| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2010 |
| Session | January |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | PDF to CDF derivation |
| Difficulty | Standard +0.3 This is a straightforward S3 question requiring standard integration to find the CDF (integrating a quadratic), sketching a CDF curve, and setting F(m)=0.5 to find the median equation. The hypothesis test for the median is a routine sign test application. All techniques are standard bookwork with no novel insight required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration5.07b Sign test: and Wilcoxon signed-rank |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F(x) = \int_2^x \frac{1}{72}(8t - t^2)\,dt\) | M1 | Correct integral with limits (which may be implied subsequently) |
| \(= \frac{1}{72}\left[4t^2 - \frac{t^3}{3}\right]_2^x\) | A1 | Correctly integrated |
| \(= \frac{1}{72}\left(4x^2 - \frac{x^3}{3} - 16 + \frac{8}{3}\right) = \frac{12x^2 - x^3 - 40}{216}\) | A1 | Limits used. Accept unsimplified form. [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct shape; nothing below \(y=0\); non-negative gradient | G1 | |
| Labels at \((2, 0)\) and \((8, 1)\) | G1 | |
| Curve (horizontal lines) shown for \(x < 2\) and \(x > 8\) | G1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F(m) = \frac{1}{2}\) \(\therefore \frac{12m^2 - m^3 - 40}{216} = \frac{1}{2}\) | M1 | Use of definition of median. Allow use of candidate's \(F(x)\) |
| \(\therefore 12m^2 - m^3 - 40 = 108\), \(\therefore m^3 - 12m^2 + 148 = 0\) | A1 | Convincingly rearranged. Beware: answer given. |
| Either: \(F(4.42) = 0.5003(977) \approx 0.5\) | ||
| Or: \(4.42^3 - 12 \times 4.42^2 + 148 = -0.0859(12) \approx 0\), \(\therefore m \approx 4.42\) | E1 | Convincingly shown, e.g. 4.418 or better seen. [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: m = 4.42\); \(H_1: m \neq 4.42\) where \(m\) is the population median | B1, B1 | Both. Accept hypotheses in words. Adequate definition of \(m\) to include "population" |
| Subtract 4.42 from each weight and rank \(\ | \)diff\(\ | \) |
| Ranks assigned correctly (1–12) | M1, A1 | For ranks; ft if ranks wrong |
| \(W_- = 2+3+4+6+7 = 22\) | B1 | \((W_+ = 1+5+8+9+10+11+12 = 56)\) |
| Refer to Wilcoxon single sample tables for \(n=12\) | M1 | No ft from here if wrong |
| Lower \(2\frac{1}{2}\%\) point is 13 (or upper is 65 if 56 used) | A1 | 2-tail test. No ft from here if wrong |
| Result is not significant | A1 | ft only candidate's test statistic |
| Evidence suggests median of 4.42 is consistent with these data | A1 | ft only candidate's test statistic [10] |
# Question 2:
## Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F(x) = \int_2^x \frac{1}{72}(8t - t^2)\,dt$ | M1 | Correct integral with limits (which may be implied subsequently) |
| $= \frac{1}{72}\left[4t^2 - \frac{t^3}{3}\right]_2^x$ | A1 | Correctly integrated |
| $= \frac{1}{72}\left(4x^2 - \frac{x^3}{3} - 16 + \frac{8}{3}\right) = \frac{12x^2 - x^3 - 40}{216}$ | A1 | Limits used. Accept unsimplified form. **[3]** |
## Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct shape; nothing below $y=0$; non-negative gradient | G1 | |
| Labels at $(2, 0)$ and $(8, 1)$ | G1 | |
| Curve (horizontal lines) shown for $x < 2$ and $x > 8$ | G1 | **[3]** |
## Part (a)(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F(m) = \frac{1}{2}$ $\therefore \frac{12m^2 - m^3 - 40}{216} = \frac{1}{2}$ | M1 | Use of definition of median. Allow use of candidate's $F(x)$ |
| $\therefore 12m^2 - m^3 - 40 = 108$, $\therefore m^3 - 12m^2 + 148 = 0$ | A1 | Convincingly rearranged. Beware: answer given. |
| Either: $F(4.42) = 0.5003(977) \approx 0.5$ | | |
| Or: $4.42^3 - 12 \times 4.42^2 + 148 = -0.0859(12) \approx 0$, $\therefore m \approx 4.42$ | E1 | Convincingly shown, e.g. 4.418 or better seen. **[3]** |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: m = 4.42$; $H_1: m \neq 4.42$ where $m$ is the population median | B1, B1 | Both. Accept hypotheses in words. Adequate definition of $m$ to include "population" |
| Subtract 4.42 from each weight and rank $\|$diff$\|$ | M1 | For subtracting 4.42 |
| Ranks assigned correctly (1–12) | M1, A1 | For ranks; ft if ranks wrong |
| $W_- = 2+3+4+6+7 = 22$ | B1 | $(W_+ = 1+5+8+9+10+11+12 = 56)$ |
| Refer to Wilcoxon single sample tables for $n=12$ | M1 | No ft from here if wrong |
| Lower $2\frac{1}{2}\%$ point is 13 (or upper is 65 if 56 used) | A1 | 2-tail test. No ft from here if wrong |
| Result is not significant | A1 | ft only candidate's test statistic |
| Evidence suggests median of 4.42 is consistent with these data | A1 | ft only candidate's test statistic **[10]** |
---2
\begin{enumerate}[label=(\alph*)]
\item A continuous random variable, $X$, has probability density function
$$f ( x ) = \begin{cases} \frac { 1 } { 72 } \left( 8 x - x ^ { 2 } \right) & 2 \leqslant x \leqslant 8 \\ 0 & \text { otherwise } \end{cases}$$
\begin{enumerate}[label=(\roman*)]
\item Find $\mathrm { F } ( x )$, the cumulative distribution function of $X$.
\item Sketch $\mathrm { F } ( x )$.
\item The median of $X$ is $m$. Show that $m$ satisfies the equation $m ^ { 3 } - 12 m ^ { 2 } + 148 = 0$. Verify that $m \approx 4.42$.
\end{enumerate}\item The random variable in part (a) is thought to model the weights, in kilograms, of lambs at birth. The birth weights, in kilograms, of a random sample of 12 lambs, given in ascending order, are as follows.
$$\begin{array} { l l l l l l l l l l l l }
3.16 & 3.62 & 3.80 & 3.90 & 4.02 & 4.72 & 5.14 & 6.36 & 6.50 & 6.58 & 6.68 & 6.78
\end{array}$$
Test at the 5\% level of significance whether a median of 4.42 is consistent with these data.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S3 2010 Q2 [19]}}