Pre-U Pre-U 9795/2 2017 June — Question 4 14 marks

Exam BoardPre-U
ModulePre-U 9795/2 (Pre-U Further Mathematics Paper 2)
Year2017
SessionJune
Marks14
TopicContinuous Probability Distributions and Random Variables
TypePiecewise PDF with k
DifficultyStandard +0.3 This is a straightforward piecewise PDF question requiring standard techniques: integrating to find the constant a (using ∫f(x)dx = 1), constructing the CDF by integrating each piece, and comparing F(0.25) to 0.75 to determine the upper quartile position. All steps are routine applications of core probability theory with no novel insight required, making it slightly easier than average.
Spec5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles

4 A continuous random variable \(X\) has probability density function $$\mathrm { f } ( x ) = \begin{cases} a & - 1 \leqslant x < 0 \\ a \left( 1 - x ^ { 2 } \right) & 0 \leqslant x \leqslant 1 \\ 0 & \text { otherwise } \end{cases}$$ where \(a\) is a constant.
  1. Find the value of \(a\).
  2. Find the cumulative distribution function of \(X\).
  3. Determine whether the upper quartile is greater than or less than 0.25 .

Part (i):
Use total area \(= 1\) with integration M1
\(a + a\left[1 - \frac{1}{3}\right] = 1\) A1 Correct integration
\(a = \frac{3}{5}\) A1 0.6 or \(\frac{3}{5}\) only
Part (ii):
Integrate for one non-zero region (\(a\) can remain, constants can be omitted) M1
\[F(x) = \begin{cases} \frac{3}{5}(x+1) & -1 \leqslant x < 0 \\ \frac{3}{5}\!\left(x - \dfrac{x^3}{3}+1\right) & 0 \leqslant x \leqslant 1 \\ 0 & x < -1 \\ 1 & x > 1 \end{cases}\]
A1 One formula correct; A1 Other formula correct; A1 Ranges \(-1 \leqslant x < 0\) and \(0 \leqslant x \leqslant 1\) correct, allow \(\leqslant\)/\(<\); B1 0 and 1. (Total 4 marks for this part)
Part (iii):
\(F(0.25) = 0.746875 < 0.75\) M1 Evaluate \(F(0.25)\) and compare with 0.75
\(UQ > 0.25\) A1 Correct conclusion from correct values. Alt method by finding UQ: \(4u^3 - 12u + 3 = 0\) (oe) M1; Correct solution (e.g. by GC \(u = 0.25556...\)) (or sign change to demonstrate root between 0.25 and 1) and correct conclusion A1. Alt method: \(u = 0.25 + u^3/3 > 0.25\) and conclusion B2.

Total: 10 marks

**Part (i):**
Use total area $= 1$ with integration **M1**

$a + a\left[1 - \frac{1}{3}\right] = 1$ **A1** Correct integration

$a = \frac{3}{5}$ **A1** 0.6 or $\frac{3}{5}$ only

**Part (ii):**
Integrate for one non-zero region ($a$ can remain, constants can be omitted) **M1**

$$F(x) = \begin{cases} \frac{3}{5}(x+1) & -1 \leqslant x < 0 \\ \frac{3}{5}\!\left(x - \dfrac{x^3}{3}+1\right) & 0 \leqslant x \leqslant 1 \\ 0 & x < -1 \\ 1 & x > 1 \end{cases}$$

**A1** One formula correct; **A1** Other formula correct; **A1** Ranges $-1 \leqslant x < 0$ and $0 \leqslant x \leqslant 1$ correct, allow $\leqslant$/$<$; **B1** 0 and 1. (Total **4** marks for this part)

**Part (iii):**
$F(0.25) = 0.746875 < 0.75$ **M1** Evaluate $F(0.25)$ and compare with 0.75

$UQ > 0.25$ **A1** Correct conclusion from correct values. Alt method by finding UQ: $4u^3 - 12u + 3 = 0$ (oe) M1; Correct solution (e.g. by GC $u = 0.25556...$) (or sign change to demonstrate root between 0.25 and 1) and correct conclusion A1. Alt method: $u = 0.25 + u^3/3 > 0.25$ and conclusion B2.

Total: **10 marks**
4 A continuous random variable $X$ has probability density function

$$\mathrm { f } ( x ) = \begin{cases} a & - 1 \leqslant x < 0 \\ a \left( 1 - x ^ { 2 } \right) & 0 \leqslant x \leqslant 1 \\ 0 & \text { otherwise } \end{cases}$$

where $a$ is a constant.\\
(i) Find the value of $a$.\\
(ii) Find the cumulative distribution function of $X$.\\
(iii) Determine whether the upper quartile is greater than or less than 0.25 .

\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2017 Q4 [14]}}