| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2020 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Multiple independent observations |
| Difficulty | Standard +0.3 This is a straightforward S2 question requiring recognition that E(X)=7.5 by symmetry of the pdf, standard variance calculation using integration, and a binomial probability calculation with two independent observations. All techniques are routine for this level with no novel problem-solving required. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(7.5\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{6}{125}\int_5^{10}(-x^4 + 15x^3 - 50x^2)\,dx\) | M1 | |
| \(\frac{6}{125}\left[-\frac{x^5}{5} + 15\frac{x^4}{4} - 50\frac{x^3}{3}\right]_5^{10} - \text{'7.5'}^2\) | M1 | |
| \(1.25\) (3 sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{6}{125}\int_5^{6}(-x^2 + 15x - 50)\,dx\) | M1 | |
| \(\frac{6}{125}\left[-\frac{x^3}{3} + 15\frac{x^2}{2} - 50x\right]_5^{6}\) | M1 | |
| \(\frac{6}{125}\left(-102 + \frac{625}{6}\right)\) oe | M1 | |
| \(0.104\) | A1 | |
| \(2 \times \text{'0.104'} \times (1 - \text{'0.104'})\) | M1 | |
| \(0.186\) (3 sf) | A1ft |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $7.5$ | B1 | |
---
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{6}{125}\int_5^{10}(-x^4 + 15x^3 - 50x^2)\,dx$ | M1 | |
| $\frac{6}{125}\left[-\frac{x^5}{5} + 15\frac{x^4}{4} - 50\frac{x^3}{3}\right]_5^{10} - \text{'7.5'}^2$ | M1 | |
| $1.25$ (3 sf) | A1 | |
---
## Question 6(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{6}{125}\int_5^{6}(-x^2 + 15x - 50)\,dx$ | M1 | |
| $\frac{6}{125}\left[-\frac{x^3}{3} + 15\frac{x^2}{2} - 50x\right]_5^{6}$ | M1 | |
| $\frac{6}{125}\left(-102 + \frac{625}{6}\right)$ oe | M1 | |
| $0.104$ | A1 | |
| $2 \times \text{'0.104'} \times (1 - \text{'0.104'})$ | M1 | |
| $0.186$ (3 sf) | A1ft | |
---6 The length, $X$ centimetres, of worms of a certain type is modelled by the probability density function
$$f ( x ) = \begin{cases} \frac { 6 } { 125 } ( 10 - x ) ( x - 5 ) & 5 \leqslant x \leqslant 10 \\ 0 & \text { otherwise } \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item State the value of $\mathrm { E } ( X )$.
\item Find $\operatorname { Var } ( X )$.
\item Two worms of this type are chosen at random.
Find the probability that exactly one of them has length less than 6 cm .
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2020 Q6 [9]}}