| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | June |
| Marks | 18 |
| 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 continuous probability distribution question requiring routine application of formulas (E(T), Var(T) = E(T²) - [E(T)]², CDF integration, percentiles, and conditional probability). While it involves piecewise functions and multiple parts (6 marks total likely), each step follows textbook procedures with no novel insight required. The algebraic integration is straightforward, and part (b) is trivial given E(T²). Slightly easier than average due to the given value simplifying the variance calculation. |
| 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 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(E(T) = \int_1^2 \frac{1}{2}t(t-1)\,dt + \int_2^4 \frac{1}{16}t(14t-3t^2-8)\,dt\) | M1 | Using \(\int tf(t)\) for both parts, attempt to multiply out and integrate |
| \(= \left[\frac{t^3}{6} - \frac{t^2}{4}\right]_1^2 + \left[\frac{14t^3}{48} - \frac{3t^4}{64} - \frac{8t^2}{32}\right]_2^4\) | A1 | Correct integration for both parts |
| \(= \frac{5}{12} + \frac{25}{12}\) | M1dep | Dep on previous M1; adding 2 parts and substituting correct limits |
| \(= 2.5\) or \(\frac{5}{2}\) | A1 | Must use algebraic integration |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\text{Var}(T) = 6.675 - (2.5)^2\) | M1 | \(\frac{267}{40} - [\text{their } E(T)]^2\); must see \(-1^2\) if \(E(T)=1\) |
| \(= \frac{17}{40}\) or \(0.425\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(F(t) = 0\), \(t \leq 1\) | ||
| \(\frac{1}{4}t^2 - \frac{1}{2}t + \frac{1}{4}\) or \(\frac{(t-1)^2}{4}\), \(1 < t \leq 2\) | M1 A1 | \(\int_1^t \frac{1}{2}(x-1)\,dx\) with correct limits; \(x^n \to x^{n+1}\) |
| \(\frac{1}{16}(7t^2 - t^3 - 8t)\), \(2 < t \leq 4\) | M1 A1 | Must be in the cdf; correct method shown |
| \(1\), \(t > 4\) | B1 | Fully correct all in terms of \(t\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{1}{4}t^2 - \frac{1}{2}t + \frac{1}{4} = 0.2\) | M1 | Their cdf for \(1 < t \leq 2 = 0.2\) |
| \(t^2 - 2t + 0.2 = 0\) | ||
| \(t = \frac{2 \pm \sqrt{2^2 - 4(1)(0.2)}}{2}\) | M1 | Correct method for solving 3-term quadratic |
| \(t = 1.894\ldots\) | A1 | awrt 1.89; allow \(\frac{5+2\sqrt{5}}{5}\) or \(1 + \frac{2}{\sqrt{5}}\); one answer only |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(1 - F(1.5) = 1 - \left(\frac{1}{4}\times1.5^2 - \frac{1}{2}\times1.5 + \frac{1}{4}\right)\) | M1 | Substitute 1.5 into line for \(1 < t \leq 2\) or \(\int_1^{1.5}\frac{1}{2}(t-1)\,dt\) |
| \(= \frac{15}{16}\) or \(0.9375\) | A1 | awrt 0.938 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(P(T>3) = 0.25\) | ||
| \(P(T>3 \mid T>1.5) = \frac{\text{"0.25"}}{\text{"0.9375"}}\) | M1 | \(0.25/\text{their (e)}\) or \([1-\text{their }F(3)]/\text{their (e)}\) |
| \(= \frac{4}{15}\) or awrt \(0.267\) | A1 |
