| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2021 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Breaking/cutting problems |
| Difficulty | Standard +0.3 This is a straightforward S2 question on continuous uniform distributions. Part (a) is trivial probability calculation, (b) is simple algebra using Pythagoras, (c) requires integrating a quadratic over uniform distribution (standard technique), and (d) involves solving a quadratic inequality then finding probability. All parts use routine methods with no novel insight required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(U[0,3]\) | M1 | Writing or using the correct distribution. Allow \(\frac{1.8}{3}\) for M1A0 |
| \(\frac{3-1.8}{3} = 0.4\) | A1 | 0.4 oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(X^2 = W^2 + (3-W)^2\) | M1 | Using Pythagoras to find the length. Note: \(X^2 = W^2 + (W-3)^2\) scores M1A0 |
| \(X^2 = W^2 + 9 + W^2 - 6W \Rightarrow X^2 = 2W^2 - 6W + 9\) | A1 | Brackets multiplied seen leading to \(X^2 = 2W^2 - 6W + 9\) with no incorrect working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(W) = 1.5\) | B1 | 1.5 |
| \(\text{Var}(W) = \frac{9}{12} = \frac{3}{4}\) | B1 | \(\text{Var}(W) = 0.75\); or using integration: \(E(W^2) = \int_0^3 \frac{1}{3}w^2\,dw\) (ignore limits) |
| \(E(W^2) = \text{"}\frac{3}{4}\text{"} + \text{"1.5"}^2\) | M1 | Writing or using \(E(W^2) = \text{Var}(W) + [E(W)]^2\); or \(\left[\frac{1}{9}w^3\right]_0^3\) (correct integration with correct limits) |
| \(E(W^2) = 3\) | A1 | 3 |
| So \(E(X^2) = 2\times\text{"3"} - 6\times\text{"1.5"} + 9 = 6\) | M1 A1 | Use of \(E(X^2) = 2E(W^2) - 6E(W) + 9\) with their values; A1: 6. An answer of 6 from correct working implies all previous marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X^2 > 5) = P(2W^2 - 6W + 4 > 0)\) | M1 | For realising they need to find the probability of \(2W^2 - 6W + 4 > 0\) (condone \(=\)) |
| \(= P((2W-2)(W-2)>0)\) | M1 | Solving their 3-term quadratic (\(W=1\) and \(W=2\) implies 1st two M marks) |
| \(= P(W>2) + P(W<1)\) | dM1 | (dep on 2nd M1) Realising they need to add the 2 outer areas |
| \(= \frac{2}{3}\) oe | A1 | awrt 0.667 |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $U[0,3]$ | M1 | Writing or using the correct distribution. Allow $\frac{1.8}{3}$ for M1A0 |
| $\frac{3-1.8}{3} = 0.4$ | A1 | 0.4 oe |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $X^2 = W^2 + (3-W)^2$ | M1 | Using Pythagoras to find the length. Note: $X^2 = W^2 + (W-3)^2$ scores M1A0 |
| $X^2 = W^2 + 9 + W^2 - 6W \Rightarrow X^2 = 2W^2 - 6W + 9$ | A1 | Brackets multiplied seen leading to $X^2 = 2W^2 - 6W + 9$ with no incorrect working |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(W) = 1.5$ | B1 | 1.5 |
| $\text{Var}(W) = \frac{9}{12} = \frac{3}{4}$ | B1 | $\text{Var}(W) = 0.75$; or using integration: $E(W^2) = \int_0^3 \frac{1}{3}w^2\,dw$ (ignore limits) |
| $E(W^2) = \text{"}\frac{3}{4}\text{"} + \text{"1.5"}^2$ | M1 | Writing or using $E(W^2) = \text{Var}(W) + [E(W)]^2$; or $\left[\frac{1}{9}w^3\right]_0^3$ (correct integration with correct limits) |
| $E(W^2) = 3$ | A1 | 3 |
| So $E(X^2) = 2\times\text{"3"} - 6\times\text{"1.5"} + 9 = 6$ | M1 A1 | Use of $E(X^2) = 2E(W^2) - 6E(W) + 9$ with their values; A1: 6. An answer of 6 from correct working implies all previous marks |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X^2 > 5) = P(2W^2 - 6W + 4 > 0)$ | M1 | For realising they need to find the probability of $2W^2 - 6W + 4 > 0$ (condone $=$) |
| $= P((2W-2)(W-2)>0)$ | M1 | Solving their 3-term quadratic ($W=1$ and $W=2$ implies 1st two M marks) |
| $= P(W>2) + P(W<1)$ | dM1 | (dep on 2nd M1) Realising they need to add the 2 outer areas |
| $= \frac{2}{3}$ oe | A1 | awrt 0.667 |
---5. A piece of wood $A B$ is 3 metres long. The wood is cut at random at a point $C$ and the random variable $W$ represents the length of the piece of wood $A C$.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the length of the piece of wood $A C$ is more than 1.8 metres.
The two pieces of wood $A C$ and $C B$ form the two shortest sides of a right-angled triangle. The random variable $X$ represents the length of the longest side of the right-angled triangle.
\item Show that $X ^ { 2 } = 2 W ^ { 2 } - 6 W + 9$\\[0pt]
[You may assume for random variables $S , T$ and $U$ and for constants $a$ and $b$ that if $S = a T + b U$ then $\mathrm { E } ( S ) = a \mathrm { E } ( T ) + b \mathrm { E } ( U ) ]$
\item Find $\mathrm { E } \left( X ^ { 2 } \right)$
\item Find $\mathrm { P } \left( X ^ { 2 } > 5 \right)$
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2021 Q5 [14]}}