| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Calculate simple probabilities |
| Difficulty | Moderate -0.8 This is a straightforward continuous uniform distribution question requiring only direct application of standard formulas. Parts (a)-(d) involve basic probability calculations, expectation, and percentiles with minimal problem-solving. Part (e) requires slightly more algebraic manipulation but still follows a routine approach. The question is easier than average A-level as it's purely procedural with no conceptual challenges. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((102-100)p = \frac{1}{3}\) or \(\frac{102-100}{k-100} = \frac{1}{3}\) or \(\frac{k-102}{k-100} = \frac{2}{3}\) or \((k-100)p = 1\) | M1 | For one of the 4 given equations |
| \(p = \frac{1}{6}\) so \((k-100)\frac{1}{6} = 1\) | dM1 | For at least one intermediate step of working |
| \(k - 100 = 6\) therefore \(k = 106\) | A1cso | No incorrect working seen leading to \(k = 106\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{5}{6}\) (or exact equivalent) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(103\) | B1 | If working is seen it must not come from a discrete distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{r-100}{6} = 0.15\) | M1 | Any correct equation or expression |
| \(= \underline{100.9}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3P(X \leqslant x-1.5) = P(X \geqslant x+1.5)\) so \(\frac{3}{6}(x-1.5-100) = \frac{1}{6}(106-x-1.5)\) | M1 | For a correct equation for \(x\); allow without \(p = \frac{1}{6}\) |
| \([3(x-1.5-100) = (106-x-1.5)]\) implies \(4x - 304.5 = 104.5\) | dM1 | For attempt to simplify; must have linear equation with \(x\) appearing only once |
| \(x = \underline{102.25}\) | A1 |
## Question 3:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(102-100)p = \frac{1}{3}$ or $\frac{102-100}{k-100} = \frac{1}{3}$ or $\frac{k-102}{k-100} = \frac{2}{3}$ or $(k-100)p = 1$ | M1 | For one of the 4 given equations |
| $p = \frac{1}{6}$ so $(k-100)\frac{1}{6} = 1$ | dM1 | For at least one intermediate step of working |
| $k - 100 = 6$ therefore $k = 106$ | A1cso | No incorrect working seen leading to $k = 106$ |
### Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{5}{6}$ (or exact equivalent) | B1 | |
### Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0$ | B1 | |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $103$ | B1 | If working is seen it must **not** come from a discrete distribution |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{r-100}{6} = 0.15$ | M1 | Any correct equation or expression |
| $= \underline{100.9}$ | A1 | |
### Part (e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3P(X \leqslant x-1.5) = P(X \geqslant x+1.5)$ so $\frac{3}{6}(x-1.5-100) = \frac{1}{6}(106-x-1.5)$ | M1 | For a correct equation for $x$; allow without $p = \frac{1}{6}$ |
| $[3(x-1.5-100) = (106-x-1.5)]$ implies $4x - 304.5 = 104.5$ | dM1 | For attempt to simplify; must have linear equation with $x$ appearing only once |
| $x = \underline{102.25}$ | A1 | |
---3. A machine pours oil into bottles. It is electronically controlled to cut off the flow of oil randomly between 100 ml and $k \mathrm { ml }$, where $k > 100$. It is equally likely to cut off the flow at any point in this range. The random variable $X$ is the volume of oil poured into a bottle.
Given that $\mathrm { P } ( 102 \leqslant X \leqslant k ) = \frac { 2 } { 3 }$
\begin{enumerate}[label=(\alph*)]
\item show that $k = 106$
\item Find the probability that the volume of oil poured into a bottle is
\begin{enumerate}[label=(\roman*)]
\item less than 105 ml ,
\item exactly 105 ml .
\end{enumerate}\item Write down the value of $\mathrm { E } ( X )$
\item Find the 15th percentile of this distribution.
\item Determine the value of $x$ such that $3 \mathrm { P } ( X \leqslant x - 1.5 ) = \mathrm { P } ( X \geqslant x + 1.5 )$
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2018 Q3 [11]}}