S2 January 2013 Q5
5. The continuous random variable \(T\) is used to model the number of days, \(t\), a mosquito survives after hatching.
The probability that the mosquito survives for more than \(t\) days is
\[\frac{225}{(t + 15)^2}, \qquad t \geqslant 0\]A large number of mosquitoes hatch on the same day.
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(T \gt t) = \dfrac{225}{(t + 15)^2}\) | |
| \(\mathrm{P}(T \leqslant t) = 1 - \mathrm{P}(T \gt t)\) \(= 1 - \dfrac{225}{(t + 15)^2}\) | |
| \(\mathrm{F}(t) = \begin{cases} 1 - \dfrac{225}{(t + 15)^2} & t \geqslant 0 \\ 0 & \text{otherwise.} \end{cases}\) | B1 |
| (1) |
Notes
B1 The line \(\mathrm{P}(T \leqslant t) = 1 - \mathrm{P}(T \gt t)\) or \(\mathrm{F}(t) = 1 - \mathrm{P}(T \gt t)\) or both of the following statements \(\mathrm{P}(T \gt t) = \dfrac{225}{(t + 15)^2}\) and \(\mathrm{P}(T \leqslant t)\) /\(\mathrm{F}(t) = 1 - \dfrac{225}{(t + 15)^2}\) must be seen and no errors. Allow equivalent in words.
Condone use of \(\lt\) instead of \(\leqslant\) or \(\gt\) instead of \(\geqslant\) and vice versa.
The cdf must be given. Allow \(t \gt 0\)
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(T \lt 3) = 1 - \dfrac{225}{(3 + 15)^2}\) | M1 |
| \(= \dfrac{11}{36}\) or \(0.30555\ldots\) awrt 0.306 | A1 |
| (2) |
Notes
M1 substituting 3 into \(\mathrm{F}(t)\)
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(T \gt 8 \mid T \gt 3) = \dfrac{P(T \gt 8)}{P(T \gt 3)}\) | M1 M1 |
| \(= \dfrac{\;\frac{225}{23^2}\;}{\;\frac{225}{18^2}\;}\) | |
| \(= \dfrac{324}{529}\) or \(0.612..\) awrt 0.612 / 0.6125 | A1 |
| (3) |
Notes
1st M1 The conditional probability must,
- be a quotient and
- have \(\mathrm{P}(T \gt 3)\) or ‘their numerical equivalent’ for the denominator and
- have \(\mathrm{P}(T \gt 8)\) or \(\mathrm{P}(T \gt 5)\) or \(\mathrm{P}(T \gt 8 \cap T \gt 3)\) or \(\mathrm{P}(T \gt 5 \cap T \gt 3)\) or ‘their numerical equivalent’ for the numerator.
Allow \(\geqslant\) in place of \(\gt\)
2nd M1 writing or using \(\mathrm{P}(T \gt 8)\) or \(\mathrm{P}(T \geqslant 8)\).
NB This is independent of the first M mark.
| Scheme | Marks |
|---|---|
| \(1 - \mathrm{F}(t) = 0.1\) | M1 |
| \(\dfrac{225}{(t + 15)^2} = 0.1\) or \(1 - \dfrac{225}{(t + 15)^2} = 0.9\) | A1 |
| \(\dfrac{225}{0.1} = (t + 15)^2\) | |
| \(t = \sqrt{\dfrac{225}{0.1}} - 15\) | M1 |
| \(t = 32.4\), also accept 32/33 | A1 |
| (4) | |
| (10 marks) |
Notes
1st M1 writing or using \(1 - \mathrm{F}(t) = 0.1\) or \(\mathrm{P}(T \geqslant t) = 0.1\) May be implied by \(\dfrac{225}{(t + 15)^2} = 0.1\) o.e.
2nd M1 either square rooting or solving a quadratic either by factorising / completing the square / using the formula - must be correct for their quadratic.
A1 awrt 32.4 or 32 or 33. Do not accept \(15\sqrt{10} - 15\)