S1 January 2013 Q2
2. The discrete random variable \(X\) can take only the values 1, 2 and 3. For these values the cumulative distribution function is defined by
\[\mathrm{F}(x) = \frac{x^3 + k}{40} \qquad x = 1, 2, 3\]Given that \(\mathrm{Var}(X) = \dfrac{259}{320}\)
| Scheme | Marks |
|---|---|
| \(\mathrm{F}(3) = 1\) gives \(\dfrac{3^3 + k}{40} = 1\) | M1 |
| So \(k = \underline{\mathbf{13}}\) | A1cso |
| (2) |
Notes
M1 for use of \(\mathrm{F}(3) = 1\). Attempt at \(\dfrac{3^3 + k}{40} = 1\) must be seen
\(27 + k = 40\) without reference to \(\mathrm{F}(3) = 1\) is M0
A1cso for no incorrect working seen and M1 scored.
Verify Allow M1 for \(\dfrac{3^3 + 13}{40} = 1\) but the A1 requires an explicit comment such as “so \(k = 13\)”
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(X = 1) = \dfrac{14}{40}\) or 0.35 (o.e.) | B1 |
| Use of \(\mathrm{P}(X = 2) = \mathrm{F}(2) - \mathrm{F}(1)\) or \(\mathrm{P}(X = 3) = \mathrm{F}(3) - \mathrm{F}(2)\) | M1 |
| \(\mathrm{P}(X = 2) = \dfrac{7}{40}\) or 0.175, \(\mathrm{P}(X = 3) = \dfrac{19}{40}\) or 0.475 | A1, A1 |
| (4) |
Notes
If a table such as this is seen then award B1M1A1A1. Ignore labels on 2nd row
| 1 | 2 | 3 | |
| \(\frac{7}{20}\) or 0.35 | \(\frac{7}{40}\) or 0.175 | \(\frac{19}{40}\) or 0.475 |
Otherwise apply the following:
B1 for \(\dfrac{14}{40}\) or 0.35 or any exact equivalent. Can be labelled \(\mathrm{F}(1)\), \(\mathrm{P}(X = 1)\) or \(\mathrm{p}(x)\) and associated with \(x = 1\) or given in a table but must have a label.
M1 for clear method showing how to obtain \(\mathrm{P}(X = \ldots)\) from \(\mathrm{F}(x)\). M1 can be implied if either \(\mathrm{P}(X = 2)\) or \(\mathrm{P}(X = 3)\) is correct
1st A1 for \(\mathrm{P}(X = 2) = \dfrac{7}{40}\) or 0.175 or exact equivalent
2nd A1 for \(\mathrm{P}(X = 3) = \dfrac{19}{40}\) or 0.475 or exact equivalent
| Scheme | Marks |
|---|---|
| \(\mathrm{Var}(4X - 5) = 4^2\,\mathrm{Var}(X)\) | M1 |
| So \(\mathrm{Var}(4X - 5) = \dfrac{259}{20}\) or 12.95 | A1 |
| (2) | |
| (8 marks) |
Notes
M1 for correct use of the variance formula (\(4^2\mathrm{Var}(X)\) alone secures M1). A value for \(\mathrm{Var}(X)\) is not required for this M1
A1 for any exact equivalent to 12.95 Correct answer only is 2/2