| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2007 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Finding binomial parameters from properties |
| Difficulty | Standard +0.3 This is a straightforward binomial distribution question testing standard formulas. Part (i) is direct substitution into the binomial probability formula, part (ii) requires solving (1-p)^11 = 0.05 using logarithms, and part (iii) uses np(1-p) = 1.76 to form a quadratic equation. All parts are routine applications of bookwork with no problem-solving insight required, making it slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| \(^{11}C_5 x (\frac{1}{3})^x x (\frac{2}{3})^5\) | M1 | or \(462 x (\frac{1}{3})^x x (\frac{2}{3})^5\) |
| \(0.0268\) (3 sfs) | A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(q^{11} = 0.05\) or \((1 - p)^{11} = 0.05\) | M1 | |
| \(\sqrt[11]{0.05}\) | M1 | oe or invlog \(\frac{\log 0.05}{11}\) |
| \(q = 0.762\) or \(0.7616 \ldots\) | A1 | ft dep M2 |
| \(p = 0.238\) (3 sfs) | A1f | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(11 x p x (1 - p) = 1.76\) | M1 | not \(11pq = 1.76\); any correct equn after mult out |
| \(11p^2 - 11p + 1.76 = 0\) or \(p^2 - p + 0.16 = 0\) | A1 | |
| \((25p^2 - 25p + 4 = 0)\) (5p - 1)(5p - 4) = 0$ | M1 | or correct fact'n or subst'n for their quad; equiv with = 0 |
| or \(p = \frac{11 \pm \sqrt{(11^2 - 4 \times 11 \times 1.76)}}{2 \times 11}\) | A1 | any correct equn after mult out |
| Answer | Marks | Guidance |
|---|---|---|
| \(p = 0.2\) or \(0.8\) | A1 | 5 marks |
**i)**
$^{11}C_5 x (\frac{1}{3})^x x (\frac{2}{3})^5$ | M1 | or $462 x (\frac{1}{3})^x x (\frac{2}{3})^5$
$0.0268$ (3 sfs) | A1 | 2 marks
**ii)**
$q^{11} = 0.05$ or $(1 - p)^{11} = 0.05$ | M1 |
$\sqrt[11]{0.05}$ | M1 | oe or invlog $\frac{\log 0.05}{11}$
$q = 0.762$ or $0.7616 \ldots$ | A1 | ft dep M2
$p = 0.238$ (3 sfs) | A1f | 4 marks
**iii)**
$11 x p x (1 - p) = 1.76$ | M1 | not $11pq = 1.76$; any correct equn after mult out
$11p^2 - 11p + 1.76 = 0$ or $p^2 - p + 0.16 = 0$ | A1 |
$(25p^2 - 25p + 4 = 0)$ (5p - 1)(5p - 4) = 0$ | M1 | or correct fact'n or subst'n for their quad; equiv with = 0
or $p = \frac{11 \pm \sqrt{(11^2 - 4 \times 11 \times 1.76)}}{2 \times 11}$ | A1 | any correct equn after mult out
**Total for Question 9:** 11 marks
$p = 0.2$ or $0.8$ | A1 | 5 marks
**TOTAL MARKS:** 72 marks9 A variable $X$ has the distribution $\mathrm { B } ( 11 , p )$.\\
(i) Given that $p = \frac { 3 } { 4 }$, find $\mathrm { P } ( X = 5 )$.\\
(ii) Given that $\mathrm { P } ( X = 0 ) = 0.05$, find $p$.\\
(iii) Given that $\operatorname { Var } ( X ) = 1.76$, find the two possible values of $p$.
\hfill \mbox{\textit{OCR S1 2007 Q9 [11]}}