| Date | May 2018 | Marks available | 3 | Reference code | 18M.2.hl.TZ1.10 |
| Level | HL only | Paper | 2 | Time zone | TZ1 |
| Command term | Show that | Question number | 10 | Adapted from | N/A |
Question
The continuous random variable X has probability density function \(f\) given by
\[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
{3ax}&,&{0 \leqslant x < 0.5} \\
{a\left( {2 - x} \right)}&,&{0.5 \leqslant x < 2} \\
0&,&{{\text{otherwise}}}
\end{array}} \right.\]
Show that \(a = \frac{2}{3}\).
Find \({\text{P}}\left( {X < 1} \right)\).
Given that \({\text{P}}\left( {s < X < 0.8} \right) = 2 \times {\text{P}}\left( {2s < X < 0.8} \right)\), and that 0.25 < s < 0.4 , find the value of s.
Markscheme
\(a\left[ {\int_0^{0.5} {3x\,{\text{d}}x} + \int_{0.5}^2 {\left( {2 - x} \right)} \,{\text{d}}x} \right] = 1\) M1
Note: Award the M1 for the total integral equalling 1, or equivalent.
\(a\left( {\frac{3}{2}} \right) = 1\) (M1)A1
\(a = \frac{2}{3}\) AG
[3 marks]
EITHER
\(\int_0^{0.5} {2x\,{\text{d}}x} + \frac{2}{3}\int_{0.5}^1 {\left( {2 - x} \right)} \,{\text{d}}x\) (M1)(A1)
\( = \frac{2}{3}\) A1
OR
\(\frac{2}{3}\int_1^2 {\left( {2 - x} \right)} \,{\text{d}}x = \frac{1}{3}\) (M1)
so \({\text{P}}\left( {X < 1} \right) = \frac{2}{3}\) (M1)A1
[3 marks]
\({\text{P}}\left( {s < X < 0.8} \right) = \int_s^{0.5} {2x\,{\text{d}}x} + \frac{2}{3}\int_{0.5}^{0.8} {\left( {2 - x} \right)} \,{\text{d}}x\) M1A1
\( = \left[ {{x^2}} \right]_s^{0.5} + 0.27\)
\(0.25 - {s^2} + 0.27\) (A1)
\({\text{P}}\left( {2s < X < 0.8} \right) = \frac{2}{3}\int_{2s}^{0.8} {\left( {2 - x} \right)} \,{\text{d}}x\) A1
\( = \frac{2}{3}\left[ {2x - \frac{{{x^2}}}{2}} \right]_{2s}^{0.8}\)
\(\frac{2}{3}\left( {1.28 - \left( {4s - 2{s^2}} \right)} \right)\)
equating
\(0.25 - {s^2} + 0.27 = \frac{4}{3}\left( {1.28 - \left( {4s - 2{s^2}} \right)} \right)\) (A1)
attempt to solve for s (M1)
s = 0.274 A1
[7 marks]
