| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2013 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find expectation E(X) |
| Difficulty | Moderate -0.3 This is a standard S3/Statistics 3 continuous probability distribution question requiring routine integration to find E(X), Var(X), and probabilities. The 'show that' format for E(X) makes it easier, and all parts follow textbook procedures with no novel problem-solving required. Slightly below average difficulty due to the straightforward polynomial pdf and guided structure. |
| Spec | 2.01c Sampling techniques: simple random, opportunity, etc5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Curve with positive gradient, through the origin, first quadrant only | G1 | Curve with positive gradient, through the origin and in the first quadrant only. |
| Correct shape for an inverted parabola ending at maximum point | G1 | Correct shape for an inverted parabola ending at maximum point. |
| End point \((2, \frac{3}{4})\) labelled | G1 | End point \((2, 3/4)\) labelled. |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X) = \dfrac{3}{16}\int_0^2(4x^2 - x^3)\,dx\) | M1 | Correct integral for \(E(X)\) with limits (which may appear later). |
| \(= \dfrac{3}{16}\left[\dfrac{4x^3}{3} - \dfrac{x^4}{4}\right]_0^2\) | M1 | Correctly integrated. Dep on previous M1. |
| \(= \dfrac{3}{16}\left\{\left(\dfrac{32}{3} - \dfrac{16}{4}\right) - 0\right\}\) | ||
| \(= \dfrac{5}{4}\) | A1 | Limits used correctly to obtain printed answer (BEWARE) convincingly. Condone absence of "\(-0\)". |
| \(E(X^2) = \dfrac{3}{16}\int_0^2(4x^3 - x^4)\,dx\) | M1 | Correct integral for \(E(X^2)\) with limits (which may appear later). |
| \(= \dfrac{3}{16}\left[x^4 - \dfrac{x^5}{5}\right]_0^2\) | M1 | Correctly integrated. Dep on previous M1. |
| \(= \dfrac{3}{16}\left\{\left(16 - \dfrac{32}{5}\right) - 0\right\}\) | ||
| \(= \dfrac{9}{5}\) | A1 | Limits used correctly to obtain result. Condone absence of "\(-0\)". |
| \(\text{Var}(X) = \dfrac{9}{5} - \left(\dfrac{5}{4}\right)^2 = \dfrac{19}{80}\) | M1 | Use of \(\text{Var}(X) = E(X^2) - E(X)^2\). |
| \(\text{sd} = \sqrt{\dfrac{19}{80}} = 0.487(3)\) | A1 | cao |
| [8] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(SE(\bar{X}) = \dfrac{0.487}{\sqrt{100}} = 0.0487\) | M1, A1 | ft c's \(\sigma/10\). |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X < 1) = \dfrac{3}{16}\int_0^1(4x - x^2)\,dx\) | M1 | Correct integral for \(P(X < 1)\) with limits (which may appear later). |
| \(= \dfrac{3}{16}\left[2x^2 - \dfrac{x^3}{3}\right]_0^1\) | ||
| \(= \dfrac{3}{16}\left\{\left(2 - \dfrac{1}{3}\right) - 0\right\}\) | ||
| \(= \dfrac{5}{16}\) | A1 | cao. Condone absence of "\(-0\)" when limits applied. |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Regard the reed beds as clusters. | E1 | NB "Clusters of reeds" scores 0 unless clearly and correctly explained. |
| Select a few clusters (maybe only one) at random. | E1 | |
| Take a (simple random) sample of reeds (or maybe all of them) from the selected cluster(s). | E1 | |
| [3] |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Curve with positive gradient, through the origin, first quadrant only | G1 | Curve with positive gradient, through the origin and in the first quadrant only. |
