S2 January 2012 Q3
3. The probability of a telesales representative making a sale on a customer call is 0.15
Find the probability that
Representatives are required to achieve a mean of at least 5 sales each day.
| Scheme | Marks |
|---|---|
| \(P(X = 0) = 0.85^{10}\) or from tables | M1 |
| \(= 0.1969\) awrt 0.197 | A1 |
| (2) |
Notes
M1 \((p)^{10}\) with \(0 \lt p \lt 1\)
| Scheme | Marks |
|---|---|
| \(P(X \gt 3) = 1 - P(X \leqslant 3)\) | M1 |
| \(= 1 - 0.6477\) \(= 0.3523\) awrt 0.352 | A1 |
| (2) |
Notes
M1 writing or using \(1 - \mathrm{P}(X \leqslant 3)\)
| Scheme | Marks |
|---|---|
| \(n \times 0.15 = 5\) | M1 |
| \(n = 33\) or 34 | A1 |
| (2) |
Notes
M1 \(np = 5\) \(0 \lt p \lt 1\)
| Scheme | Marks |
|---|---|
| \(1 - \mathrm{P}(X = 0) \gt 0.95\) | M1 |
| \(1 - (0.85)^n \gt 0.95.\) \(0.85^n \lt 0.05\) \(n \gt 18.4\) | A1 |
| \(n = 19\) | A1 |
| (3) | |
| (9 marks) |
Notes
M1 writing or using \(1 - \mathrm{P}(X = 0) \gt 0.95\) or \(\mathrm{P}(X = 0) \lt 0.05\) (also accepted are = or \(\geqslant\) instead of > and = or \(\leqslant\) instead of or <) \(\mathrm{P}(X \leqslant 0)\) is equivalent to \(\mathrm{P}(X = 0)\)
A1 writing or using \(1 - (0.85)^n \gt 0.95\) or \((0.85)^n \lt 0.05\) (also accepted are \(\geqslant\) instead of > and \(\leqslant\) instead of or <). Any value of \(n\) may be used
A1 cao
NB an answer of 18.4 gets M1 A1 A0
An answer of 19 gets M1 A1 A1 unless it follows from clearly incorrect working.