| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | October |
| Paper | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | E(X) and Var(X) with probability calculations |
| Difficulty | Standard +0.3 This is a straightforward binomial distribution question with standard applications. Part (a) requires stating B(25, 0.2), part (b) involves a simple linear transformation and expectation calculation, part (c) is a direct binomial probability computation, and part (d) requires using cumulative binomial tables. All parts are routine S2 exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02d Binomial: mean np and variance np(1-p)5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(B\left(25, \frac{1}{5}\right)\) | B1 |
| (b)(i) | \([M =] 4X - (25-X) \quad [= 5X - 25]\) | B1 |
| (ii) | \(E(M) = '5' E(X) - '25'\) | M1 |
| \(E(X) = np = 25 \times \frac{1}{5} = 5\) | M1 | |
| \(E(M) = 5 \times 5 - 25 = 0\) * | A1* | |
| (4) | ||
| (c) | \(M ...30 \Rightarrow '5'X - '25' ...30 [\Rightarrow X ...11]\) | M1 |
| \([P(X ...11') =] 1 - P(X_r, '10') = 1 - '0.9944'\) | M1 | |
| \(= 0.0056\) awrt 0.0056 | A1 | (3) |
| (d) | \(Y \sim B(50, 0.5)\) | M1 |
| \(P(n < Y_r, 30) = 0.9328\) | M1 | |
| \(P(Y_r, 30) - P(Y_r, n) = 0.9328\) | M1 | |
| \(P(Y_r, n) = 0.0077\) | M1 | |
| \(n = 16\) | A1 | (4) |
| Total | 12 |
| (a) | $B\left(25, \frac{1}{5}\right)$ | B1 | (1) |
|---|---|---|---|
| (b)(i) | $[M =] 4X - (25-X) \quad [= 5X - 25]$ | B1 | (1) |
| (ii) | $E(M) = '5' E(X) - '25'$ | M1 | |
| | $E(X) = np = 25 \times \frac{1}{5} = 5$ | M1 | |
| | $E(M) = 5 \times 5 - 25 = 0$ * | A1* | |
| | | | (4) |
| (c) | $M ...30 \Rightarrow '5'X - '25' ...30 [\Rightarrow X ...11]$ | M1 | |
| | $[P(X ...11') =] 1 - P(X_r, '10') = 1 - '0.9944'$ | M1 | |
| | $= 0.0056$ awrt 0.0056 | A1 | (3) |
| (d) | $Y \sim B(50, 0.5)$ | M1 | |
| | $P(n < Y_r, 30) = 0.9328$ | M1 | |
| | $P(Y_r, 30) - P(Y_r, n) = 0.9328$ | M1 | |
| | $P(Y_r, n) = 0.0077$ | M1 | |
| | $n = 16$ | A1 | (4) |
| | **Total** | **12** | |
**Notes for Question 2:**
- **(a) B1:** Correct distribution fully specified. Allow in words e.g. Binomial with $n = 25$ and $p = 0.2$; Must be seen in part (a)
- **(b)(i) B1:** For a correct expression for $M$ Allow unsimplified
- For either '$5'E(X) - '25'$ or $E(M) = 5 \times \left(25 \times \frac{1}{5}\right) - 25$ or '$4'E(X) - 1'(25 - E(X))$
- This must be an **expectation statement** with the expectation stated in symbol or in words.
- $5 \times 5 - 25 = 0$ or $4 \times 5 - 1 \times 20 = 0$ on its own is M0
- **(b) M1:** For sight of $25 \times \frac{1}{5}$ or stating $E(X) = 5$
- **(b) A1*:** Fully correct solution with $E(M) = 0$ stated. This may be stated in words. The answer is given so no incorrect working can be seen
- **SC:** M1M1 [Expected number of marks (per question) =] $4 \times - 1 \times \frac{4}{5}$ A1 therefore $E(M) = 0$
- **(c) M1:** For substitution of their $M$ into a linear inequality in terms of $X$ implied by $X ...11$
- **(c) M1:** For use of correct probability statement from their '11'
- **(c) A1:** awrt 0.0056 (calc 0.0055549...)
- **(d) M1:** For a correct probability equation (implied by 2nd M1)
- **(d) M1:** For $P(Y_r, 30) - P(Y_r, n) = 0.9328$ or 0.9405 − P(Y_r, n) = 0.9328
- **(d) M1:** For $P(Y_r, n) = 0.0077$
- **(d) A1:** Cao. $P(n < Y_r, 30) = 0.9328$, $P(Y_r, 30) - P(Y_r, n-1) = 0.9328$, $P(Y_r, n-1) = 0.0077$
- **SC:** scores M1M0M1A0
---\begin{enumerate}
\item A multiple-choice test consists of 25 questions, each having 5 responses, only one of which is correct.
\end{enumerate}
Each correct answer gains 4 marks but each incorrect answer loses 1 mark.\\
Sam answers all 25 questions by choosing at random one response for each question.\\
Let $X$ be the number of correct answers that Sam achieves.\\
(a) State the distribution of $X$
Let $M$ be the number of marks that Sam achieves.\\
(b) (i) State the distribution of $M$ in terms of $X$\\
(ii) Hence, show clearly that the number of marks that Sam is expected to achieve is zero.
In order to pass the test at least 30 marks are required.\\
(c) Find the probability that Sam will pass the test.
Past records show that when the test is done properly, the probability that a student answers the first question correctly is 0.5
A random sample of 50 students that did the test properly was taken.\\
Given that the probability that more than $n$ but at most 30 students answered the first question correctly was 0.9328 to 4 decimal places,\\
(d) find the value of $n$
\hfill \mbox{\textit{Edexcel S2 2024 Q2}}