| Correct shape for an inverted parabola ending at maximum point | G1 | Correct shape for an inverted parabola ending at maximum point. |
| End point $(2, \frac{3}{4})$ labelled | G1 | End point $(2, 3/4)$ labelled. |
| **[3]** | | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X) = \dfrac{3}{16}\int_0^2(4x^2 - x^3)\,dx$ | M1 | Correct integral for $E(X)$ with limits (which may appear later). |
| $= \dfrac{3}{16}\left[\dfrac{4x^3}{3} - \dfrac{x^4}{4}\right]_0^2$ | M1 | Correctly integrated. Dep on previous M1. |
| $= \dfrac{3}{16}\left\{\left(\dfrac{32}{3} - \dfrac{16}{4}\right) - 0\right\}$ | | |
| $= \dfrac{5}{4}$ | A1 | Limits used correctly to obtain printed answer (BEWARE) convincingly. Condone absence of "$-0$". |
| $E(X^2) = \dfrac{3}{16}\int_0^2(4x^3 - x^4)\,dx$ | M1 | Correct integral for $E(X^2)$ with limits (which may appear later). |
| $= \dfrac{3}{16}\left[x^4 - \dfrac{x^5}{5}\right]_0^2$ | M1 | Correctly integrated. Dep on previous M1. |
| $= \dfrac{3}{16}\left\{\left(16 - \dfrac{32}{5}\right) - 0\right\}$ | | |
| $= \dfrac{9}{5}$ | A1 | Limits used correctly to obtain result. Condone absence of "$-0$". |
| $\text{Var}(X) = \dfrac{9}{5} - \left(\dfrac{5}{4}\right)^2 = \dfrac{19}{80}$ | M1 | Use of $\text{Var}(X) = E(X^2) - E(X)^2$. |
| $\text{sd} = \sqrt{\dfrac{19}{80}} = 0.487(3)$ | A1 | cao |
| **[8]** | | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $SE(\bar{X}) = \dfrac{0.487}{\sqrt{100}} = 0.0487$ | M1, A1 | ft c's $\sigma/10$. |
| **[2]** | | |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X < 1) = \dfrac{3}{16}\int_0^1(4x - x^2)\,dx$ | M1 | Correct integral for $P(X < 1)$ with limits (which may appear later). |
| $= \dfrac{3}{16}\left[2x^2 - \dfrac{x^3}{3}\right]_0^1$ | | |
| $= \dfrac{3}{16}\left\{\left(2 - \dfrac{1}{3}\right) - 0\right\}$ | | |
| $= \dfrac{5}{16}$ | A1 | cao. Condone absence of "$-0$" when limits applied. |
| **[2]** | | |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Regard the reed beds as clusters. | E1 | NB "Clusters of reeds" scores 0 unless clearly and correctly explained. |
| Select a few clusters (maybe only one) at random. | E1 | |
| Take a (simple random) sample of reeds (or maybe all of them) from the selected cluster(s). | E1 | |
| **[3]** | | |
---2 A particular species of reed that grows up to 2 metres in length is used for thatching. The lengths in metres of the reeds when harvested are modelled by the random variable $X$ which has the following probability density function, $\mathrm { f } ( x )$.
$$f ( x ) = \begin{cases} \frac { 3 } { 16 } \left( 4 x - x ^ { 2 } \right) & \text { for } 0 \leqslant x \leqslant 2 \\ 0 & \text { elsewhere } \end{cases}$$
(i) Sketch $\mathrm { f } ( x )$.\\
(ii) Show that $\mathrm { E } ( X ) = \frac { 5 } { 4 }$ and find the standard deviation of the lengths of the harvested reeds.\\
(iii) Find the standard error of the mean length for a random sample of 100 reeds.
Once the harvested reeds have been collected, any that are shorter than 1 metre are discarded.\\
(iv) Find the proportion of reeds that should be discarded according to the model.\\
(v) Reeds are harvested from a large area which is divided into several reed beds. A sample of the harvested reeds is required for quality control. How might the method of cluster sampling be used to obtain it?
\hfill \mbox{\textit{OCR MEI S3 2013 Q2 [18]}}