| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2020 |
| Session | October |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Direct variance calculation from pdf |
| Difficulty | Standard +0.3 This is a standard S2 question requiring routine application of variance formula (E(T²) - [E(T)]²) with piecewise integration, followed by straightforward CDF construction and probability calculations. The mean is given, reducing computational burden. All techniques are textbook exercises 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 pdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| VIXV SIHIANI III IM IONOO | VIAV SIHI NI JYHAM ION OO | VI4V SIHI NI JLIYM ION OO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(T^2) = \int_0^3 \frac{1}{50}(18t^2 - 2t^3)\,dt + \int_3^5 \frac{1}{20}t^2\,dt\) | M1 | Intention to find \(E(T^2)\) correctly; must add 2 integrals and attempt to integrate at least one term \(x^n \to x^{n+1}\); algebraic integration must be seen |
| \(= \left[\frac{1}{50}\left(6t^3 - \frac{t^4}{2}\right)\right]_0^3 + \left[\frac{t^3}{60}\right]_3^5\) | A1 | Correct integration |
| \(= \frac{1}{50}\left(6\times3^3 - \frac{3^4}{2}\right) + \left(\frac{125}{60} - \frac{27}{60}\right)\) | M1d | Dep on previous M; correct limits and attempt to substitute |
| \(= \frac{1219}{300} = 4.063...\) | — | Allow 1219/300 or 243/100 or 49\30 oe |
| \(\text{Var}(T) = \text{"4.063..."} - (1.66)^2\) | M1 | For their \(E(T^2) - 1.66^2\) |
| \(= 1.3077...\) | A1 | awrt 1.31; allow 2452/1875 oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int_3^t \frac{1}{20}\,dx + C\) where \(C = 0.9\) or \(\int_0^3 \frac{1}{50}(18-2t)\,dt\) or using \(F(5)=1\) to find \(C\) | M1 | For a correct method to find the 3rd line including limits unless using \(F(5)=1\) method |
| \(F(t) = 0\) for \(t < 0\) | B1 | 2nd line correct — any letter; ignore missing inequality |
| \(F(t) = \frac{1}{50}(18t - t^2)\) or \(1.62 - \frac{(18-2t)^2}{200}\), \(\quad 0 \leq t \leq 3\) | A1 | 3rd line correct — any letter; ignore missing inequality |
| \(F(t) = \frac{1}{20}t + 0.75\), \(\quad 3 < t \leq 5\) | A1 | |
| \(F(t) = 1\) for \(t > 5\) | — | Fully correct CDF; allow \(<\) instead of \(\leq\) and vice versa |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(T > 2) = 1 - \frac{1}{50}(18\times2 - 2^2)\) or \(1 - \int_0^2 \frac{1}{50}(18-2t)\,dt\) | M1 | For finding \(1 - F(2)\) using their second line or starting again; must substitute in 2 |
| \(= \frac{9}{25}\) or \(0.36\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(0 < T < 3.66) = F(3.66)\) | M1 | For realising they need \(F(3.66)\); allow \(F(3.66)[-F(0)]\); allow \(F(\text{"their mean"}+2")[-F(0)]\) |
| \(= 0.933\) | A1 | cao; allow answer as a fraction |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(T^2) = \int_0^3 \frac{1}{50}(18t^2 - 2t^3)\,dt + \int_3^5 \frac{1}{20}t^2\,dt$ | M1 | Intention to find $E(T^2)$ correctly; must add 2 integrals and attempt to integrate at least one term $x^n \to x^{n+1}$; algebraic integration must be seen |
| $= \left[\frac{1}{50}\left(6t^3 - \frac{t^4}{2}\right)\right]_0^3 + \left[\frac{t^3}{60}\right]_3^5$ | A1 | Correct integration |
| $= \frac{1}{50}\left(6\times3^3 - \frac{3^4}{2}\right) + \left(\frac{125}{60} - \frac{27}{60}\right)$ | M1d | Dep on previous M; correct limits and attempt to substitute |
| $= \frac{1219}{300} = 4.063...$ | — | Allow 1219/300 or 243/100 or 49\30 oe |
| $\text{Var}(T) = \text{"4.063..."} - (1.66)^2$ | M1 | For their $E(T^2) - 1.66^2$ |
| $= 1.3077...$ | A1 | awrt 1.31; allow 2452/1875 oe |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_3^t \frac{1}{20}\,dx + C$ where $C = 0.9$ or $\int_0^3 \frac{1}{50}(18-2t)\,dt$ or using $F(5)=1$ to find $C$ | M1 | For a correct method to find the 3rd line including limits unless using $F(5)=1$ method |
| $F(t) = 0$ for $t < 0$ | B1 | 2nd line correct — any letter; ignore missing inequality |
| $F(t) = \frac{1}{50}(18t - t^2)$ or $1.62 - \frac{(18-2t)^2}{200}$, $\quad 0 \leq t \leq 3$ | A1 | 3rd line correct — any letter; ignore missing inequality |
| $F(t) = \frac{1}{20}t + 0.75$, $\quad 3 < t \leq 5$ | A1 | |
| $F(t) = 1$ for $t > 5$ | — | Fully correct CDF; allow $<$ instead of $\leq$ and vice versa |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(T > 2) = 1 - \frac{1}{50}(18\times2 - 2^2)$ or $1 - \int_0^2 \frac{1}{50}(18-2t)\,dt$ | M1 | For finding $1 - F(2)$ using their second line or starting again; must substitute in 2 |
| $= \frac{9}{25}$ or $0.36$ | A1 | cao |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(0 < T < 3.66) = F(3.66)$ | M1 | For realising they need $F(3.66)$; allow $F(3.66)[-F(0)]$; allow $F(\text{"their mean"}+2")[-F(0)]$ |
| $= 0.933$ | A1 | cao; allow answer as a fraction |
---5. The waiting time, $T$ minutes, of a customer to be served in a local post office has probability density function
$$\mathrm { f } ( t ) = \begin{cases} \frac { 1 } { 50 } ( 18 - 2 t ) & 0 \leqslant t \leqslant 3 \\ \frac { 1 } { 20 } & 3 < t \leqslant 5 \\ 0 & \text { otherwise } \end{cases}$$
Given that the mean number of minutes a customer waits to be served is 1.66
\begin{enumerate}[label=(\alph*)]
\item use algebraic integration to find $\operatorname { Var } ( T )$, giving your answer to 3 significant figures.
\item Find the cumulative distribution function $\mathrm { F } ( t )$ for all values of $t$.
\item Calculate the probability that a randomly chosen customer's waiting time will be more than 2 minutes.
\item Calculate $\mathrm { P } ( [ \mathrm { E } ( T ) - 2 ] < T < [ \mathrm { E } ( T ) + 2 ] )$\\
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VIXV SIHIANI III IM IONOO & VIAV SIHI NI JYHAM ION OO & VI4V SIHI NI JLIYM ION OO \\
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\hfill \mbox{\textit{Edexcel S2 2020 Q5 [13]}}