| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Basic E(X) and Var(X) calculation |
| Difficulty | Moderate -0.3 This is a straightforward S2 binomial question requiring standard calculations: (a) finding a probability distribution using binomial probabilities with p=1/5, and (b) solving P(Y≥1)>0.95 using complement rule. Both parts are routine textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Values: \(8, 11, 14, 17, 20\) | M1 | For at least 2 scores correct and no more than 3 incorrect |
| \(\left[P(\text{even})=\right]\frac{1}{5}\) and \(\left[P(\text{odd})=\right]\frac{4}{5}\) | M1 | For writing or using \(\frac{4}{5}\) and \(\frac{1}{5}\). May be implied by a correct probability |
| \(\left[P(X=8)=\right]\left(\frac{4}{5}\right)^4\) or \(\left[P(X=20)=\right]\left(\frac{1}{5}\right)^4\) | M1 | For \(p^4\) where \(0 < p < 1\) |
| \(\left[P(X=11)=\right]4\times\left(\frac{1}{5}\right)\left(\frac{4}{5}\right)^3\) or \(\left[P(X=17)=\right]4\times\left(\frac{4}{5}\right)\left(\frac{1}{5}\right)^3\) | M1 | For \(4\times(1-p)p^3\) where \(0 < p < 1\) |
| \(\left[P(X=14)=\right]{}^4C_2\times\left(\frac{1}{5}\right)^2\left(\frac{4}{5}\right)^2\) | M1 | For \(6\times(1-p)^2p^2\) where \(0 |
| \(X\): \(8, 11, 14, 17, 20\) with \(P(X=x)\): \(\frac{256}{625}, \frac{256}{625}, \frac{96}{625}, \frac{16}{625}, \frac{1}{625}\) (i.e. \(0.4096, 0.4096, 0.1536, 0.0256, 0.0016\)) | A1 | For all 5 probabilities correct and associated with the correct values. Need not be in a table but probabilities must be attached to the correct total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(1-(1-\text{"0.1536"})^n > 0.95\) or \((\text{"0.8464"})^n < 0.05\) | M1 | For using \(1-(1-P(Y=0))^n > 0.95\); allow \(=\) instead of \(>/\geqslant\); condone \( |
| \(n > 17.96\) or \(n > \frac{\log(0.05)}{\log(\text{"0.8464"})}\) or \(n > \log_{\text{"0.8464"}}(0.05)\) | M1 | For \(n >\) awrt 17.96 or \(n > \frac{\log(0.05)}{\log(\text{"0.8464"})}\) ft their 0.8464 or \(n > \log_{\text{"0.8464"}}(0.05)\) ft their 0.8464 or for the two trials for \(n=17\) and \(18\). Allow \(=\) instead of \(>/\geqslant\); condone \( |
| \(n = 18\) | A1 | Cao (Do not allow any inequality for this mark) |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Values: $8, 11, 14, 17, 20$ | M1 | For at least 2 scores correct and no more than 3 incorrect |
| $\left[P(\text{even})=\right]\frac{1}{5}$ and $\left[P(\text{odd})=\right]\frac{4}{5}$ | M1 | For writing or using $\frac{4}{5}$ and $\frac{1}{5}$. May be implied by a correct probability |
| $\left[P(X=8)=\right]\left(\frac{4}{5}\right)^4$ or $\left[P(X=20)=\right]\left(\frac{1}{5}\right)^4$ | M1 | For $p^4$ where $0 < p < 1$ |
| $\left[P(X=11)=\right]4\times\left(\frac{1}{5}\right)\left(\frac{4}{5}\right)^3$ or $\left[P(X=17)=\right]4\times\left(\frac{4}{5}\right)\left(\frac{1}{5}\right)^3$ | M1 | For $4\times(1-p)p^3$ where $0 < p < 1$ |
| $\left[P(X=14)=\right]{}^4C_2\times\left(\frac{1}{5}\right)^2\left(\frac{4}{5}\right)^2$ | M1 | For $6\times(1-p)^2p^2$ where $0<p<1$ or probabilities that add to 1 (at least 2 but not more than 5) |
| $X$: $8, 11, 14, 17, 20$ with $P(X=x)$: $\frac{256}{625}, \frac{256}{625}, \frac{96}{625}, \frac{16}{625}, \frac{1}{625}$ (i.e. $0.4096, 0.4096, 0.1536, 0.0256, 0.0016$) | A1 | For all 5 probabilities correct and associated with the correct values. Need not be in a table but probabilities must be attached to the correct total |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1-(1-\text{"0.1536"})^n > 0.95$ or $(\text{"0.8464"})^n < 0.05$ | M1 | For using $1-(1-P(Y=0))^n > 0.95$; allow $=$ instead of $>/\geqslant$; condone $</\leqslant$ or allow for at least 2 trials for $n$ between 10 and 20 ft their $P(X=14)$ |
| $n > 17.96$ or $n > \frac{\log(0.05)}{\log(\text{"0.8464"})}$ or $n > \log_{\text{"0.8464"}}(0.05)$ | M1 | For $n >$ awrt 17.96 or $n > \frac{\log(0.05)}{\log(\text{"0.8464"})}$ ft their 0.8464 or $n > \log_{\text{"0.8464"}}(0.05)$ ft their 0.8464 or for the two trials for $n=17$ and $18$. Allow $=$ instead of $>/\geqslant$; condone $</\leqslant$. May be implied by a correct answer ft their 0.8464 |
| $n = 18$ | A1 | Cao (Do not allow any inequality for this mark) |
---\begin{enumerate}
\item A bag contains a large number of counters with an odd number or an even number written on each.
\end{enumerate}
Odd and even numbered counters occur in the ratio $4 : 1$\\
In a game a player takes a random sample of 4 counters from the bag.\\
The player scores\\
5 points for each counter taken that has an even number written on it\\
2 points for each counter taken that has an odd number written on it\\
The random variable $X$ represents the total score, in points, from the 4 counters.\\
(a) Find the sampling distribution of $X$
A random sample of $n$ sets of 4 counters is taken. The random variable $Y$ represents the number of these $n$ sets that have a total score of exactly 14\\
(b) Calculate the minimum value of $n$ such that $\mathrm { P } ( Y \geqslant 1 ) > 0.95$
\hfill \mbox{\textit{Edexcel S2 2024 Q6 [9]}}