| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | October |
| Paper | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Find parameters from given statistics |
| Difficulty | Standard +0.3 This is a straightforward S2 question on continuous uniform distributions requiring standard formulas for probability and variance, plus basic algebraic manipulation. Part (i) uses P(X>27) and Var(X) formulas to find a,b via simultaneous equations, part (b) solves a linear equation, and part (ii) applies the uniform distribution to a simple geometric context. All steps are routine applications of known techniques with no novel insight required. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| (i)(a) | \(\frac{b-27}{b-a} = \frac{3}{4}\) or \(\frac{27-a}{b-a} = \frac{1}{4}\) and \(\frac{(b-a)^2}{12} = 300\) | M1M1 |
| \(a = 12\) and \(b = 72\) | A1 | (3) |
| (b) | \([4P(X < k-10) = P(X > k+20) \Rightarrow]\) \(4\left(\frac{k-10-'12'}{72-'(k+20)'}\right) = \frac{'72'-(k+20)}{72-'12'}\) | M1 |
| \(4(k-22) = 52-k \Rightarrow k = 28\) | A1 | (2) |
| (ii) | \(L \sim U(21, 42)\) or \(L \sim U(0, 42)\) | M1 |
| \(\frac{L}{4}\left(\frac{42-L}{4}\right) > 2\) or \(\left(\frac{42-L}{4}\right) \frac{L}{4} > 2\) or \(S-(10.5-S) > 2\) | M1 | |
| \(L > 25\) or \(L < 17\) or \(L > 25\) | A1 | |
| \(= (42-'25') \times \frac{1}{21}\) | M1 | |
| \(= ('17'-0) \times \frac{1}{42} + (42-'25') \times \frac{1}{42}\) or \(\frac{17}{21}\) oe | A1 | (4) |
| Total | 9 |
| (i)(a) | $\frac{b-27}{b-a} = \frac{3}{4}$ or $\frac{27-a}{b-a} = \frac{1}{4}$ and $\frac{(b-a)^2}{12} = 300$ | M1M1 | |
| | $a = 12$ and $b = 72$ | A1 | (3) |
| (b) | $[4P(X < k-10) = P(X > k+20) \Rightarrow]$ $4\left(\frac{k-10-'12'}{72-'(k+20)'}\right) = \frac{'72'-(k+20)}{72-'12'}$ | M1 | |
| | $4(k-22) = 52-k \Rightarrow k = 28$ | A1 | (2) |
| (ii) | $L \sim U(21, 42)$ or $L \sim U(0, 42)$ | M1 | |
| | $\frac{L}{4}\left(\frac{42-L}{4}\right) > 2$ or $\left(\frac{42-L}{4}\right) \frac{L}{4} > 2$ or $S-(10.5-S) > 2$ | M1 | |
| | $L > 25$ or $L < 17$ or $L > 25$ | A1 | |
| | $= (42-'25') \times \frac{1}{21}$ | M1 | |
| | $= ('17'-0) \times \frac{1}{42} + (42-'25') \times \frac{1}{42}$ or $\frac{17}{21}$ oe | A1 | (4) |
| | **Total** | **9** | |
**Notes for Question 4:**
- **(i)(a) M1:** For setting up a correct equation for the probability or the variance
- **(i)(a) M1:** For setting up a correct equation for the probability **and** the variance
- **(i)(a) A1:** For $a = 12$ and $b = 72$ (correct answers score 3 out of 3)
- **(b) M1:** For an unsimplified equation ft their $a$ and their $b$
- **(b) A1:** Cao
- **(ii) M1:** For $\frac{L}{4}\left(\frac{42-L}{4}\right) > 2$ or $\left(\frac{42-L}{4}\right) - \frac{L}{4} > 2$ or $S-(10.5-S) > 2$ may be seen in a probability statement; may be implied by $L > 25$ or $L < 17$ or $S > 6.25$
- **(ii) A1:** $L > 25$ or $L < 17$ or $S > 6.25$ may be seen in a probability statement or implied by 2nd M1
- **(ii) M1:** For use of $(42-'25') \times \frac{1}{21}$ or $('17'-0) \times \frac{1}{42} + (42-'25') \times \frac{1}{42}$ or $(10.5-'6.25') \times \frac{1}{5.25}$
- **(ii) A1:** Allow awrt 0.81
---\begin{enumerate}
\item (i) The continuous random variable $X$ is uniformly distributed over the interval $[ a , b ]$
\end{enumerate}
Given that
\begin{itemize}
\item $\mathrm { P } ( X > 27 ) = \frac { 3 } { 4 }$
\item $\operatorname { Var } ( X ) = 300$\\
(a) find the value of $a$ and the value of $b$
\end{itemize}
Given also that
$$4 \times \mathrm { P } ( X < k - 10 ) = \mathrm { P } ( X > k + 20 )$$
(b) find the value of $k$\\
(ii) A piece of wire of length 42 cm is cut into 2 pieces at a random point.
Each of the two pieces of the wire is bent to form the outline of a square.\\
Find the probability that the side length of the larger square minus the side length of the smaller square will be greater than 2 cm .
\hfill \mbox{\textit{Edexcel S2 2024 Q4}}