Pre-U Pre-U 9795/2 2020 Specimen — Question 6 2 marks

Exam BoardPre-U
ModulePre-U 9795/2 (Pre-U Further Mathematics Paper 2)
Year2020
SessionSpecimen
Marks2
TopicContinuous Probability Distributions and Random Variables
TypeCalculate and compare mean, median, mode
DifficultyStandard +0.3 This is a standard continuous probability distribution question requiring routine calculus (integration for mean, differentiation for mode) and numerical methods for quartiles. While it involves multiple parts, each step follows textbook procedures without requiring novel insight or particularly challenging manipulation.
Spec1.09d Newton-Raphson method5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration

6 The lengths of time, in years, that sales representatives for a certain company keep their company cars may be modelled by the distribution with probability density function \(\mathrm { f } ( x )\), where $$f ( x ) = \left\{ \begin{array} { c c } \frac { 4 } { 27 } x ^ { 2 } ( 3 - x ) & 0 \leqslant x \leqslant 3 \\ 0 & \text { otherwise. } \end{array} \right.$$
  1. Draw a sketch of this probability density function.
  2. Calculate the mean and the mode of \(X\).
  3. Comment briefly on the values obtained in part (b) in relation to the sketch in part (a).
  4. Show that the lower quartile \(\mathrm { Q } _ { 1 }\) of \(X\) satisfies the equation \(\mathrm { Q } _ { 1 } { } ^ { 4 } - 4 \mathrm { Q } _ { 1 } { } ^ { 3 } + 6.75 = 0\), and use an appropriate numerical method to find the value of \(\mathrm { Q } _ { 1 }\) correct to 2 decimal places, showing full details of your method.

Question 6(a):
Above \(x\)-axis between \((0, 0)\) to \((3, 0)\) [B1]
Correct concavity. (Do not condone parabolas) [B1]
Total: 2 marks
Question 6(b):
\(\mu = \dfrac{4}{27}\int_0^3 (3x^3 - x^4)\,\text{d}x\) (Limits required) [M1]
\(= \dfrac{4}{27}\left[\dfrac{3x^4}{4} - \dfrac{x^5}{5}\right]_0^3 = 1.8\) [A1A1]
\(f'(x) = \dfrac{4}{27}(6x - 3x^2) = 0\) [M1A1]
\(\Rightarrow x = 0, 2 \quad \therefore\) Mode \(= 2\) [A1]
Total: 6 marks
Question 6(c):
Mean less than mode in (b) matches negative skew in sketch. [B1]
Total: 1 mark
Question 6(d):
\(\left[\dfrac{4}{27}\left(x^3 - \dfrac{x^4}{4}\right)\right]_0^l = \dfrac{1}{4} \Rightarrow Q_1^4 - 4Q_1^3 + 6.75 = 0\) AG [B1]
Use e.g. \(Q_{1_{n+1}} = Q_{1_n} - \dfrac{Q_{1_n}^4 - 4Q_{1_n}^3 + 6.75}{4Q_{1_n}^3 - 12Q_{1_n}^2}\) or decimal search etc. [M1A1]
Convincingly obtain \(1.37\) [A1]
Total: 4 marks
**Question 6(a):**

Above $x$-axis between $(0, 0)$ to $(3, 0)$ [B1]

Correct concavity. (Do not condone parabolas) [B1]

**Total: 2 marks**

**Question 6(b):**

$\mu = \dfrac{4}{27}\int_0^3 (3x^3 - x^4)\,\text{d}x$ (Limits required) [M1]

$= \dfrac{4}{27}\left[\dfrac{3x^4}{4} - \dfrac{x^5}{5}\right]_0^3 = 1.8$ [A1A1]

$f'(x) = \dfrac{4}{27}(6x - 3x^2) = 0$ [M1A1]

$\Rightarrow x = 0, 2 \quad \therefore$ Mode $= 2$ [A1]

**Total: 6 marks**

**Question 6(c):**

Mean less than mode in **(b)** matches negative skew in sketch. [B1]

**Total: 1 mark**

**Question 6(d):**

$\left[\dfrac{4}{27}\left(x^3 - \dfrac{x^4}{4}\right)\right]_0^l = \dfrac{1}{4} \Rightarrow Q_1^4 - 4Q_1^3 + 6.75 = 0$ **AG** [B1]

Use e.g. $Q_{1_{n+1}} = Q_{1_n} - \dfrac{Q_{1_n}^4 - 4Q_{1_n}^3 + 6.75}{4Q_{1_n}^3 - 12Q_{1_n}^2}$ or decimal search etc. [M1A1]

Convincingly obtain $1.37$ [A1]

**Total: 4 marks**
6 The lengths of time, in years, that sales representatives for a certain company keep their company cars may be modelled by the distribution with probability density function $\mathrm { f } ( x )$, where

$$f ( x ) = \left\{ \begin{array} { c c } 
\frac { 4 } { 27 } x ^ { 2 } ( 3 - x ) & 0 \leqslant x \leqslant 3 \\
0 & \text { otherwise. }
\end{array} \right.$$
\begin{enumerate}[label=(\alph*)]
\item Draw a sketch of this probability density function.
\item Calculate the mean and the mode of $X$.
\item Comment briefly on the values obtained in part (b) in relation to the sketch in part (a).
\item Show that the lower quartile $\mathrm { Q } _ { 1 }$ of $X$ satisfies the equation $\mathrm { Q } _ { 1 } { } ^ { 4 } - 4 \mathrm { Q } _ { 1 } { } ^ { 3 } + 6.75 = 0$, and use an appropriate numerical method to find the value of $\mathrm { Q } _ { 1 }$ correct to 2 decimal places, showing full details of your method.
\end{enumerate}

\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2020 Q6 [2]}}