S1 June 2016 Q2
2. The discrete random variable \(X\) has the following probability distribution, where \(p\) and \(q\) are constants.
| \(x\) | \(-2\) | \(-1\) | \(\dfrac{1}{2}\) | \(\dfrac{3}{2}\) | 2 |
|---|---|---|---|---|---|
| \(\mathrm{P}(X = x)\) | \(p\) | \(q\) | 0.2 | 0.3 | \(p\) |
Given that \(\mathrm{E}(X) = 0.4\)
Given also that \(\mathrm{E}(X^2) = 2.275\)
Sarah and Rebecca play a game.
A computer selects a single value of \(X\) using the probability distribution above.
Sarah’s score is given by the random variable \(S = X\) and Rebecca’s score is given by the random variable \(R = \dfrac{1}{X}\)
Sarah and Rebecca work out their scores and the person with the higher score is the winner. If the scores are the same, the game is a draw.
| Scheme | Marks |
|---|---|
| \(p + q + 0.2 + 0.3 + p = 1\) or \(2p + q = 0.5\) (o.e.) | B1 |
| (1) |
Notes
B1 for any correct equation based on sum of probs. = 1
Correct answer only in (b), (c), (d), (e) or (f) scores full marks for that part.
| Scheme | Marks |
|---|---|
| [E(\(X\)) =] \(-2p - q + \tfrac{1}{2} \times 0.2 + \tfrac{3}{2} \times 0.3 + 2p\ [= 0.4]\) or \(-q + 0.1 + 0.45\ [= 0.4]\) | M1A1 |
| \(q = 0.15\) | A1 |
| (3) |
Notes
M1 for an attempt at an expression based on E(\(X\)). At most 2 errors or omissions.
1st A1 for a correct equation [May be implied by a correct answer]
2nd A1 for \(q = 0.15\) or exact equivalent e.g. \(\frac{6}{40}\)
| Scheme | Marks |
|---|---|
| \(2p +\) “0.15” \(= 0.5\) (o.e) | M1 |
| \(p = 0.175\) | A1 |
| (2) |
Notes
M1 for correct equation or using their equation from (a) with their \(q\), provided \(q \in [0, 1]\)
A1 for \(p = 0.175\) or exact equivalent e.g. \(\frac{7}{40}\)
| Scheme | Marks |
|---|---|
| [Var(\(X\)) =] \(2.275 - (0.4)^2\) | M1 |
| \(= \)2.115 (Accept 2.12) | A1 |
| (2) |
Notes
M1 for a correct numerical expression but M0 if followed by division by \(k\) (e.g. \(k = 5\))
A1 for 2.115 or accept awrt 2.12 (also accept exact equivalent e.g. \(\frac{423}{200}\))
| Scheme | Marks | ||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| M1 | ||||||||||||
| \(\mathrm{E}(R) = -\tfrac{1}{2}p - q + 0.4 + 0.2 + \tfrac{1}{2}p\) | dM1 | ||||||||||||
| \(= 0.6 - q = \)0.45 ( or \(\frac{9}{20}\) ) | A1ft | ||||||||||||
| (3) |
Notes
1st M1 for correct values for \(R\), allow 1 error, and allow unsimplified. Condone no label if not used as probabilities. If seen in table on QP allow, but must be labelled.
Just writing the sum \(\Sigma r\) is M0 but adding later can score 1st M1
2nd dM1 dependent on 1st M1 for an attempt at an expression based on E(\(R\)), ft \(p\) and \(q\), (if probabilities) ft their \(r\) values. At least 3 correct (or correct ft) products seen.
A1ft for 0.45 or ( 0.6 – their \(q\)) provided \(q\) is a probability
| Scheme | Marks |
|---|---|
| (i) \(S \gt R\) when \(x = 1.5\) and 2 | M1 |
| P(Sarah wins) = 0.3 + \(p\) = 0.475 (or \(\frac{19}{40}\)) | A1ft |
| (ii) \(R \gt S\) when \(x = -2\) and \(\frac{1}{2}\) or \(r = -\frac{1}{2}\) and 2 | M1 |
| P(Rebecca wins) = 0.2 + \(p\) = 0.375 (or \(\frac{15}{40}\)) | A1ft |
| (4) | |
| (15 marks) |
Notes
Answers for (f) must be clearly labelled or take 1st as (i) and 2nd as (ii)
(i) M1 for identifying the correct values of \(X\)
A1ft for 0.475 or 0.3 + their \(p\) , provided answer is a probability
(ii) M1 for identifying the correct values of \(X\) or \(R\)
A1ft for 0.375 or 0.2 + their \(p\) or 1 – their 0.475 – their \(q\), provided ans. is a probability
SC1 \(X_1, X_2\) They use two values of \(X\): (i) for \(\mathrm{P}(S \gt R) = 0.445\) (B1) (ii) for \(\mathrm{P}(R \gt S) = 0.4625\) (B1). No ft
SC2 swap Answers wrong way round: (i) \(\mathrm{P}(S \gt R) = 0.375\) and (ii) \(\mathrm{P}(R \gt S) = 0.475\) (B1) No ft
Epen On epen record SC1 as: (i) M0A1 (ii) M0A1 and SC2 as M0A0M0A1