| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | Mode from PDF |
| Difficulty | Standard +0.3 This is a straightforward S2 question testing standard CDF/PDF concepts. Parts (a)-(b) require simple substitution into the given CDF, part (c) is routine differentiation, part (d) involves differentiating the PDF and solving a quadratic, and part (e) is a qualitative comment. All techniques are standard with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(= P(T > 2) = 1 - F(2) = 1 - \frac{1}{135}(108 + 36 - 32) = \frac{23}{135}\) | M1, M1, A1 | |
| (b) \(F(m) = \frac{1}{2}\); \(F(1.1) = 0.4812\); \(F(1.2) = 0.5248\) \(\therefore 1.1 < m < 1.2\) \(\therefore\) median between 11 and 12 minutes | M1, M1, A1 | |
| (c) \(f(t) = F'(t) = \frac{1}{135}(54 + 18t - 12t^2)\); \(f(t) = \begin{cases} \frac{2}{45}(9 + 3t - 2t^2), & 0 \leq t \leq 3, \\ 0, & \text{otherwise}. \end{cases}\) | M1, A1, A1 | |
| (d) \(t'(t) = \frac{2}{45}(3 - 4t)\); S.P. when \(t'(t) = 0\) \(\therefore t = \frac{3}{4}\); some justification e.g. –ve quadratic \(\therefore\) mode \(= \frac{3}{4} \times 10 = 7.5\) mins | M1, M1, M1, A1 | |
| (e) e.g. assumes patients never wait for more than 30 mins | B1 | (14 marks) |
**(a)** $= P(T > 2) = 1 - F(2) = 1 - \frac{1}{135}(108 + 36 - 32) = \frac{23}{135}$ | M1, M1, A1 |
**(b)** $F(m) = \frac{1}{2}$; $F(1.1) = 0.4812$; $F(1.2) = 0.5248$ $\therefore 1.1 < m < 1.2$ $\therefore$ median between 11 and 12 minutes | M1, M1, A1 |
**(c)** $f(t) = F'(t) = \frac{1}{135}(54 + 18t - 12t^2)$; $f(t) = \begin{cases} \frac{2}{45}(9 + 3t - 2t^2), & 0 \leq t \leq 3, \\ 0, & \text{otherwise}. \end{cases}$ | M1, A1, A1 |
**(d)** $t'(t) = \frac{2}{45}(3 - 4t)$; S.P. when $t'(t) = 0$ $\therefore t = \frac{3}{4}$; some justification e.g. –ve quadratic $\therefore$ mode $= \frac{3}{4} \times 10 = 7.5$ mins | M1, M1, M1, A1 |
**(e)** e.g. assumes patients never wait for more than 30 mins | B1 | (14 marks)The length of time, in tens of minutes, that patients spend waiting at a doctor's surgery is modelled by the continuous random variable $T$, with the following cumulative distribution function:
$$F(t) = \begin{cases}
0, & t < 0, \\
\frac{1}{135}(54t + 9t^2 - 4t^3), & 0 \leq t \leq 3, \\
1, & t > 3.
\end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Find the probability that a patient waits for more than 20 minutes. [3]
\item Show that the median waiting time is between 11 and 12 minutes. [3]
\item Define fully the probability density function f(t) of $T$. [3]
\item Find the modal waiting time in minutes. [4]
\item Give one reason why this model may need to be refined. [1]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q6 [14]}}