| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | November |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Other continuous |
| Difficulty | Standard +0.8 This is a multi-part chi-squared goodness of fit question requiring: (1) calculating the theoretical mean from a pdf using integration, (2) calculating expected frequencies from a continuous pdf by integration, and (3) performing a chi-squared test. While the individual techniques are standard Further Maths content, the combination of calculus integration with a quadratic pdf and statistical testing makes this moderately challenging, though still within typical FM2/FP2 scope. |
| Spec | 5.02a Discrete probability distributions: general5.06b Fit prescribed distribution: chi-squared test |
| Interval | \(2 \leqslant x < 2.4\) | \(2.4 \leqslant x < 2.8\) | \(2.8 \leqslant x < 3.2\) | \(3.2 \leqslant x < 3.6\) | \(3.6 \leqslant x < 4\) |
| Observed frequency | 19 | 17 | 16 | 8 | 0 |
| Interval | \(2 \leqslant x < 2.4\) | \(2.4 \leqslant x < 2.8\) | \(2.8 \leqslant x < 3.2\) | \(3.2 \leqslant x < 3.6\) | \(3.6 \leqslant x < 4\) |
| Expected frequency | 15.456 | 16.032 | 14.304 | 10.272 | 3.936 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Stating or implying reactions \(R_P\), \(R_S\) same as for \(B\) | B1 | |
| Stating or implying \(F_P = F_S\) by moments about \(O_A\) | B1 | |
| Stating or implying 3 independent equations for \(F\), \(R_P\), \(R_S\) | \(3 \times\) M1 A1 | |
| \(2R_P = 3W\) | Up to 2 resolutions of forces | |
| \(R_P = W + R_S \cos\theta + F_S \sin\theta\) | \(\uparrow\) for \(A\) | |
| \(2R_S \cos\theta + 2F_S \sin\theta = W\) | \(\uparrow\) for \(C\) | |
| \(R_P \cos\theta + F_P \sin\theta = R_S + W\cos\theta\) | \(O_A \rightarrow O_C\) for \(A\) | |
| Moments about \(S\) for \(A\): \(F_P(r + r\cos\theta) + Wr\sin\theta = R_P r\sin\theta\) | ||
| \(R_P = \frac{3W}{2}\) | A1 | |
| \(R_S = \frac{W}{2}\) | A1 | |
| \(F = \frac{(W\sin\theta)}{2(1+\cos\theta)}\) | A1 | Find \(F\) at \(P\) and/or \(S\) |
| \(\mu \geq \frac{\sin\theta}{3(1+\cos\theta)}\) | A.G. M1 A1 | Use \(F \leq \mu R_P\) to find bound for \(\mu\) |
| \(\mu' \geq \frac{\sin\theta}{(1+\cos\theta)}\) | A.G. M1 | Use \(F \leq \mu' R_S\) to find bound for \(\mu'\) |
| 14 | [14] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(X) = \int_2^4 (5x^2 - x^3 - 4x)/10 \, dx\) | M1 A1 | Find \(E(X)\) using \(\int xf(x)\,dx\) |
| \(= \frac{1}{2}(4^3 - 2^3) - 3(4^4 - 2^4)/40 - 3(4^2 - 2^2)/5\) | ||
| \(= 28 - 18 - 7.2 = 2.8\) | *A1 | |
| \((E(X) - 2.69)/2.69 = 0.041 < 0.1\) *or* \(1.1 \times 2.69 = 2.96 > E(X)\) | M1 A1 | Verify \(E(X)\) within 10% of 2.69; A1 dep *A1 |
| 5 | ||
| \(60\int_{2}^{3}(5x - x^2 - 4)/10\,dx\) | M1 | Show derivation of tabular entry |
| \(= 60[3(5x^2/2 - x^3/3 - 4x)/10]_2^3\) *or* \([45x^2 - 6x^3 - 72x]_2^3\) | ||