# Question 3:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $E(T) = \int_1^2 \frac{1}{2}t(t-1)\,dt + \int_2^4 \frac{1}{16}t(14t-3t^2-8)\,dt$ | M1 | Using $\int tf(t)$ for both parts, attempt to multiply out and integrate |
| $= \left[\frac{t^3}{6} - \frac{t^2}{4}\right]_1^2 + \left[\frac{14t^3}{48} - \frac{3t^4}{64} - \frac{8t^2}{32}\right]_2^4$ | A1 | Correct integration for both parts |
| $= \frac{5}{12} + \frac{25}{12}$ | M1dep | Dep on previous M1; adding 2 parts and substituting correct limits |
| $= 2.5$ or $\frac{5}{2}$ | A1 | Must use algebraic integration |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\text{Var}(T) = 6.675 - (2.5)^2$ | M1 | $\frac{267}{40} - [\text{their } E(T)]^2$; must see $-1^2$ if $E(T)=1$ |
| $= \frac{17}{40}$ or $0.425$ | A1 | |
## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $F(t) = 0$, $t \leq 1$ | | |
| $\frac{1}{4}t^2 - \frac{1}{2}t + \frac{1}{4}$ or $\frac{(t-1)^2}{4}$, $1 < t \leq 2$ | M1 A1 | $\int_1^t \frac{1}{2}(x-1)\,dx$ with correct limits; $x^n \to x^{n+1}$ |
| $\frac{1}{16}(7t^2 - t^3 - 8t)$, $2 < t \leq 4$ | M1 A1 | Must be in the cdf; correct method shown |
| $1$, $t > 4$ | B1 | Fully correct all in terms of $t$ |
## Part (d):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{1}{4}t^2 - \frac{1}{2}t + \frac{1}{4} = 0.2$ | M1 | Their cdf for $1 < t \leq 2 = 0.2$ |
| $t^2 - 2t + 0.2 = 0$ | | |
| $t = \frac{2 \pm \sqrt{2^2 - 4(1)(0.2)}}{2}$ | M1 | Correct method for solving 3-term quadratic |
| $t = 1.894\ldots$ | A1 | awrt 1.89; allow $\frac{5+2\sqrt{5}}{5}$ or $1 + \frac{2}{\sqrt{5}}$; one answer only |
## Part (e):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $1 - F(1.5) = 1 - \left(\frac{1}{4}\times1.5^2 - \frac{1}{2}\times1.5 + \frac{1}{4}\right)$ | M1 | Substitute 1.5 into line for $1 < t \leq 2$ or $\int_1^{1.5}\frac{1}{2}(t-1)\,dt$ |
| $= \frac{15}{16}$ or $0.9375$ | A1 | awrt 0.938 |
## Part (f):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $P(T>3) = 0.25$ | | |
| $P(T>3 \mid T>1.5) = \frac{\text{"0.25"}}{\text{"0.9375"}}$ | M1 | $0.25/\text{their (e)}$ or $[1-\text{their }F(3)]/\text{their (e)}$ |
| $= \frac{4}{15}$ or awrt $0.267$ | A1 | |
---\begin{enumerate}
\item The length of time, $T$, minutes, spent completing a particular task has probability density function
\end{enumerate}
$$f ( t ) = \left\{ \begin{array} { c c }
\frac { 1 } { 2 } ( t - 1 ) & 1 < t \leqslant 2 \\
\frac { 1 } { 16 } \left( 14 t - 3 t ^ { 2 } - 8 \right) & 2 < t \leqslant 4 \\
0 & \text { otherwise }
\end{array} \right.$$
(a) Use algebraic integration to find $\mathrm { E } ( T )$
Given that $\mathrm { E } \left( T ^ { 2 } \right) = \frac { 267 } { 40 }$\\
(b) find $\operatorname { Var } ( T )$\\
(c) Find the cumulative distribution function $\mathrm { F } ( t )$\\
(d) Find the 20th percentile of the time taken to complete the task.\\
(e) Find the probability that the time spent completing the task is more than 1.5 minutes.
Given that a person has already spent 1.5 minutes on the task,\\
(f) find the probability that this person takes more than 3 minutes to complete the task.
\hfill \mbox{\textit{Edexcel S2 2018 Q3 [18]}}