| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Sampling distribution of median |
| Difficulty | Standard +0.3 This is a straightforward sampling distribution problem requiring systematic enumeration of outcomes and basic probability calculations. While it involves the median concept and multiple parts, the sample size is small (n=3), making enumeration manageable, and the probability calculations are direct applications of the multiplication rule with given probabilities. It's slightly above average difficulty due to the sampling distribution concept, but remains accessible to S2 students. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((5,5,5)\) or \((1,5,5)\) or \((2,5,5)\) | B1 | For two of the given triples, any order |
| \((5,5,5)\ (5,5,1)\ (5,1,5)\ (1,5,5)\ (5,5,2)\ (5,2,5)\ (2,5,5)\) or \((5,5,5)\) and \((5,5,1)(\times3)\) and \((5,5,2)(\times3)\) | B1 | For all 7 cases, no incorrect extras |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((5,5,5)\ \ \left(\frac{3}{10}\right)^3 = \frac{27}{1000} = 0.027\) | B1 | \(\left(\frac{3}{10}\right)^3\) or \(0.027\) oe. Can be a single term in a summation |
| \((5,5,1)\ \ 3\times\frac{1}{2}\times\left(\frac{3}{10}\right)^2 = \frac{135}{1000}\) or \(\frac{27}{200} = 0.135\) | M1 | Either "3"\(\times\frac{1}{2}\times\left(\frac{3}{10}\right)^2\) or "3"\(\times\frac{1}{5}\times\left(\frac{3}{10}\right)^2\) oe. May omit the \(3\times\) or have another positive integer in place of the 3 |
| \((5,5,2)\ \ 3\times\frac{1}{5}\times\left(\frac{3}{10}\right)^2 = \frac{54}{1000} = \frac{27}{500} = 0.054\) | ||
| \(P(M=5) = \left(\frac{3}{10}\right)^3 + 3\times\frac{1}{2}\times\left(\frac{3}{10}\right)^2 + 3\times\frac{1}{5}\times\left(\frac{3}{10}\right)^2 = \frac{27}{125} = 0.216\) oe | A1A1 | \(\left(\frac{3}{10}\right)^3 + 3\times\frac{1}{2}\times\left(\frac{3}{10}\right)^2 + 3\times\frac{1}{5}\times\left(\frac{3}{10}\right)^2\) oe; then \(0.216\) oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(M=1) = (0.5)^3 + 3(0.5)^2(0.2) + 3(0.5)^2(0.3)\) | M1 | Correct calculation for \(P(M=1)\) or \(P(M=2)\), working must be shown and not implied by a correct answer |
| \(= 0.5\) | A1 | Either \(P(M=1)\) or \(P(M=2)\) correct |
| \(P(M=2) = \left(\frac{1}{5}\right)^3 + 3\times\left(\frac{1}{5}\right)^2\times\frac{1}{2} + 3\times\left(\frac{1}{5}\right)^2\times\frac{3}{10} + 6\times\frac{1}{2}\times\frac{1}{5}\times\frac{3}{10}\) | M1 | Correct calculation for both \(P(M=1)\) and \(P(M=2)\), or their probabilities adding to 1, but do not allow probabilities of \(0.5\), \(0.2\) and \(0.3\) |
| \(= 0.284\) or \(\frac{71}{250}\) oe | A1 | Both \(P(M=1)\) and \(P(M=2)\) correct |
| Table with \(m\): 1, 2, 5 and \(P(M=m)\): 0.5, 0.284, 0.216 | A1 | All three values written down with correct probabilities (need not be in a table). NB A fully correct table with no working gets M0 A0 M1 A1 A0 |
# Question 1:
## Part 1(a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(5,5,5)$ or $(1,5,5)$ or $(2,5,5)$ | B1 | For two of the given triples, any order |
| $(5,5,5)\ (5,5,1)\ (5,1,5)\ (1,5,5)\ (5,5,2)\ (5,2,5)\ (2,5,5)$ or $(5,5,5)$ **and** $(5,5,1)(\times3)$ **and** $(5,5,2)(\times3)$ | B1 | For all 7 cases, no incorrect extras |
## Part 1(b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(5,5,5)\ \ \left(\frac{3}{10}\right)^3 = \frac{27}{1000} = 0.027$ | B1 | $\left(\frac{3}{10}\right)^3$ or $0.027$ oe. Can be a single term in a summation |
| $(5,5,1)\ \ 3\times\frac{1}{2}\times\left(\frac{3}{10}\right)^2 = \frac{135}{1000}$ or $\frac{27}{200} = 0.135$ | M1 | Either "3"$\times\frac{1}{2}\times\left(\frac{3}{10}\right)^2$ **or** "3"$\times\frac{1}{5}\times\left(\frac{3}{10}\right)^2$ oe. May omit the $3\times$ or have another positive integer in place of the 3 |
| $(5,5,2)\ \ 3\times\frac{1}{5}\times\left(\frac{3}{10}\right)^2 = \frac{54}{1000} = \frac{27}{500} = 0.054$ | | |
| $P(M=5) = \left(\frac{3}{10}\right)^3 + 3\times\frac{1}{2}\times\left(\frac{3}{10}\right)^2 + 3\times\frac{1}{5}\times\left(\frac{3}{10}\right)^2 = \frac{27}{125} = 0.216$ oe | A1A1 | $\left(\frac{3}{10}\right)^3 + 3\times\frac{1}{2}\times\left(\frac{3}{10}\right)^2 + 3\times\frac{1}{5}\times\left(\frac{3}{10}\right)^2$ oe; then $0.216$ oe |
## Part 1(c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(M=1) = (0.5)^3 + 3(0.5)^2(0.2) + 3(0.5)^2(0.3)$ | M1 | Correct calculation for $P(M=1)$ **or** $P(M=2)$, working must be shown and **not** implied by a correct answer |
| $= 0.5$ | A1 | Either $P(M=1)$ **or** $P(M=2)$ correct |
| $P(M=2) = \left(\frac{1}{5}\right)^3 + 3\times\left(\frac{1}{5}\right)^2\times\frac{1}{2} + 3\times\left(\frac{1}{5}\right)^2\times\frac{3}{10} + 6\times\frac{1}{2}\times\frac{1}{5}\times\frac{3}{10}$ | M1 | Correct calculation for both $P(M=1)$ **and** $P(M=2)$, or their probabilities adding to 1, but do not allow probabilities of $0.5$, $0.2$ and $0.3$ |
| $= 0.284$ or $\frac{71}{250}$ oe | A1 | Both $P(M=1)$ **and** $P(M=2)$ correct |
| Table with $m$: 1, 2, 5 and $P(M=m)$: 0.5, 0.284, 0.216 | A1 | All three values written down with correct probabilities (need not be in a table). **NB** A fully correct table with no working gets M0 A0 M1 A1 A0 |
---\begin{enumerate}
\item A bag contains a large number of $1 \mathrm { p } , 2 \mathrm { p }$ and 5 p coins.\\
$50 \%$ are 1 p coins\\
$20 \%$ are $2 p$ coins\\
30\% are 5p coins\\
A random sample of 3 coins is chosen from the bag.\\
(a) List all the possible samples of size 3 with median 5p.\\
(b) Find the probability that the median value of the sample is 5 p .\\
(c) Find the sampling distribution of the median of samples of size 3\\
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2013 Q1 [11]}}