| Date | May 2017 | Marks available | 5 | Reference code | 17M.2.hl.TZ2.10 |
| Level | HL only | Paper | 2 | Time zone | TZ2 |
| Command term | Show that | Question number | 10 | Adapted from | N/A |
Question
A continuous random variable \(X\) has probability density function \(f\) given by
\(f(x) = \left\{ {\begin{array}{*{20}{l}} {\frac{{{x^2}}}{a} + b,}&{0 \leqslant x \leqslant 4} \\ 0&{{\text{otherwise}}} \end{array}} \right.{\text{where }}a{\text{ and }}b{\text{ are positive constants.}}\)
It is given that \({\text{P}}(X \geqslant 2) = 0.75\).
Eight independent observations of \(X\) are now taken and the random variable \(Y\) is the number of observations such that \(X \geqslant 2\).
Show that \(a = 32\) and \(b = \frac{1}{{12}}\).
Find \({\text{E}}(X)\).
Find \({\text{Var}}(X)\).
Find the median of \(X\).
Find \({\text{E}}(Y)\).
Find \({\text{P}}(Y \geqslant 3)\).
Markscheme
\(\int\limits_0^4 {\left( {\frac{{{x^2}}}{a} + b} \right){\text{d}}x = 1 \Rightarrow \left[ {\frac{{{x^3}}}{{3a}} + bx} \right]_0^4 = 1 \Rightarrow \frac{{64}}{{3a}} + 4b = 1} \) M1A1
\(\int\limits_2^4 {\left( {\frac{{{x^2}}}{a} + b} \right){\text{d}}x = 0.75 \Rightarrow \frac{{56}}{{3a}} + 2b = 0.75} \) M1A1
Note: \(\int\limits_0^2 {\left( {\frac{{{x^2}}}{a} + b} \right)} \,dx = 0.25 \Rightarrow \frac{8}{{3a}} + 2b = 0.25\) could be seen/used in place of either of the above equations.
evidence of an attempt to solve simultaneously (or check given a,b values are consistent) M1
\(a = 32,{\text{ }}b = \frac{1}{{12}}\) AG
[5 marks]
\({\text{E}}\left( X \right) = \int\limits_0^4 {x\left( {\frac{{{x^2}}}{{32}} + \frac{1}{{12}}} \right){\text{d}}x} \) (M1)
\({\text{E}}(X) = \frac{8}{3}\,\,\,( = 2.67)\) A1
[2 marks]
\({\text{E}}\left( {{X^2}} \right) = \int\limits_0^4 {{x^2}\left( {\frac{{{x^2}}}{{32}} + \frac{1}{{12}}} \right){\text{d}}x} \) (M1)
\({\text{Var}}(X) = {\text{E}}({X^2}) - {[{\text{E}}(X)]^2} = \frac{{16}}{{15}}\,\,\,( = 1.07)\) A1
[2 marks]
\(\int\limits_0^m {\left( {\frac{{{x^2}}}{{32}} + \frac{1}{{12}}} \right){\text{d}}x = 0.5} \) (M1)
\(\frac{{{m^3}}}{{96}} + \frac{m}{{12}} = 0.5\,\,\,( \Rightarrow {m^3} + 8m - 48 = 0)\) (A1)
\(m = 2.91\) A1
[3 marks]
\(Y \sim B(8,{\text{ }}0.75)\) (M1)
\({\text{E}}(Y) = 8 \times 0.75 = 6\) A1
[2 marks]
\({\text{P}}(Y \geqslant 3) = 0.996\) A1
[1 mark]
