| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Probability of range of values |
| Difficulty | Moderate -0.8 This is a straightforward binomial probability question requiring only table lookups and basic probability calculations (complementary probability and subtraction). All three parts are routine applications with no problem-solving or conceptual challenges—easier than the average A-level question which typically requires more technique integration. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| 3(i) | Use of 5 or 6 instead of 5.5 for last value of x: all M-marks can be scored, but no A-marks. (ans: 5 gives 2.32 and 1.23; 6 gives 2.39 and 1.40); \(\frac{\sum fx}{\sum f}\) attempted \(= \frac{662000}{280900} = 2.36\) (3 sf); \(\frac{\sum fx^2}{\sum f}\) attempted \(= \frac{2042350}{280900} = 7.270737\) | M1, A1, M1 |
| \(= "\)2.36"\(^{1/2}\) = 1.70 to 1.72, 3 sf | M1, A1 | dep +ve result; \(\div\) 5 or \(\div\) 6 M0M0A0; allow 1.3 |
| 3(ii) | 2, 3 | B1, B1 |
| 3(iii) | Binomial stated, or seen or implied with any n & p; \({}^{11}C_4 \times 0.8^7 \times 0.2^4 = 0.111\) (3 sf) | B1, M1, A1 |
| 3(iv) | Binomial stated or seen or implied; \(0.6228 - 0.3497 = 0.273\) (3 sf) | M1, M1, A1 |
**3(i)** | Use of 5 or 6 instead of 5.5 for last value of x: all M-marks can be scored, but no A-marks. (ans: 5 gives 2.32 and 1.23; 6 gives 2.39 and 1.40); $\frac{\sum fx}{\sum f}$ attempted $= \frac{662000}{280900} = 2.36$ (3 sf); $\frac{\sum fx^2}{\sum f}$ attempted $= \frac{2042350}{280900} = 7.270737$ | M1, A1, M1 | 3 terms of $\sum fx$ correct, and $\div \sum f$; Allow incorrect $\sum f$ NOT if $\sum f = \sum x$; 3 terms of $\sum fx^2$ correct and $\div \sum f$; M2 (86900×1.36$^2$ + 92500×0.36$^2$ + 45000×0.64$^2$ + 37100×1.64$^2$ + 1940×3.1$^2$), 482210.64, 280900 |
| $= "$2.36"$^{1/2}$ = 1.70 to 1.72, 3 sf | M1, A1 | dep +ve result; $\div$ 5 or $\div$ 6 M0M0A0; allow 1.3 |
**3(ii)** | 2, 3 | B1, B1 | allow IQR = 3 – 1 = 2, ie UQ = 3 implied; NB 3, 2 B0B0 unless labelled correctly |
**3(iii)** | Binomial stated, or seen or implied with any n & p; ${}^{11}C_4 \times 0.8^7 \times 0.2^4 = 0.111$ (3 sf) | B1, M1, A1 | eg by 0.8$^y$ × 0.2$^7$ ($r,s>1$) not just by "$C_r$; Correct method, Correct answer, no working M1M1A1 |
**3(iv)** | Binomial stated or seen or implied; $0.6228 - 0.3497 = 0.273$ (3 sf) | M1, M1, A1 | by use of table or ${}^6C_r$ or $(\frac{2}{3})^l(\frac{1}{3})^p$ ($p + q = 9$); "$C_6({\frac{1}{3}})(\frac{2}{3})^6$; NB 0.3498 (from 0.6228 - 0.273) rounds to 0.350 so B1 |3 In a supermarket the proportion of shoppers who buy washing powder is denoted by $p$. 16 shoppers are selected at random.\\
(i) Given that $p = 0.35$, use tables to find the probability that the number of shoppers who buy washing powder is
\begin{enumerate}[label=(\alph*)]
\item at least 8,
\item between 4 and 9 inclusive.\\
(ii) Given instead that $p = 0.38$, find the probability that the number of shoppers who buy washing powder is exactly 6 .
\section*{June 2005}
\end{enumerate}
\hfill \mbox{\textit{OCR S1 Q3 [8]}}