S1 June 2015 Q5
5. In a quiz, a team gains 10 points for every question it answers correctly and loses 5 points for every question it does not answer correctly. The probability of answering a question correctly is 0.6 for each question. One round of the quiz consists of 3 questions.
The discrete random variable \(X\) represents the total number of points scored in one round.
The table shows the incomplete probability distribution of \(X\)
| \(x\) | 30 | 15 | 0 | \(-15\) |
|---|---|---|---|---|
| \(\mathrm{P}(X = x)\) | 0.216 | 0.064 |
In a bonus round of 3 questions, a team gains 20 points for every question it answers correctly and loses 5 points for every question it does not answer correctly.
| Scheme | Marks |
|---|---|
| To score 15 points, 2 correct and 1 not correct \([0.6 \times 0.6 \times 0.4] + [0.6 \times 0.4 \times 0.6] + [0.4 \times 0.6 \times 0.6]\) or \(3 \times (0.6 \times 0.6 \times 0.4)\) | M1 |
| \(= 0.432\) (*) | A1cso |
| (2) |
Notes
M1 for \(0.6^2 \times 0.4\) may be \(\Rightarrow\) by tree diagram with 0.6 & 0.4 but just \(3 \times 0.144\) or \(2 \times 0.216\) is M0
A1 cso for \(3 \times 0.6^2 \times 0.4\) (seen) and no incorrect working seen
| Scheme | Marks |
|---|---|
| 1 – (0.216 + 0.432 + 0.064) = 0.288 or \(3 \times 0.6 \times (0.4)^2\) | B1 |
| (1) |
Notes
0.288 or \(\dfrac{36}{125}\) answer may be seen in table. [NB Fractions: \(\dfrac{27}{125}, \dfrac{54}{125}, \dfrac{36}{125}\) and \(\dfrac{8}{125}\)]
Correct answers to (c), (d) and (e) score full marks for these parts.
| Scheme | Marks |
|---|---|
| [(30, 0), (0, 30) or (15, 15)] \(0.216 \times \text{'}0.288\text{'} + \text{'}0.288\text{'} \times 0.216 + 0.432 \times 0.432\) | M1 A1ft |
| awrt 0.311 | A1 |
| (3) |
Notes
M1 for either \(0.216 \times \text{'}0.288\text{'} = (0.062208)\) or \(0.432 \times 0.432 = 0.186624\)
(ft (b) provided their (b) is a probability)
1st A1ft for a fully correct expression 2nd A1 for awrt 0.311 or \(\dfrac{972}{3125}\)
SC 6 questions 4 correct Award M1&1st A1 for \({}^6\mathrm{C}_4 \times 0.6^4 \times 0.4^2\) or \(15 \times 0.6^4 \times 0.4^2\)
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X) = [30 \times 0.216] + [15 \times 0.432] + [0 \times 0.288] + [(-15) \times 0.064]\) | M1 |
| \(\mathrm{E}(X) = 12\) 12 (only) | A1 |
| (2) |
Notes
M1 for a correct expression for E(\(X\)) (0 term not required, ft their (b))
NB alt: \(3 \times (10 \times 0.6 + (-5) \times 0.4)\). E(\(X\)) = 12 scores M1A1 if (b) is a probability.
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X^2) = 30^2 \times 0.216 + 15^2 \times 0.432 + 0^2 \times 0.288 + (-15)^2 \times 0.064\ (= 306)\) | M1 |
| \(\mathrm{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2 = \text{'}306\text{'} - \text{'}12\text{'}^2 =,\) 162 | M1, A1 |
| (3) |
Notes
1st M1 for correct expres’ for \(\mathrm{E}(X^2)\) (0 term not required, ft their(b))Condone \(-15^2\)
Ignore label so Var(\(X\)) = [E(\(X^2\))] = 306 can score M1M0A0
2nd M1 for correct expression for \(\mathrm{Var}(X)\) (may follow through their values)
ALT 1st M1 for \([10^2 \times 0.6 + (-5)^2 \times 0.4 = 70]\) 2nd M1 for \(3 \times (70 - 4^2) = 54\) and A1 for 162
| Scheme | Marks | ||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|
Let \(Y\) = number of points scored in bonus round
| M1 | ||||||||||
| \(\mathrm{E}(Y) = 60 \times 0.216 + 35 \times 0.432 + 10 \times 0.288 + (-15) \times 0.064\) | dM1 | ||||||||||
| \(= \)30 | A1 | ||||||||||
| (3) | |||||||||||
| (14 marks) |
Notes
1st M1 for correct distribution for \(Y\) (ft(b)) or \(20 \times 0.6 + (-5) \times 0.4\) or \(Y = \frac{5}{3}X + 10\)
2nd dM1 for correct expres’ for E(\(Y\)) or \(3 \times (20 \times 0.6 + (-5) \times 0.4)\) or \(\mathrm{E}(Y) = \frac{5}{3}\mathrm{E}(X) + 10\)
Dep. on 1st M1 but can ft their (b) or their E(\(X\)). Correct expres’ (line 2) scores M1M1
A1 for 30 with at least 1 M mark scored. Answer only is 0/3 but 30 after M1 is 3/3