| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2009 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Standard two-outcome diagnostic test |
| Difficulty | Standard +0.3 This is a standard conditional probability question using Bayes' theorem with clearly defined probabilities. Part (i) requires straightforward application of the formula P(Disease|Positive) = P(Positive|Disease)×P(Disease)/P(Positive), part (ii) extends this to two tests, and part (iii) is a simple expected value calculation. While it requires careful bookkeeping and multiple steps, the techniques are routine for S4 level with no novel insight required. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(-2) = -16 + 36 - 22 - 8 = -10\) | M1 | Attempt \(f(-2)\), or equiv |
| A1 2 | Obtain -10 |
| Answer | Marks | Guidance |
|---|---|---|
| \(f'(\frac{1}{2}) = \frac{1}{4} + 2\frac{1}{4} + 5\frac{1}{2} - 8 = 0\) AG | M1 | Attempt \(f'(\frac{1}{2})\) (no other method allowed) |
| A1 2 | Confirm \(f'(\frac{1}{2}) = 0\), extra line of working required |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x) = (2x - 1)(x^2 + 5x + 8)\) | M1 | Attempt complete division by \((2x - 1)\) or \((x - \frac{1}{2})\) or equiv |
| A1 3 | Obtain \(x^2 + 5x + c\) or \(2x^2 + 10x + c\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x)\) has one real root (\(x = \frac{1}{2}\)) because \(b^2 - 4ac = 25 - 32 = -7\) hence quadratic has no real roots as \(-7 < 0\). | B1√ | State 1 root, following their quotient, ignore reason |
| B1√ 2 | Correct calculation, eg discriminant or quadratic formula, following their quotient, or cubic has max at (-2.15, -9.9) |
### (i)
$f(-2) = -16 + 36 - 22 - 8 = -10$ | M1 | Attempt $f(-2)$, or equiv |
| A1 2 | Obtain -10 |
### (ii)
$f'(\frac{1}{2}) = \frac{1}{4} + 2\frac{1}{4} + 5\frac{1}{2} - 8 = 0$ AG | M1 | Attempt $f'(\frac{1}{2})$ (no other method allowed) |
| A1 2 | Confirm $f'(\frac{1}{2}) = 0$, extra line of working required |
### (iii)
$f(x) = (2x - 1)(x^2 + 5x + 8)$ | M1 | Attempt complete division by $(2x - 1)$ or $(x - \frac{1}{2})$ or equiv |
| A1 3 | Obtain $x^2 + 5x + c$ or $2x^2 + 10x + c$ |
### (iv)
$f(x)$ has one real root ($x = \frac{1}{2}$) because $b^2 - 4ac = 25 - 32 = -7$ hence quadratic has no real roots as $-7 < 0$. | B1√ | State 1 root, following their quotient, ignore reason |
| B1√ 2 | Correct calculation, eg discriminant or quadratic formula, following their quotient, or cubic has max at (-2.15, -9.9) |7 A particular disease occurs in a proportion $p$ of the population of a town. A diagnostic test has been developed, in which a positive result indicates the presence of the disease. It has a probability 0.98 of giving a true positive result, i.e. of indicating the presence of the disease when it is actually present. The test will give a false positive result with probability 0.08 when the disease is not present. A randomly chosen person is given the test.\\
(i) Find, in terms of $p$, the probability that
\begin{enumerate}[label=(\alph*)]
\item the person has the disease when the result is positive,
\item the test will lead to a wrong conclusion.
It is decided that if the result of the test on someone is positive, that person is tested again. The result of the second test is independent of the result of the first test.\\
(ii) Find the probability that the person has the disease when the result of the second test is positive.\\
(iii) The town has 24000 children and plans to test all of them at a cost of $\pounds 5$ per test. Assuming that $p = 0.001$, calculate the expected total cost of carrying out these tests.
\end{enumerate}
\hfill \mbox{\textit{OCR S4 2009 Q7 [11]}}