| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2017 |
| Session | October |
| 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 question covering routine pdf techniques: finding k by integration, calculating E(S) and Var(S), finding a probability, then applying binomial distribution to calculate expected value. All steps are textbook exercises with no novel insight required. The symmetry of the pdf makes part (b) trivial. Slightly easier than average due to straightforward structure and symmetric function. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| \(\boldsymbol { X }\) | Bonus Earned |
| \(X \leqslant 5\) | \(\pounds 0\) |
| \(X = 6\) | \(\pounds 1000\) |
| \(X \geqslant 7\) | \(\pounds 5000\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(k\int_{2}^{10}(12s-20-s^2)\,ds[=1]\) | M1 | Attempting to integrate, at least one \(s^n \to s^{n+1}\), ignore limits, does not need to equal 1 |
| \(k\left[6s^2-20s-\frac{s^3}{3}\right]_{2}^{10}[=1]\) | A1 | Correct integration, ignore limits, does not need to equal 1 |
| \(k\left(\frac{200}{3}+\frac{56}{3}\right)=1\) | dM1 | Dependent on first M1; use of both limits and setting equal to 1; must see intermediate line |
| \(k=\frac{3}{256}\) | A1cso | Condone use of \(x\) instead of \(s\), condone missing \(ds\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(E(S)=6\) | B1 | Ignore £6000 if 6 is seen |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(E(S^2)=k\int_{2}^{10}(12s^3-20s^2-s^4)\,ds\) | M1 | Attempting to integrate \(s^2\cdot f(s)\); \(s^n \to s^{n+1}\) |
| \(=\frac{3}{256}\left[3s^4-\frac{20s^3}{3}-\frac{s^5}{5}\right]_{2}^{10}\) | A1ft | Correct integration (or ft correct integration of \(s^2\cdot f(s)\)) |
| \(=39.2\) | ||
| \(\text{Var}(S)=39.2-6^2=3.2\) | M1 | Using \(E(S^2)-[E(S)]^2\) |
| \(\text{s.d.}(S)=\sqrt{3.2}=1.7888\) | dM1 A1 | Dependent on previous M1; square rooting \(\text{Var}(S)\); awrt 1.79 |
| standard deviation \(= £1788.85\), awrt £1790 | A1ft | Dependent on all method marks; \(1000\times\) their s.d. |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\frac{3}{256}\int_{7.1}^{10}(12s-20-s^2)\,ds\) | M1 | Correct expression with correct limits; ft their \(f(s)\) |
| \(=0.2989=0.3\) (1 dp) | A1 | awrt 0.3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(P(X\leq 5)=0.8822\) | ||
\(P(5| M1 A1ft |
M1 writing/using \(P(X\leq 6)-P(X\leq 5)\) or correct expression for \(P(X=6)\); A1ft awrt 0.079 |
|
\(P(6| M1 |
Using \(1-P(X\leq 6)\), \(X\sim B(12,\) "their ans to (d)") |
|
| Bonus \(=1000\times 0.0792+5000\times 0.0386\) | M1 | \(1000\times\) "their 0.0792" \(+5000\times\) "their 0.0386" |
| \(=£79.20+£193.00=£272.20\), awrt £270 | A1 | awrt £270 (2sf) |
## Question 2:
### Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $k\int_{2}^{10}(12s-20-s^2)\,ds[=1]$ | M1 | Attempting to integrate, at least one $s^n \to s^{n+1}$, ignore limits, does not need to equal 1 |
| $k\left[6s^2-20s-\frac{s^3}{3}\right]_{2}^{10}[=1]$ | A1 | Correct integration, ignore limits, does not need to equal 1 |
| $k\left(\frac{200}{3}+\frac{56}{3}\right)=1$ | dM1 | Dependent on first M1; use of both limits and setting equal to 1; must see intermediate line |
| $k=\frac{3}{256}$ | A1cso | Condone use of $x$ instead of $s$, condone missing $ds$ |
### Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $E(S)=6$ | B1 | Ignore £6000 if 6 is seen |
### Part (c):
| Working | Marks | Guidance |
|---------|-------|----------|
| $E(S^2)=k\int_{2}^{10}(12s^3-20s^2-s^4)\,ds$ | M1 | Attempting to integrate $s^2\cdot f(s)$; $s^n \to s^{n+1}$ |
| $=\frac{3}{256}\left[3s^4-\frac{20s^3}{3}-\frac{s^5}{5}\right]_{2}^{10}$ | A1ft | Correct integration (or ft correct integration of $s^2\cdot f(s)$) |
| $=39.2$ | | |
| $\text{Var}(S)=39.2-6^2=3.2$ | M1 | Using $E(S^2)-[E(S)]^2$ |
| $\text{s.d.}(S)=\sqrt{3.2}=1.7888$ | dM1 A1 | Dependent on previous M1; square rooting $\text{Var}(S)$; awrt 1.79 |
| standard deviation $= £1788.85$, awrt **£1790** | A1ft | Dependent on all method marks; $1000\times$ their s.d. |
### Part (d):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\frac{3}{256}\int_{7.1}^{10}(12s-20-s^2)\,ds$ | M1 | Correct expression with correct limits; ft their $f(s)$ |
| $=0.2989=0.3$ (1 dp) | A1 | awrt 0.3 |
### Part (e):
| Working | Marks | Guidance |
|---------|-------|----------|
| $P(X\leq 5)=0.8822$ | | |
| $P(5<X\leq 6)=P(X\leq 6)-P(X\leq 5)=0.9614-0.8822=$ awrt $0.079$ | M1 A1ft | M1 writing/using $P(X\leq 6)-P(X\leq 5)$ or correct expression for $P(X=6)$; A1ft awrt 0.079 |
| $P(6<X\leq 12)=1-P(X\leq 6)=0.0386$ | M1 | Using $1-P(X\leq 6)$, $X\sim B(12,$ "their ans to (d)") |
| Bonus $=1000\times 0.0792+5000\times 0.0386$ | M1 | $1000\times$ "their 0.0792" $+5000\times$ "their 0.0386" |
| $=£79.20+£193.00=£272.20$, awrt **£270** | A1 | awrt £270 (2sf) |
---2. The weekly sales, $S$, in thousands of pounds, of a small business has probability density function
$$\mathrm { f } ( s ) = \left\{ \begin{array} { c c }
k ( s - 2 ) ( 10 - s ) & 2 < s < 10 \\
0 & \text { otherwise }
\end{array} \right.$$
\begin{enumerate}[label=(\alph*)]
\item Use algebraic integration to show that $k = \frac { 3 } { 256 }$
\item Write down the value of $\mathrm { E } ( S )$
\item Use algebraic integration to find the standard deviation of the weekly sales.
A week is selected at random.
\item Showing your working, find the probability that this week's sales exceed $\pounds 7100$ Give your answer to one decimal place.
A quarter is defined as 12 consecutive weeks.
The discrete random variable $X$ is the number of weeks in a quarter in which the weekly sales exceed £7100
The manager earns a bonus at the following rates:
\begin{center}
\begin{tabular}{ | c | c | }
\hline
$\boldsymbol { X }$ & Bonus Earned \\
\hline
$X \leqslant 5$ & $\pounds 0$ \\
\hline
$X = 6$ & $\pounds 1000$ \\
\hline
$X \geqslant 7$ & $\pounds 5000$ \\
\hline
\end{tabular}
\end{center}
\item Using your answer to part (d), calculate the manager's expected bonus per quarter.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2017 Q2 [18]}}