| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | Verify CDF properties |
| Difficulty | Moderate -0.8 This is a straightforward S2 question testing basic understanding of CDFs. Part (a) requires recognizing that F(t) = 1 - P(T > t), which is given directly. Parts (b-d) involve simple substitution into the CDF formula and basic algebraic manipulation. No novel insight or complex problem-solving is required—just routine application of standard CDF properties and conditional probability formulas. |
| Spec | 2.03d Calculate conditional probability: from first principles5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F(t) = \begin{cases} 1 - \dfrac{225}{(t+15)^2} & t \geq 0 \\ 0 & \text{otherwise} \end{cases}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(T < 3) = 1 - \dfrac{225}{(3+15)^2}\) | M1 | |
| \(= \dfrac{11}{36}\) or \(0.30555\ldots\) | A1 | awrt 0.306 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(T > 8 \mid T > 3) = \dfrac{P(T > 8)}{P(T > 3)}\) | M1 | |
| \(= \dfrac{\dfrac{225}{23^2}}{\dfrac{225}{18^2}}\) | M1 | |
| \(= \dfrac{324}{529}\) or \(0.612\ldots\) | A1 | awrt 0.612 / 0.6125 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1 - F(t) = 0.1\) | 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 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(T>t) = \frac{225}{(t+15)^2}\) and \(F(t) = 1 - \frac{225}{(t+15)^2}\), \(t>0\) | B1 | Both expressions must be seen with no errors. Allow equivalent in words. The cdf must be given. Condone \(<\) instead of \(\leq\) or \(>\) instead of \(\geq\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Substitute \(t=3\) into \(F(t)\) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Conditional probability as a quotient with \(P(T>3)\) or numerical equivalent in denominator | M1 | Must be a quotient |
| \(P(T>8)\) or \(P(T>5)\) or \(P(T>8 \cap T>3)\) or \(P(T>5 \cap T>3)\) in numerator | M1 | 2nd M1 independent of first. Writing or using \(P(T>8)\) or \(P(T \geq 8)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(1-F(t)=0.1\) or \(P(T \geq t)=0.1\) | M1 | May be implied by \(\frac{225}{(t+15)^2}=0.1\) |
| Square rooting or solving quadratic by factorising/completing the square/formula | M1 | Must be correct for their quadratic |
| \(t \approx 32.4\) (or 32 or 33) | A1 | Do not accept \(15\sqrt{10}-15\) |
## Question 5:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F(t) = \begin{cases} 1 - \dfrac{225}{(t+15)^2} & t \geq 0 \\ 0 & \text{otherwise} \end{cases}$ | B1 | |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(T < 3) = 1 - \dfrac{225}{(3+15)^2}$ | M1 | |
| $= \dfrac{11}{36}$ or $0.30555\ldots$ | A1 | awrt 0.306 |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(T > 8 \mid T > 3) = \dfrac{P(T > 8)}{P(T > 3)}$ | M1 | |
| $= \dfrac{\dfrac{225}{23^2}}{\dfrac{225}{18^2}}$ | M1 | |
| $= \dfrac{324}{529}$ or $0.612\ldots$ | A1 | awrt 0.612 / 0.6125 |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1 - F(t) = 0.1$ | 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 | |
# Mark Scheme Extraction
---
## Question 5 (CDF Question):
**(a)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(T>t) = \frac{225}{(t+15)^2}$ and $F(t) = 1 - \frac{225}{(t+15)^2}$, $t>0$ | B1 | Both expressions must be seen with no errors. Allow equivalent in words. The cdf must be given. Condone $<$ instead of $\leq$ or $>$ instead of $\geq$ |
**(b)**
| Answer | Mark | Guidance |
|--------|------|----------|
| Substitute $t=3$ into $F(t)$ | M1 | |
**(c)**
| Answer | Mark | Guidance |
|--------|------|----------|
| Conditional probability as a quotient with $P(T>3)$ or numerical equivalent in denominator | M1 | Must be a quotient |
| $P(T>8)$ or $P(T>5)$ or $P(T>8 \cap T>3)$ or $P(T>5 \cap T>3)$ in numerator | M1 | 2nd M1 independent of first. Writing or using $P(T>8)$ or $P(T \geq 8)$ |
**(d)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $1-F(t)=0.1$ or $P(T \geq t)=0.1$ | M1 | May be implied by $\frac{225}{(t+15)^2}=0.1$ |
| Square rooting or solving quadratic by factorising/completing the square/formula | M1 | Must be correct for their quadratic |
| $t \approx 32.4$ (or 32 or 33) | A1 | Do not accept $15\sqrt{10}-15$ |
---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
$$\frac { 225 } { ( t + 15 ) ^ { 2 } } , \quad t \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item Show that the cumulative distribution function of $T$ is given by
$$\mathrm { F } ( t ) = \begin{cases} 1 - \frac { 225 } { ( t + 15 ) ^ { 2 } } & t \geqslant 0 \\ 0 & \text { otherwise } \end{cases}$$
\item Find the probability that a randomly selected mosquito will die within 3 days of hatching.
\item Given that a mosquito survives for 3 days, find the probability that it will survive for at least 5 more days.
A large number of mosquitoes hatch on the same day.
\item Find the number of days after which only $10 \%$ of these mosquitoes are expected to survive.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2013 Q5 [10]}}