| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2017 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | Conditional probability with CDF |
| Difficulty | Standard +0.3 This is a straightforward S2 question requiring standard CDF manipulations: P(T>4) = 1-F(4), conditional probability using P(A|B) = P(A∩B)/P(B), solving F(t)=0.75 numerically, and integrating tf(t) for the mean. All techniques are routine for S2 with no novel insight required, making it slightly easier than average. |
| Spec | 2.03d Calculate conditional probability: from first principles5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(P(\text{customer waits} > 4 \text{ minutes}) = 1 - F(4) = 1 - [0.3(4) - 0.004(4^3)] = \frac{7}{125}\) or 0.056 | M1, A1 | M1 for writing or using \(1 - F(4)\) [Just writing \(F(4) = 0.944\) alone is M0] |
| (b) \(P(\text{customer waits} > 4 \text{ minutes} \mid \text{customer waits} > 2 \text{ minutes}) = \frac{P(T > 4)}{P(T > 2)} = \frac{\text{"0.056"}}{\text{1 - F(2)}} = \frac{\text{"0.056"}}{1 - [0.3(2) - 0.004(2^2)]} = \frac{\text{"0.056"}}{0.432} = \frac{7}{54}\) or awrt 0.130 | M1, A1ft, A1 | M1 for a correct ratio expression \(\frac{P(T > 4)}{P(T > 2)}\) or \(\frac{\text{"(a)"}}{\text{P(T > 2)}}\) or better. Ignore e.g. \(P(T > 2 |
| (c) \(F(2.7) = 0.73(1268\ldots)\) or \(F(2.8) = 0.752(192\ldots)\) OR \(F(2.7) - 0.75 = -0.02\ldots\), \(F(2.8) - 0.75 = (+)0.002\ldots\) | M1, A1 cso | M1 for attempting to find \(F(2.7)\) and \(F(2.8)\). A1 cso for both \(F(2.7) = \text{awrt } 0.73 < 0.75 < F(2.8) = \text{awrt } 0.752\) and correct conclusion. May use \(F(x) - 0.75\) and look for a change of sign. There must be sight of 0.75. |
| \(F(2.7) < 0.75 < F(2.8)\) therefore \(2.7 <\) upper quartile \(< 2.8\) | ||
| (d) \(f(t) = 0.3 - 0.012t^2\) | M1 | |
| \(E(T) = \int_0^5 t(0.3 - 0.012t^2) dt = \left[\frac{0.3t^2}{2} - \frac{0.012t^4}{4}\right]_0^5 = \frac{0.3 \times 5^2}{2} - \frac{0.012 \times 5^4}{4}\) | M1, A1 | 1st M1 for differentiating \(F(t)\) to find \(f(t)\) [Just 2 terms with at least 1 correct]. 2nd M1 for attempting to integrate \(t \times\) their \(f(t)\) ignore limits [at least one \(t^n \to t^{n+1}\)]. 1st A1 for correct integration of correct \(f(t)\) & use of limits (must see some correct use of 5) |
| \(= \frac{15}{8}\) or 1.875 | A1 | 2nd A1 \(\frac{15}{8}\) or 1.875 (condone 1.88) [Correct answer only of \(\frac{15}{8}\) or 1.875 scores 4/4] |
**(a)** $P(\text{customer waits} > 4 \text{ minutes}) = 1 - F(4) = 1 - [0.3(4) - 0.004(4^3)] = \frac{7}{125}$ or **0.056** | M1, A1 | M1 for writing or using $1 - F(4)$ [Just writing $F(4) = 0.944$ alone is M0]
**(b)** $P(\text{customer waits} > 4 \text{ minutes} \mid \text{customer waits} > 2 \text{ minutes}) = \frac{P(T > 4)}{P(T > 2)} = \frac{\text{"0.056"}}{\text{1 - F(2)}} = \frac{\text{"0.056"}}{1 - [0.3(2) - 0.004(2^2)]} = \frac{\text{"0.056"}}{0.432} = \frac{7}{54}$ or awrt **0.130** | M1, A1ft, A1 | M1 for a correct ratio expression $\frac{P(T > 4)}{P(T > 2)}$ or $\frac{\text{"(a)"}}{\text{P(T > 2)}}$ or better. Ignore e.g. $P(T > 2|T > 4)$. 1st A1ft for a correct ratio expression with 'their (a)' on numerator and correct numerical expression or 0.43 or better on denominator. 2nd A1 awrt 0.130
**(c)** $F(2.7) = 0.73(1268\ldots)$ or $F(2.8) = 0.752(192\ldots)$ OR $F(2.7) - 0.75 = -0.02\ldots$, $F(2.8) - 0.75 = (+)0.002\ldots$ | M1, A1 cso | M1 for attempting to find $F(2.7)$ and $F(2.8)$. A1 cso for both $F(2.7) = \text{awrt } 0.73 < 0.75 < F(2.8) = \text{awrt } 0.752$ and correct conclusion. May use $F(x) - 0.75$ and look for a change of sign. There must be sight of 0.75.
| $F(2.7) < 0.75 < F(2.8)$ therefore $2.7 <$ upper quartile $< 2.8$ | |
**(d)** $f(t) = 0.3 - 0.012t^2$ | M1 |
| $E(T) = \int_0^5 t(0.3 - 0.012t^2) dt = \left[\frac{0.3t^2}{2} - \frac{0.012t^4}{4}\right]_0^5 = \frac{0.3 \times 5^2}{2} - \frac{0.012 \times 5^4}{4}$ | M1, A1 | 1st M1 for differentiating $F(t)$ to find $f(t)$ [Just 2 terms with at least 1 correct]. 2nd M1 for attempting to integrate $t \times$ their $f(t)$ ignore limits [at least one $t^n \to t^{n+1}$]. 1st A1 for correct integration of correct $f(t)$ & use of limits (must see some correct use of 5)
| $= \frac{15}{8}$ or **1.875** | A1 | 2nd A1 $\frac{15}{8}$ or 1.875 (condone 1.88) [Correct answer only of $\frac{15}{8}$ or 1.875 scores 4/4]
**Total: 11**
---5. A call centre records the length of time, $T$ minutes, its customers wait before being connected to an agent. The random variable $T$ has a cumulative distribution function given by
$$\mathrm { F } ( t ) = \left\{ \begin{array} { l r }
0 & t < 0 \\
0.3 t - 0.004 t ^ { 3 } & 0 \leqslant t \leqslant 5 \\
1 & t > 5
\end{array} \right.$$
\begin{enumerate}[label=(\alph*)]
\item Find the proportion of customers waiting more than 4 minutes to be connected to an agent.
\item Given that a customer waits more than 2 minutes to be connected to an agent, find the probability that the customer waits more than 4 minutes.
\item Show that the upper quartile lies between 2.7 and 2.8 minutes.
\item Find the mean length of time a customer waits to be connected to an agent.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2017 Q5 [11]}}