S2 June 2013 (R) Q7
7. As part of a selection procedure for a company, applicants have to answer all 20 questions of a multiple choice test. If an applicant chooses answers at random the probability of choosing a correct answer is 0.2 and the number of correct answers is represented by the random variable \(X\).
Each applicant gains 4 points for each correct answer but loses 1 point for each incorrect answer. The random variable \(S\) represents the final score, in points, for an applicant who chooses answers to this test at random.
An applicant who achieves a score of at least 20 points is invited to take part in the final stage of the selection process.
Cameron is taking the final stage of the selection process which is a multiple choice test consisting of 100 questions. He has been preparing for this test and believes that his chance of answering each question correctly is 0.4
| Scheme | Marks |
|---|---|
| \(X \sim \mathrm{B}(20, 0.2)\) | M1 A1 |
| (2) |
Notes
M1 for "binomial" or B(...
A1 for \(n = 20\) and \(p = 0.2\)
| Scheme | Marks |
|---|---|
| \(S = 4X - 1(20 - X)\) | M1 |
| \(S = 5X - 20\) | A1cso |
| (2) |
Notes
NB this is a ‘show that’ so working must be shown
M1 for attempt at any correct expression for \(S\) that uses 4 and – 1 (1 may not be seen)
A1cso for correct expression derived. No incorrect working seen and M1 scored.
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X) = 4, \quad \mathrm{Var}(X) = 3.2\) | B1, B1 |
| \(\mathrm{E}(S) = 5 \times 4 - 20 = 0, \quad \mathrm{Var}(S) = 5^2\,\mathrm{Var}(X) = 80\) | M1 A1 |
| (4) |
Notes
1st B1 for E(\(X\)) = 4 seen. Condone E(S) = 4. May be implied by correct E(S) or be seen in the calculation for E(S)
2nd B1 for Var(\(X\)) = 3.2 seen. Condone Var(S) = 3.2. May be implied by correct Var(S) or be seen in the calculation for Var(S)
M1 for a correct formula for \(\mathrm{E}(S)\) or \(\mathrm{Var}(S)\) – follow through their \(\mathrm{E}(X)\) and \(\mathrm{Var}(X)\) may be implied by either answer being correct
A1 for 0 and 80 correctly assigned.
| Scheme | Marks |
|---|---|
| \(S \geqslant 20\) implies \(5X - 20 \geqslant 20\) | M1 |
| [So \(5X \geqslant 40\)] \(X \geqslant 8\) | A1 |
| \(\mathrm{P}(S \geqslant 20) = \mathrm{P}(X \geqslant 8) = 1 - \mathrm{P}(X \leqslant 7)\) | M1 |
| \(= 1 - 0.9679 = \underline{\mathbf{0.0321}}\) | A1 |
| (4) |
Notes
1st M1 for an attempt to solve the inequality for \(X\)
2nd M1 for \(1 - \mathrm{P}(X \leqslant 7)\)
| Scheme | Marks |
|---|---|
| [Let \(C\) = no. Cameron gets correct. \(C \sim \mathrm{B}(100, 0.4)\)] \(Y \sim \mathrm{N}\left(40, \sqrt{24}^{\,2}\right)\) | M1A1 |
| \(\mathrm{P}(C \gt 50) \simeq \mathrm{P}(Y \gt 50.5)\) | |
| \(= \mathrm{P}\left(Z \gt \dfrac{50.5 - 40}{\sqrt{24}}\right)\) | M1 M1 |
| \(= \mathrm{P}(Z \gt 2.14\ldots) = 1 - 0.9838 = 0.0162\) or \(0.016044..\) (awrt 0.016) N.B. exact Bin (0.01676...) Poisson approx (0.0526...) | A1 |
| (5) | |
| (17 marks) |
Notes
1st M1 for use of normal approx. and mean = 40
1st A1 for Var = 24 or st. dev = \(\sqrt{24}\) May be implied by later work
2nd M1 49.5 or 50.5
3rd M1 Standardising using their mean and their sd, If they have not written down a mean and sd then these need to be correct here to award the mark. They must also use 50.5, 49.5 or 50 and find the correct area ie using \(1 - \mathrm{P}(Z \leqslant\) “their 2.14”),
2nd A1 for awrt 0.016