| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | June |
| 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 distribution question requiring only table lookups and basic probability calculations. Part (i) uses cumulative binomial tables directly with no manipulation beyond P(X≥8)=1-P(X≤7), and part (ii) is a single probability calculation. All values are standard and require no problem-solving or insight beyond recognizing the binomial model. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| (i) (a) \(B(16, 0.35)\) stated | B1 | Or implied by use of tables or \(0.35^n \times 0.65^n\) (\(a+b = 16\)) in (a) or (b). Allow \(1 - 0.9329\) or \(0.0671\). Or complete method using formula, \(P(r = 8-16\) or \(9-16)\) or \(1-P(r = 7\) or \(0-8)\) |
| \(1 - 0.8406\) | M1 | |
| \(= 0.159\) (3 sfs) | A1 | 3 |
| (b) \(0.9771 - 0.1339\) | M1 | Allow \(0.9771 - 0.2892\). Or complete method using formula (\(r = 4-9)\) |
| \(= 0.843\) (3 sfs) | A1 | 2 |
| (ii) \(^{16}C_0(0.38)^{16}(0.62)^0\) | M2 | Absent or incorrect coeff: M1 or \(^{16}C_0(0.38)^0(0.62)^{16}\): M1 |
| \(= 0.202\) (3 sfs) | A1 | 3 |
(i) (a) $B(16, 0.35)$ stated | B1 | Or implied by use of tables or $0.35^n \times 0.65^n$ ($a+b = 16$) in (a) or (b). Allow $1 - 0.9329$ or $0.0671$. Or complete method using formula, $P(r = 8-16$ or $9-16)$ or $1-P(r = 7$ or $0-8)$
$1 - 0.8406$ | M1 |
$= 0.159$ (3 sfs) | A1 | 3 |
(b) $0.9771 - 0.1339$ | M1 | Allow $0.9771 - 0.2892$. Or complete method using formula ($r = 4-9)$
$= 0.843$ (3 sfs) | A1 | 2 |
(ii) $^{16}C_0(0.38)^{16}(0.62)^0$ | M2 | Absent or incorrect coeff: M1 or $^{16}C_0(0.38)^0(0.62)^{16}$: M1
$= 0.202$ (3 sfs) | A1 | 3 |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 .
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2005 Q3 [8]}}