| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Sampling distribution of mean or linear combination |
| Difficulty | Standard +0.3 This is a straightforward S2 sampling distribution question requiring systematic enumeration of outcomes, basic probability calculations with sampling without replacement, and a simple binomial probability inequality. The methods are standard and well-practiced, though the multi-part structure and careful bookkeeping elevate it slightly above average difficulty. |
| Spec | 2.04a Discrete probability distributions2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{10}{50} \times \frac{9}{49} = \frac{9}{245}\) | B1cso | A correct equivalent expression |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Number of counters: numbered \(4 = 10\), numbered \(7 = 15\), numbered \(10 = 25\) | M1 | For 10, 15 and 25; may be seen in (a) or in probability expressions |
| \(M = 4, 5.5, 7, 8.5, 10\) | B1 | All means correct with no incorrect extra unless probability of 0 |
| \(P(M=5.5) = 2 \times \frac{10}{50} \times \frac{15}{49} = \frac{6}{49}\) | M1 | One correct probability (not including 9/245) |
| \(P(M=7) = 2 \times \frac{10}{50} \times \frac{25}{49} + \frac{15}{50} \times \frac{14}{49} = \frac{71}{245}\) | M1 | Two correct probabilities (not including 9/245) |
| \(P(M=8.5) = 2 \times \frac{15}{50} \times \frac{25}{49} = \frac{15}{49}\) | M1 | Three correct probabilities (not including 9/245) |
| \(P(M=10) = \frac{25}{50} \times \frac{24}{49} = \frac{12}{49}\) | ||
| Complete correct table with all values | A1 | Fully correct; need not be in table but must have correct probability associated with correct mean |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\left(1 - \frac{9}{245}\right)^n < 0.15\) | M1 | Setting up a correct inequality (allow any inequality or equal sign) |
| \(n = 50.689...\) or \(n=50\) is \(0.1539...\) or \(n=51\) is \(0.148...\) or \([n >] \frac{\log 0.15}{\log(1 - \frac{9}{245})}\) | M1 | For a value \(n =\) awrt 50.7 or awrt 0.154 or awrt 0.148 or correct log expression for \(n\) |
| \(n = 51\) | A1 | 51 cao; do not allow \(n \geq 51\) |
# Question 4:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{10}{50} \times \frac{9}{49} = \frac{9}{245}$ | B1cso | A correct equivalent expression |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Number of counters: numbered $4 = 10$, numbered $7 = 15$, numbered $10 = 25$ | M1 | For 10, 15 and 25; may be seen in (a) or in probability expressions |
| $M = 4, 5.5, 7, 8.5, 10$ | B1 | All means correct with no incorrect extra unless probability of 0 |
| $P(M=5.5) = 2 \times \frac{10}{50} \times \frac{15}{49} = \frac{6}{49}$ | M1 | One correct probability (not including 9/245) |
| $P(M=7) = 2 \times \frac{10}{50} \times \frac{25}{49} + \frac{15}{50} \times \frac{14}{49} = \frac{71}{245}$ | M1 | Two correct probabilities (not including 9/245) |
| $P(M=8.5) = 2 \times \frac{15}{50} \times \frac{25}{49} = \frac{15}{49}$ | M1 | Three correct probabilities (not including 9/245) |
| $P(M=10) = \frac{25}{50} \times \frac{24}{49} = \frac{12}{49}$ | | |
| Complete correct table with all values | A1 | Fully correct; need not be in table but must have correct probability associated with correct mean |
## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\left(1 - \frac{9}{245}\right)^n < 0.15$ | M1 | Setting up a correct inequality (allow any inequality or equal sign) |
| $n = 50.689...$ or $n=50$ is $0.1539...$ or $n=51$ is $0.148...$ or $[n >] \frac{\log 0.15}{\log(1 - \frac{9}{245})}$ | M1 | For a value $n =$ awrt 50.7 or awrt 0.154 or awrt 0.148 or correct log expression for $n$ |
| $n = 51$ | A1 | 51 cao; do not allow $n \geq 51$ |
---4 A bag contains 50 counters, each with one of the numbers 4,7 or 10 written on it in the ratio $2 : 3 : 5$ respectively.
A random sample of 2 counters is taken from the bag. The numbers on the 2 counters are recorded as $D _ { 1 }$ and $D _ { 2 }$
The random variable $M$ represents the mean of $D _ { 1 }$ and $D _ { 2 }$
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { P } ( M = 4 ) = \frac { 9 } { 245 }$
\item Find the sampling distribution of $M$
A random sample of $n$ sets of 2 counters is taken. The random variable $T$ represents the number of these $n$ sets of 2 counters that have a mean of 4
Given that each set of 2 counters is replaced after it is drawn,
\item calculate the minimum value of $n$ such that $\mathrm { P } ( T = 0 ) < 0.15$
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2024 Q4 [10]}}