }}6).
[1 mark]
median = 3 A1
Note: Award A1A0A0 for three equal answers in (0,{\text{ }}6).
[1 mark]
mode = 3 A1
Note: Award A1A0A0 for three equal answers in (0,{\text{ }}6).
[1 mark]
\frac{\pi }{{12}}\int\limits_0^2 {\sin } \left( {\frac{{\pi x}}{6}} \right){\text{d}}x M1
= \frac{\pi }{{12}}\left[ { - \frac{6}{\pi }\cos \left( {\frac{{\pi x}}{6}} \right)} \right]_0^2 A1
Note: Accept without the \frac{\pi }{{12}} at this stage if it is added later.
\frac{\pi }{{12}}\left[ { - \frac{6}{\pi }\left( {\cos \frac{\pi }{3} - 1} \right)} \right] M1
= \frac{1}{4} AG
[4 marks]
from (c)(i) {Q_1} = 2 (A1)
as the graph is symmetrical about the middle value x = 3 \Rightarrow {Q_3} = 4 (A1)
so interquartile range is
4 - 2
= 2 A1
[3 marks]
P(X \leqslant 4|X \geqslant 3) = \frac{{P(3 \leqslant X \leqslant 4)}}{{P(X \geqslant 3)}}
= \frac{{\frac{1}{4}}}{{\frac{1}{2}}} (M1)
= \frac{1}{2} A1
[2 marks]