S1 June 2018 Q5
5. The score when a spinner is spun is given by the discrete random variable \(X\) with the following probability distribution, where \(a\) and \(b\) are probabilities.
| \(x\) | \(-1\) | 0 | 2 | 4 | 5 |
|---|---|---|---|---|---|
| \(\mathrm{P}(X = x)\) | \(b\) | \(a\) | \(a\) | \(a\) | \(b\) |
Given that \(\mathrm{Var}(X) = 7.1\)
The discrete random variable \(Y = 10 - 3X\)
The spinner is spun once.
| Scheme | Marks |
|---|---|
| The distribution is symmetric about the value 2 (o.e.) [ “data” is B0] | B1 cso |
| (1) |
Notes
B1 for argument using symmetry “distribution is symmetric” B1
“probs are symmetric” B0 “it is symmetric” is B0
or a correct expression (\(6a + 4b\)) and use of sum of probs = 1
| Scheme | Marks |
|---|---|
| Sum of probs = 1 (or use of E(\(X\)) = 2) leading to \(3a + 2b = 1\) | B1 |
| (1) |
Notes
B1 for \(3a + 2b = 1\) (o.e.) (any equivalent correct equation, needn’t be simplified)
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X^2) = (-1)^2 b + 2^2 a + 4^2 a + 5^2 b\) [\(= 20a + 26b\) …condone \(24b\)] | M1 |
| \(7.1 = 20a + \text{“}26\text{”}b - 2^2\) or \(7.1 = 20a + \text{“}26\text{”}b - (6a + 4b)^2\) or \(7.1 = 8a + 18b\) | M1 |
| \(11.1 = 20a + 26b\) | A1 |
| (3) |
Notes
1st M1 for a full expression for E(\(X^2\)). Condone \(-1^2 b\) … or \(20a + 26b\) or \(20a + 24b\)
Allow Var(\(X\)) called E(\(X^2\)). M0 for \(\frac{20a + 26b}{5}\) unless you see E(\(X^2\)) = \(20a + 26b\) (o.e.) first.
2nd M1 for use of the correct formula to form an equation for \(a\) and \(b\). ft their E(\(X^2\))
A1 for \(11.1 = 20a + 26b\) (or equivalent but must be only 3 non-zero terms)
| Scheme | Marks |
|---|---|
| e.g. (b)\(\times 13\) and subtract (c) yielding: \(1.9 = 19a\) | M1 |
| \(a = 0.1\) and \(b = 0.35\) | A1, A1 |
| (3) |
Notes
M1 for solving their 2 linear equations in \(a\) and \(b\) and reducing to an equ’n in one variable
Condone 1 arithmetic or sign error
1st A1 for \(a = 0.10\) or an exact equivalent
2nd A1 for \(b = 0.35\) or an exact equivalent
Ans only One correct value scores M1 and the relevant A1 and both correct scores 3/3
| Scheme | Marks |
|---|---|
| (i) [E(\(Y\)) = 10 – 3E(\(X\)) = \(10 - 3 \times 2\) ] = 4 | B1 |
| (ii) \([\mathrm{Var}(Y)] = (-3)^2\,\mathrm{Var}(X)\) | M1 |
| \(= \)63.9 | A1 |
| (3) |
Notes
(ii) M1 for correct use of the Var(\(aX + b\)) formula. Condone \(-3^2\) if it later becomes +9
or [E(\(Y^2\))] = 79.9 and [Var(\(Y\))] = 79.9 – their (E(\(Y\)))\(^2\)
A1 for 63.9
| Scheme | Marks |
|---|---|
| \(Y \gt X\) gives: \(10 - 3X \gt X\) leading to \(10 \gt 3X + X\) or \(X \lt 2.5\) | M1 |
| \(X \lt 2.5\) means \(X = -1\), 0 and 2 | A1 |
| \(\mathrm{P}(Y \gt X) = 2a + b = \)0.55 or \(\frac{11}{20}\) (o.e.) | A1ft |
| (3) | |
| (14 marks) |
Notes
M1 for an attempt to solve the linear inequality leading to \(10 \gt 3X + X\) or \(Y \gt 2.5\) or \(Y \geqslant 4\)
A1 for the correct 3 values of \(X\) or prob. dist. for \(Y\) and \(y\) = 4, 10, 13 or P(\(X\) < 2.5) = \(2a + b\)
A1ft for an answer = their \(2a + b\) provided \(a\), \(b\) and \(2a + b\) are probabilities. Must be a value
NB(e/f)
| \(x\) | \(-1\) | 0 | 2 | 4 | 5 |
|---|---|---|---|---|---|
| \(y\) | 13 | 10 | 4 | \(-2\) | \(-5\) |
Correct answer only for their \(a\) and \(b\) is 3/3 BUT \(2a + b\) only is M0