| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | E(X) and Var(X) with probability calculations |
| Difficulty | Easy -1.3 This is a straightforward binomial distribution question requiring only direct application of standard formulas from the formula booklet. All parts involve routine calculations with no problem-solving or conceptual insight needed—students simply substitute values into P(X=r) = nCr × p^r × (1-p)^(n-r) and use E(X)=np, Var(X)=np(1-p). This is easier than average A-level content. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.02d Binomial: mean np and variance np(1-p) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: 0.299 (3 sf) or \(\frac{0.2991 - 0.1040}{1}\) = 0.195 (3 sf) or \(\frac{1280}{oe}\) | M1, A1 2 | Must subtract correct pair from table |
| \(^{15}C_x(1-0.22)^{15}×0.22^x\) = 0.208 (3 sf) | M1, A1 2 | Allow M1 for \(^{15}C_x×0.88^{15}×0.22^x\) |
| (15 × 0.22 × (1-0.22)) or \(3.3×(1-0.22)\) = 2.57 (3 sf) | B1, M1, A1 3 | Allow M1 for 15 × 0.22 × 0.88 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: \(\frac{1}{2}×\frac{1}{3}\) or \(\frac{2}{3}×\frac{1}{3}\) or \(\frac{-1}{C_2}\) or \(\frac{2}{12}\) (= \(\frac{1}{6}\) AG) | B1 | or 1 out of 6 or 2 out of 12 or \(\frac{2}{11}×2\) |
| \(\frac{1}{4}×\frac{2}{3}\) or \(2×\frac{1}{4}×\frac{1}{3}\) or \(\frac{1}{3}×\frac{1}{3}\) or \(\frac{2}{3}×\frac{1}{3}\) | B1 | or \(\frac{2}{12}\) or \(\frac{1}{6}\) or \(\frac{1}{C_2}\) or \(\frac{2}{11}×2\) |
| Add two of these or double one (= \(\frac{1}{3}\) AG) | B1 3 | or \(\frac{2}{C_2}\) or \(4×\frac{1}{4}×\frac{1}{3}\) or \(\frac{2}{3}×\frac{2}{3}\) or \(\frac{3}{12}\) or \(\frac{11}{4}\) B1B1 or \(\frac{2}{3}\) or \(2×\frac{1}{3}\) or \(\frac{3}{12}\) B1B1 |
| X = 3, 4, 5, 6 only, stated or used | B1 | Allow repetitions, Allow other values with zero probabilities. |
| P(X=5) wking as for P(X=4) above or \(1-(^{\frac{1}{6}}+\frac{1}{3}+\frac{1}{6})\) or \(\frac{1}{3}\) | M1 | or M1 for total of their probs = 1, dep B1 |
| P(X=3) wking as for P(X=6) above or \(1-(\frac{1}{3}+\frac{1}{3}+\frac{1}{6})\) or \(\frac{1}{6}\) | M1 | or M1 for total of their probs = 1, dep B1 |
| \(\begin{array}{cccc}3&4&5&6\\\frac{1}{6}&\frac{1}{3}&\frac{1}{3}&\frac{1}{6}\end{array}\) oe | A1 4 | Complete list of values linked to probs |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: \(\Sigma xp\) = \(4\frac{1}{2}\) | M1, A1 | \(\geq 2\) terms correct ft |
| \(\Sigma x^2p\) (= \(21\frac{1}{6}\)) \(-4 \frac{1}{2}^{2}\) = \(\frac{11}{12}\) or 0.917 (3 sf) | M1, M1, A1 5 | \(\geq 2\) terms correct ft, Independent except dependent on +ve result |
## Part i:
**Answer/Working:** 0.299 (3 sf) or $\frac{0.2991 - 0.1040}{1}$ = 0.195 (3 sf) or $\frac{1280}{oe}$ | **M1, A1 2** | Must subtract correct pair from table
$^{15}C_x(1-0.22)^{15}×0.22^x$ = 0.208 (3 sf) | **M1, A1 2** | Allow M1 for $^{15}C_x×0.88^{15}×0.22^x$
(15 × 0.22 × (1-0.22)) or $3.3×(1-0.22)$ = 2.57 (3 sf) | **B1, M1, A1 3** | Allow M1 for 15 × 0.22 × 0.88
**Total: 8 marks**
## Part ii:
**Answer/Working:** $\frac{1}{2}×\frac{1}{3}$ or $\frac{2}{3}×\frac{1}{3}$ or $\frac{-1}{C_2}$ or $\frac{2}{12}$ (= $\frac{1}{6}$ AG) | **B1** | or 1 out of 6 or 2 out of 12 or $\frac{2}{11}×2$
$\frac{1}{4}×\frac{2}{3}$ or $2×\frac{1}{4}×\frac{1}{3}$ or $\frac{1}{3}×\frac{1}{3}$ or $\frac{2}{3}×\frac{1}{3}$ | **B1** | or $\frac{2}{12}$ or $\frac{1}{6}$ or $\frac{1}{C_2}$ or $\frac{2}{11}×2$
Add two of these or double one (= $\frac{1}{3}$ AG) | **B1 3** | or $\frac{2}{C_2}$ or $4×\frac{1}{4}×\frac{1}{3}$ or $\frac{2}{3}×\frac{2}{3}$ or $\frac{3}{12}$ or $\frac{11}{4}$ B1B1 or $\frac{2}{3}$ or $2×\frac{1}{3}$ or $\frac{3}{12}$ B1B1
X = 3, 4, 5, 6 only, stated or used | **B1** | Allow repetitions, Allow other values with zero probabilities.
P(X=5) wking as for P(X=4) above or $1-(^{\frac{1}{6}}+\frac{1}{3}+\frac{1}{6})$ or $\frac{1}{3}$ | **M1** | or M1 for total of their probs = 1, dep B1
P(X=3) wking as for P(X=6) above or $1-(\frac{1}{3}+\frac{1}{3}+\frac{1}{6})$ or $\frac{1}{6}$ | **M1** | or M1 for total of their probs = 1, dep B1
$\begin{array}{cccc}3&4&5&6\\\frac{1}{6}&\frac{1}{3}&\frac{1}{3}&\frac{1}{6}\end{array}$ oe | **A1 4** | Complete list of values linked to probs
## Part iii:
**Answer/Working:** $\Sigma xp$ = $4\frac{1}{2}$ | **M1, A1** | $\geq 2$ terms correct ft
$\Sigma x^2p$ (= $21\frac{1}{6}$) $-4 \frac{1}{2}^{2}$ = $\frac{11}{12}$ or 0.917 (3 sf) | **M1, M1, A1 5** | $\geq 2$ terms correct ft, Independent except dependent on +ve result
**Total: 12 marks**
---\begin{enumerate}[label=(\roman*)]
\item The random variable $W$ has the distribution B$(10, \frac{1}{4})$. Find
\begin{enumerate}[label=(\alph*)]
\item P$(W \leq 2)$, [1]
\item P$(W = 2)$. [2]
\end{enumerate}
\item The random variable $X$ has the distribution B$(15, 0.22)$.
\begin{enumerate}[label=(\alph*)]
\item Find P$(X = 4)$. [2]
\item Find E$(X)$ and Var$(X)$. [3]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2010 Q4 [8]}}