| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2018 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | General probability threshold |
| Difficulty | Standard +0.8 This is a multi-part binomial distribution question requiring normal approximation with continuity correction, algebraic manipulation of binomial coefficients, and finding the mode through ratio analysis. While the techniques are standard A-level material, the question demands careful reasoning across multiple connected parts, particularly the algebraic proof in (ii) and the inequality work in (iii), making it moderately challenging but not exceptional. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n2.04c Calculate binomial probabilities2.04d Normal approximation to binomial2.05f Pearson correlation coefficient |
| Answer | Marks | Guidance |
|---|---|---|
| \(N(450 \times 0.15, 450 \times 0.15 \times 0.85)\) or \(N(67.5, 57.375)\) oe | M1 | AO 3.1b |
| \(P(Y > \mu + \sigma) \approx \frac{1}{6}\) or \(\Phi^{-1}(\frac{1}{6}) = 0.9674\) | ||
| \('67.5' + \sqrt{57.375}\) or \('67.5' + 0.9674 \times \sqrt{57.375}\) | M1 | AO 1.2 |
| \(= 74\) or \(75\) or \(76\) | A1 [3] | AO 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{50!}{(r+1)!(50-(r+1))!} \times 0.15^{r+1} \times 0.85^{50-(r+1)}\) oe | M1 | AO 1.1a |
| eg \(\frac{\frac{1}{50-r} \times 0.85}{\frac{1}{r+1} \times 0.15}\) or \(\frac{0.85}{50-r} \times \frac{r+1}{0.15}\) oe | A1 | AO 2.1 |
| \(= \frac{17(r+1)}{3(50-r)}\) AG | A1 [3] | AO 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{17(r+1)}{3(50-r)} \leq 1\) oe | M1 | AO 3.1b |
| \(17r + 17 \leq 150 - 3r\) | ||
| \(20r \leq 133\) oe | A1 | AO 1.1 |
| \(r \leq 6.65\) | A1 | AO 1.1 |
| \(r\) is an integer so \(r \leq 6\) | A1 [4] | AO 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{P(X=6)}{P(X=7)} = \frac{17(6+1)}{3(50-6)} = 0.902 < 1\) | B1 | AO 2.1 |
| Most likely value is 7 | B1 [2] | AO 3.2a |
# Question 13:
## Part (i):
$N(450 \times 0.15, 450 \times 0.15 \times 0.85)$ or $N(67.5, 57.375)$ oe | M1 | AO 3.1b | seen or implied | $B(450, 0.15)$ with T & I method using $\geq$ one of 74, 75, 76, 61, 60, 59 |
$P(Y > \mu + \sigma) \approx \frac{1}{6}$ or $\Phi^{-1}(\frac{1}{6}) = 0.9674$ | | | $P(Y < a) = \frac{5}{6}$ | $P(X > 74) = 0.177$; $P(X > 75) = 0.145$ both |
$'67.5' + \sqrt{57.375}$ or $'67.5' + 0.9674 \times \sqrt{57.375}$ | M1 | AO 1.2 | or 74.83 seen; ft their $\mu$ & $\sigma$ for M1 only |
$= 74$ or $75$ or $76$ | A1 [3] | AO 1.1 | Integer. No ft Dep M1M1; Correct ans, inadequate wking: M0M0A0; **NB** $450/6 = 75$ **M0M0A0** | $a = 74$ or $75$ or $76$ |
## Part (ii):
$\frac{50!}{r!(50-r)!} \times 0.15^r \times 0.85^{50-r}$
$\frac{50!}{(r+1)!(50-(r+1))!} \times 0.15^{r+1} \times 0.85^{50-(r+1)}$ oe | M1 | AO 1.1a | $\frac{{}^{50}C_r \times 0.15^r \times 0.85^{50-r}}{{}^{50}C_{r+1} \times 0.15^{r+1} \times 0.85^{50-(r+1)}}$ | Fully correct |
eg $\frac{\frac{1}{50-r} \times 0.85}{\frac{1}{r+1} \times 0.15}$ or $\frac{0.85}{50-r} \times \frac{r+1}{0.15}$ oe | A1 | AO 2.1 | Any correct simplification without factorials **OR** without indices | or $\frac{17}{20} \times \frac{20}{3} \times \frac{r+1}{50-r}$ |
$= \frac{17(r+1)}{3(50-r)}$ **AG** | A1 [3] | AO 1.1 | Any correct simplification without factorials **AND** without indices and correctly obtain result |
## Part (iii)(a):
$\frac{17(r+1)}{3(50-r)} \leq 1$ oe | M1 | AO 3.1b | $\frac{1}{50-r} \times 0.85 \leq \frac{1}{r+1} \times 0.15$ oe | No factorials or indices |
$17r + 17 \leq 150 - 3r$ | | | $0.85(r+1) \leq 0.15(50-r)$ |
$20r \leq 133$ oe | A1 | AO 1.1 | $r \leq 50 \times 0.15 - 0.85$ | Correct, in form $ar \leq b$ or $r <$ correct expr'n |
$r \leq 6.65$ | A1 | AO 1.1 | |
$r$ is an integer so $r \leq 6$ | A1 [4] | AO 1.1 | SC: $P(X=6)=0.142$, $P(X=7)=0.157$, $P(X=8)=0.149$; B1 (must be these three) hence $r \leq 6$ B1dep | No wking B0B0 |
## Part (iii)(b):
$P(X=r) \leq P(X=r+1)$ for $r \leq 6$
Hence most likely value is $r$ is 6 or 7
$\frac{P(X=6)}{P(X=7)} = \frac{17(6+1)}{3(50-6)} = 0.902 < 1$ | B1 | AO 2.1 | or $P(X=6) = 0.142$ & $P(X=7) = 0.157$ | NOT 6.65 rounds to 7 B0B0 |
Most likely value is 7 | B1 [2] | AO 3.2a | indep, but dep on some reasonable explanation | No expl'n: B0B0 |
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To extract mark scheme content, please share the **interior pages** of the document that contain the actual marking guidance.13 In this question you must show detailed reasoning.
The probability that Paul's train to work is late on any day is 0.15 , independently of other days.\\
(i) The number of days on which Paul's train to work is late during a 450-day period is denoted by the random variable $Y$. Find a value of $a$ such that $\mathrm { P } ( Y > a ) \approx \frac { 1 } { 6 }$.
In the expansion of $( 0.15 + 0.85 ) ^ { 50 }$, the terms involving $0.15 ^ { r }$ and $0.15 ^ { r + 1 }$ are denoted by $T _ { r }$ and $T _ { r + 1 }$ respectively.\\
(ii) Show that $\frac { T _ { r } } { T _ { r + 1 } } = \frac { 17 ( r + 1 ) } { 3 ( 50 - r ) }$.\\
(iii) The number of days on which Paul's train to work is late during a 50-day period is modelled by the random variable $X$.
\begin{enumerate}[label=(\alph*)]
\item Find the values of $r$ for which $\mathrm { P } ( X = r ) \leqslant \mathrm { P } ( X = r + 1 )$.
\item Hence find the most likely number of days on which the train will be late during a 50-day period.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/02 2018 Q13 [12]}}