| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2024 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Repeated binomial experiments |
| Difficulty | Standard +0.3 This is a straightforward application of binomial distribution followed by normal approximation. Part (a) is routine binomial calculation, (b) uses binomial again with probability from (a), (c) is simple expectation (60×10×1/6), and (d) is standard normal approximation with continuity correction. All steps are textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| \(X \sim B(10, \frac{1}{6})\) [Allow 0.167 or better for \(\frac{1}{6}\)] | M1 | Must have B, or Bin or Bpd or Bcd and correct values for \(n\) and \(p\); just \(n=10\), \(p=\frac{1}{6}\) is M0 |
| Answer | Marks | Guidance |
|---|---|---|
| \([P(X=3)=]\ 0.155045\ldots\) awrt 0.155 | A1 | 1st A1 for awrt 0.155 |
| Answer | Marks | Guidance |
|---|---|---|
| \([P(X<3) = P(X \leq 2) =]\ 0.775226\ldots\) awrt 0.775 | A1 | 2nd A1 for awrt 0.775 |
| Answer | Marks | Guidance |
|---|---|---|
| [Let \(D\) = no. of days when \(X=3\)] \(D \sim B(60, \text{"0.155"})\) | M1 | 1st M1 for selecting correct model; sight or use of correct binomial, ft their (a)(i) |
| \(P(D \geq 12) = 1 - P(D \leq 11)\) [Allow \(1 - P(D < 12)\)] | M1 | 2nd M1 for correct interpretation of "at least 12" and writing or using \(1 - P(D \leq 11)\) |
| \(= 1 - 0.78819\ldots\) awrt 0.212 | A1 | For awrt 0.212 [Answer only 3/3] |
| Answer | Marks | Guidance |
|---|---|---|
| \([n = 600,\ p = \frac{1}{6}]\) estimate = 100 | B1 | For 100 but must be seen in part (c) i.e. between (b) and (d) |
| Answer | Marks | Guidance |
|---|---|---|
| \([S = \text{total no. of sixes over 60 days.}]\ S \approx T \sim N\!\left(\text{"100"},\ \sqrt{\frac{5}{6} \times 100}\ ^2\right)\) | M1A1 | 1st M1 for attempting normal with mean = 100 or ft their answer to (c); 1st A1 for correctly labelled s.d. \((\sigma)\) or var \((\sigma^2)\); allow \(\sqrt{\frac{250}{3}} = \sqrt{83.3\ldots} = 9.1(28\ldots)\) or correctly labelled variance \(N\!\left(\mu, \frac{250}{3}\right)\) |
| \(P(S > 95) \approx P\!\left(\left[Z >\right] \frac{95.5 - \text{"100"}}{\text{"9.128..."}}\right)\) or \(P([Z>]-0.49\ldots)\) | M1 | 2nd M1 for attempt at continuity correction i.e. sight of \(95 \pm 0.5\) |
| \(= 0.688976\ldots\) awrt 0.689 | A1 | 2nd A1 for awrt 0.689 [Answer only 4/4] |
# Question 1:
## Part (a)
$X \sim B(10, \frac{1}{6})$ [Allow 0.167 or better for $\frac{1}{6}$] | M1 | Must have B, or Bin or Bpd or Bcd **and** correct values for $n$ and $p$; just $n=10$, $p=\frac{1}{6}$ is M0
### Part (a)(i)
$[P(X=3)=]\ 0.155045\ldots$ **awrt 0.155** | A1 | 1st A1 for awrt 0.155
### Part (a)(ii)
$[P(X<3) = P(X \leq 2) =]\ 0.775226\ldots$ **awrt 0.775** | A1 | 2nd A1 for awrt 0.775
**(3 marks)**
---
## Part (b)
[Let $D$ = no. of days when $X=3$] $D \sim B(60, \text{"0.155"})$ | M1 | 1st M1 for selecting correct model; sight or use of correct binomial, ft their (a)(i)
$P(D \geq 12) = 1 - P(D \leq 11)$ [Allow $1 - P(D < 12)$] | M1 | 2nd M1 for correct interpretation of "at least 12" and writing or using $1 - P(D \leq 11)$
$= 1 - 0.78819\ldots$ **awrt 0.212** | A1 | For awrt 0.212 [Answer only 3/3]
**(3 marks)**
---
## Part (c)
$[n = 600,\ p = \frac{1}{6}]$ estimate = **100** | B1 | For 100 but must be seen in part (c) i.e. between (b) and (d)
**(1 mark)**
---
## Part (d)
$[S = \text{total no. of sixes over 60 days.}]\ S \approx T \sim N\!\left(\text{"100"},\ \sqrt{\frac{5}{6} \times 100}\ ^2\right)$ | M1A1 | 1st M1 for attempting normal **with mean = 100** or ft their answer to (c); 1st A1 for correctly labelled s.d. $(\sigma)$ or var $(\sigma^2)$; allow $\sqrt{\frac{250}{3}} = \sqrt{83.3\ldots} = 9.1(28\ldots)$ or correctly labelled variance $N\!\left(\mu, \frac{250}{3}\right)$
$P(S > 95) \approx P\!\left(\left[Z >\right] \frac{95.5 - \text{"100"}}{\text{"9.128..."}}\right)$ or $P([Z>]-0.49\ldots)$ | M1 | 2nd M1 for attempt at continuity correction i.e. sight of $95 \pm 0.5$
$= 0.688976\ldots$ **awrt 0.689** | A1 | 2nd A1 for awrt 0.689 [Answer only 4/4]
**(4 marks)**
---
**Total: 11 marks**
### Additional Guidance Notes for (d):
- **NB:** If model not stated for 1st M1 but probabilities given: 1st M1 implied by $P(Y > 94.5) = 0.52(63\ldots)$, $P(Y>95)=0.52(39\ldots)$, $P(Y>95.5)=0.52(15\ldots)$
- **No cc:** 1st M1 1st A1 implied by $P(T>95) = 0.70(805\ldots)$
- **Exact binomial** gives $0.68567\ldots$ and will likely score 0/4\begin{enumerate}
\item Xian rolls a fair die 10 times.
\end{enumerate}
The random variable $X$ represents the number of times the die lands on a six.\\
(a) Using a suitable distribution for $X$, find\\
(i) $\mathrm { P } ( X = 3 )$\\
(ii) $\mathrm { P } ( X < 3 )$
Xian repeats this experiment each day for 60 days and records the number of days when $X = 3$\\
(b) Find the probability that there were at least 12 days when $X = 3$\\
(c) Find an estimate for the total number of sixes that Xian will roll during these 60 days.\\
(d) Use a normal approximation to estimate the probability that Xian rolls a total of more than 95 sixes during these 60 days.
\hfill \mbox{\textit{Edexcel Paper 3 2024 Q1 [11]}}