S1 June 2018 Q1
1. The discrete random variable \(X\) has the following probability distribution
| \(x\) | 2 | 4 | 7 | 10 |
|---|---|---|---|---|
| \(\mathrm{P}(X = x)\) | \(a\) | \(b\) | 0.1 | \(c\) |
where \(a\), \(b\) and \(c\) are probabilities.
The cumulative distribution function of \(X\) is \(\mathrm{F}(x)\) and \(\mathrm{F}(3) = 0.2\) and \(\mathrm{F}(6) = 0.8\)
(a) Find the value of \(a\), the value of \(b\) and the value of \(c\). (3)
(b) Write down the value of \(\mathrm{F}(7)\). (1)
| Scheme | Marks |
|---|---|
| \(\mathrm{F}(3) = \mathrm{P}(X = 2)\) so \(a = 0.2\) | B1 |
| \(\mathrm{F}(6) = \mathrm{P}(X = 2) + \mathrm{P}(X = 4)\) so \(a + b = 0.8\) so \(b = 0.6\) | B1 |
| Sum of probs = 1 implies \(c = 0.1\) | B1ft |
| (3) |
Notes
1st B1 for \(a = 0.2\)
2nd B1 for \(b = 0.6\)
3rd B1 ft for \(c = 0.1\)
or a value of \(c\) so that their \(a + b + c = 0.9\) provided \(a\), \(b\) and \(c\) are probabilities
The labels may not be explicit but it must be clear which is which
| Scheme | Marks |
|---|---|
| \(\mathrm{F}(7) = \mathrm{F}(6) + 0.1\) or \(a + b + 0.1\) or \(1 - c\) = 0.9 | B1 |
| (1) | |
| (4 marks) |
Notes
B1 for 0.9 only (no ft)
If their answer is based on their values of \(a\), \(b\) or \(c\), these values must be probabilities and have \(a + b = 0.8\) or \(c = 0.1\)
Just stating 0.9 with no justification is B1