| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Single batch expected count |
| Difficulty | Moderate -0.8 This is a straightforward AS-level statistics question requiring basic binomial probability calculations, simple estimation from data, and elementary model comparison. All parts involve routine procedures: calculating P(X≥4) for n=5, multiplying by 50, pooling data for better estimates, and comparing expected values to observed data. No novel insight or complex reasoning required. |
| Spec | 2.01a Population and sample: terminology2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| \(50 \times \frac{6}{32}\) or \(50 \times 0.1875\) | M1 | Condone \(50 \times \frac{5}{32} = 7.8125\). Must be a genuine attempt at using binomial distribution. Coeff of 5 must be seen or implied or cumulative binomial found. \(p(4H) = 5 \times \frac{1}{2} \times \left(\frac{1}{2}\right)^4 = \frac{5}{32}\); \(p(5H) = \left(\frac{1}{2}\right)^5 = \frac{1}{32}\); \(p(X \geq 4) = 0.1875\) |
| \(\frac{75}{8}\) oe or 9 games | A1 | If they round to 10 then withhold this mark. |
| Answer | Marks | Guidance |
|---|---|---|
| Combined/larger sample because unbiased samples become more representative of theoretical distributions as sample size increases | B1 | Needs to refer to the idea that experimental data only mimics theoretical distributions if samples are large and representative. Accept 'increasing the sample size/combining the samples gives a more accurate estimate of the value of \(p\)'. Accept 'the larger the sample, the more accurate the data'. |
| Answer | Marks | Guidance |
|---|---|---|
| \(40p = 28\) so \(p = 0.7\) | B1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| \(50 \times (5 \times 0.7^4 \times 0.3 + 0.7^5)\) or \(50 \times p(X \geq 4)\) using \(X \sim B(5, 0.7)\); gives \(50 \times 0.52822\) | M1 | Makes a valid attempt at calculating using \(X \sim B(5, 0.7)\). Must be a genuine attempt at using binomial distribution. Coeff of 5 must be seen or implied or cumulative binomial found. Condone \(50 \times 5 \times 0.7^4 \times 0.3\) |
| \(26.4\ldots\) or \(26\) | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Ali's model better fit since 25 much closer to 26 than to 9 | B1 FT | Correct statement FT their values from (a) and (d). Must compare 25 with the expected number of wins for both Ali and Sam for the reasoning. The comparison can be indirect e.g. Ali's better as number of expected wins closer to the true value. |
## Question 15(a):
$50 \times \frac{6}{32}$ or $50 \times 0.1875$ | **M1** | Condone $50 \times \frac{5}{32} = 7.8125$. Must be a genuine attempt at using binomial distribution. Coeff of 5 must be seen or implied or cumulative binomial found. $p(4H) = 5 \times \frac{1}{2} \times \left(\frac{1}{2}\right)^4 = \frac{5}{32}$; $p(5H) = \left(\frac{1}{2}\right)^5 = \frac{1}{32}$; $p(X \geq 4) = 0.1875$
$\frac{75}{8}$ oe or 9 games | **A1** | If they round to 10 then withhold this mark.
**Total: [2]**
---
## Question 15(b):
**Combined/larger** sample because unbiased samples become **more representative** of theoretical distributions as **sample size increases** | **B1** | Needs to refer to the idea that experimental data only mimics theoretical distributions if samples are large and representative. Accept 'increasing the sample size/combining the samples gives a more accurate estimate of the value of $p$'. Accept 'the larger the sample, the more accurate the data'.
**Total: [1]**
---
## Question 15(c):
$40p = 28$ so $p = 0.7$ | **B1** | —
**Total: [1]**
---
## Question 15(d):
$50 \times (5 \times 0.7^4 \times 0.3 + 0.7^5)$ or $50 \times p(X \geq 4)$ using $X \sim B(5, 0.7)$; gives $50 \times 0.52822$ | **M1** | Makes a valid attempt at calculating using $X \sim B(5, 0.7)$. Must be a genuine attempt at using binomial distribution. Coeff of 5 must be seen or implied or cumulative binomial found. Condone $50 \times 5 \times 0.7^4 \times 0.3$
$26.4\ldots$ or $26$ | **A1** | —
**Total: [2]**
---
## Question 15(e):
Ali's model better fit since 25 much closer to 26 than to 9 | **B1 FT** | Correct statement FT their values from (a) and (d). Must compare 25 with the expected number of wins for both Ali and Sam for the reasoning. The comparison can be indirect e.g. Ali's better as number of expected wins closer to the true value.
**Total: [1]**15 Ali and Sam are playing a game in which Ali tosses a coin 5 times. If there are 4 or 5 heads, Ali wins the game. Otherwise Sam wins the game. They decide to play the game 50 times.
\begin{enumerate}[label=(\alph*)]
\item Initially Sam models the situation by assuming the coin is fair. Determine the number of games Ali is expected to win according to this model.
Ali thinks the coin may be biased, with probability $p$ of obtaining heads when the coin is tossed. Before playing the game, Ali and Sam decide to collect some data to estimate the value of $p$. Sam tosses the coin 15 times and records the number of heads obtained. Ali tosses the coin 25 times and records the number of heads obtained.
\item Explain why it is better to use the combined data rather than just Sam's data or just Ali's data to estimate the value of $p$.
Ali records 20 heads and Sam records 8 heads.
\item Use the combined data to estimate the value of $p$.
Ali now models the situation using the value of $p$ found in part (c) as the probability of obtaining heads when the coin is tossed.
\item Determine how many games Ali expects to win using this value of $p$ to model the situation.
\item Ali wins 25 of the 50 games. Explain whether Sam's model or Ali's model is a better fit for the data.
\section*{END OF QUESTION PAPER}
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\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 2 2024 Q15 [7]}}