| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Probability distribution table |
| Difficulty | Moderate -0.8 This is a straightforward S2 question requiring systematic enumeration of outcomes, basic probability calculations using binomial distribution (p^3, 3p^2q, etc.), and simple arithmetic. All steps are routine applications of standard techniques with no conceptual challenges or novel insights required. |
| Spec | 2.01a Population and sample: terminology5.01a Permutations and combinations: evaluate probabilities5.02b Expectation and variance: discrete random variables |
| \(\bar { x }\) | 20 | \(a\) | \(b\) | 50 |
| \(\mathrm { P } ( \bar { X } = \bar { x } )\) | \(\frac { 4913 } { 8000 }\) | \(c\) | \(d\) | \(\frac { 27 } { 8000 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| 20, 20, 20 \ | 20, 20, 50 (×3) \ | 20, 50, 50 (×3) \ |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(a = 30\) and \(b = 40\) | B1 | For \(a = 30\) and \(b = 40\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(p^3 = \frac{4913}{8000}\) or \(q^3 = \frac{27}{8000}\) | M1 | Either \(p^3 = \frac{4913}{8000}\) or \(q^3 = \frac{27}{8000}\) |
| \(p = \frac{17}{20}\ (0.85)\) and \(q = \frac{3}{20}\ (0.15)\) | A1 | \(p = 0.85\) oe and \(q = 0.15\) oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([P(30)] = 3 \times p^2 \times q\) | M1 | \([P(30)] = 3 \times (\text{their } p)^2 \times (\text{their } q)\); must see values substituted |
| \([P(40)] = 3 \times p \times q^2\) | M1 | \([P(40)] = 3 \times (\text{their } p) \times (\text{their } q)^2\); or use of sum of probabilities \(= 1\) i.e. \(c + d = \frac{153}{400}\) |
| \(c = \frac{2601}{8000}\), \(d = \frac{459}{8000}\) | A1 | For \(c = \frac{2601}{8000}\ (= 0.325125)\) and \(d = \frac{459}{8000}\ (= 0.057375)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(M = 20\) and \(M = 50\) only | B1 | For 20 and 50 only (ignore notation used for \(M\)) |
| \(P(M=20) = \frac{3757}{4000}\), \(P(M=50) = \frac{243}{4000}\) | M1, A1ft | Either \(\frac{4913}{8000} +\) their \(c\) or \(\frac{27}{8000} +\) their \(d\). Follow through their values for \(c\) and \(d\) but \(P(M=20) + P(M=50)\) must sum to 1. NB: If \(a\) and \(b\) reversed then \(c = \frac{459}{8000}\), \(d = \frac{2601}{8000}\) |
# Question 2:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| 20, 20, 20 \| 20, 20, 50 (×3) \| 20, 50, 50 (×3) \| 50, 50, 50 | B2 | For all 4 correct combinations. B1 for 3 correct combinations. Ignore extraneous repetitions of any of the given combinations |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $a = 30$ and $b = 40$ | B1 | For $a = 30$ and $b = 40$ |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $p^3 = \frac{4913}{8000}$ or $q^3 = \frac{27}{8000}$ | M1 | Either $p^3 = \frac{4913}{8000}$ or $q^3 = \frac{27}{8000}$ |
| $p = \frac{17}{20}\ (0.85)$ and $q = \frac{3}{20}\ (0.15)$ | A1 | $p = 0.85$ oe and $q = 0.15$ oe |
## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| $[P(30)] = 3 \times p^2 \times q$ | M1 | $[P(30)] = 3 \times (\text{their } p)^2 \times (\text{their } q)$; must see values substituted |
| $[P(40)] = 3 \times p \times q^2$ | M1 | $[P(40)] = 3 \times (\text{their } p) \times (\text{their } q)^2$; or use of sum of probabilities $= 1$ i.e. $c + d = \frac{153}{400}$ |
| $c = \frac{2601}{8000}$, $d = \frac{459}{8000}$ | A1 | For $c = \frac{2601}{8000}\ (= 0.325125)$ and $d = \frac{459}{8000}\ (= 0.057375)$ |
## Part (e)
| Answer | Mark | Guidance |
|--------|------|----------|
| $M = 20$ and $M = 50$ only | B1 | For 20 and 50 only (ignore notation used for $M$) |
| $P(M=20) = \frac{3757}{4000}$, $P(M=50) = \frac{243}{4000}$ | M1, A1ft | Either $\frac{4913}{8000} +$ their $c$ or $\frac{27}{8000} +$ their $d$. Follow through their values for $c$ and $d$ but $P(M=20) + P(M=50)$ must sum to 1. NB: If $a$ and $b$ reversed then $c = \frac{459}{8000}$, $d = \frac{2601}{8000}$ |
---\begin{enumerate}
\item A bag contains a large number of coins. It only contains 20 p and 50 p coins. A random sample of 3 coins is taken from the bag.\\
(a) List all the possible combinations of 3 coins that might be taken.
\end{enumerate}
Let $\bar { X }$ represent the mean value of the 3 coins taken.\\
Part of the sampling distribution of $\bar { X }$ is given below.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$\bar { x }$ & 20 & $a$ & $b$ & 50 \\
\hline
$\mathrm { P } ( \bar { X } = \bar { x } )$ & $\frac { 4913 } { 8000 }$ & $c$ & $d$ & $\frac { 27 } { 8000 }$ \\
\hline
\end{tabular}
\end{center}
(b) Write down the value of $a$ and the value of $b$
The probability of taking a 20p coin at random from the bag is $p$ The probability of taking a 50p coin at random from the bag is $q$\\
(c) Find the value of $p$ and the value of $q$\\
(d) Hence, find the value of $c$ and the value of $d$
Let $M$ represent the mode of the 3 coins taken at random from the bag.\\
(e) Find the sampling distribution of $M$
\hfill \mbox{\textit{Edexcel S2 2023 Q2 [11]}}