| \(= 122.4 - 83.328 - 28.8\) *or* \(60 \times 0.1712\) | ||
| \(= 10.272\) | A.G. A1 | |
| 2 | ||
| \(H_0\): \(f(x)\) fits data (A.E.F.) | B1 | State null hypothesis |
| \(O\): … 8 | Combine last 2 cells since exp. value \(< 5\) | |
| \(E\): … 14.208 | B1 | |
| \(\chi^2 = 0.8126 + 0.0584 + 0.2011 + 2.7135 = 3.78[47]\) | M1 *A1 | Calculate \(\chi^2\) to 2 d.p. |
| \(\chi^2_{3,\,0.9} = 6.25[1]\) | ||
| \(\chi^2_{4,\,0.9} = 7.78\) | *B1 | [if no cells combined] |
| Accept \(H_0\) if \(\chi^2 <\) tabular value | M1 | Valid method for conclusion |
| \(3.78 < 6.25\) so \(f(x)\) does fit | A1 | Conclusion (A.E.F., dep *A1, *B1) |
| 7 | [14] |
# Question 10:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Stating or implying reactions $R_P$, $R_S$ same as for $B$ | B1 | |
| Stating or implying $F_P = F_S$ by moments about $O_A$ | B1 | |
| Stating or implying 3 independent equations for $F$, $R_P$, $R_S$ | $3 \times$ M1 A1 | |
| $2R_P = 3W$ | | Up to 2 resolutions of forces |
| $R_P = W + R_S \cos\theta + F_S \sin\theta$ | | $\uparrow$ for $A$ |
| $2R_S \cos\theta + 2F_S \sin\theta = W$ | | $\uparrow$ for $C$ |
| $R_P \cos\theta + F_P \sin\theta = R_S + W\cos\theta$ | | $O_A \rightarrow O_C$ for $A$ |
| Moments about $S$ for $A$: $F_P(r + r\cos\theta) + Wr\sin\theta = R_P r\sin\theta$ | | |
| $R_P = \frac{3W}{2}$ | A1 | |
| $R_S = \frac{W}{2}$ | A1 | |
| $F = \frac{(W\sin\theta)}{2(1+\cos\theta)}$ | A1 | Find $F$ at $P$ and/or $S$ |
| $\mu \geq \frac{\sin\theta}{3(1+\cos\theta)}$ | A.G. M1 A1 | Use $F \leq \mu R_P$ to find bound for $\mu$ |
| $\mu' \geq \frac{\sin\theta}{(1+\cos\theta)}$ | A.G. M1 | Use $F \leq \mu' R_S$ to find bound for $\mu'$ |
| | **14** | **[14]** |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X) = \int_2^4 (5x^2 - x^3 - 4x)/10 \, dx$ | M1 A1 | Find $E(X)$ using $\int xf(x)\,dx$ |
| $= \frac{1}{2}(4^3 - 2^3) - 3(4^4 - 2^4)/40 - 3(4^2 - 2^2)/5$ | | |
| $= 28 - 18 - 7.2 = 2.8$ | *A1 | |
| $(E(X) - 2.69)/2.69 = 0.041 < 0.1$ *or* $1.1 \times 2.69 = 2.96 > E(X)$ | M1 A1 | Verify $E(X)$ within 10% of 2.69; A1 dep *A1 |
| | **5** | |
| $60\int_{2}^{3}(5x - x^2 - 4)/10\,dx$ | M1 | Show derivation of tabular entry |
| $= 60[3(5x^2/2 - x^3/3 - 4x)/10]_2^3$ *or* $[45x^2 - 6x^3 - 72x]_2^3$ | | |
| $= 122.4 - 83.328 - 28.8$ *or* $60 \times 0.1712$ | | |
| $= 10.272$ | A.G. A1 | |
| | **2** | |
| $H_0$: $f(x)$ fits data (A.E.F.) | B1 | State null hypothesis |
| $O$: … 8 | | Combine last 2 cells since exp. value $< 5$ |
| $E$: … 14.208 | B1 | |
| $\chi^2 = 0.8126 + 0.0584 + 0.2011 + 2.7135 = 3.78[47]$ | M1 *A1 | Calculate $\chi^2$ to 2 d.p. |
| $\chi^2_{3,\,0.9} = 6.25[1]$ | | |
| $\chi^2_{4,\,0.9} = 7.78$ | *B1 | [if no cells combined] |
| Accept $H_0$ if $\chi^2 <$ tabular value | M1 | Valid method for conclusion |
| $3.78 < 6.25$ so $f(x)$ does fit | A1 | Conclusion (A.E.F., dep *A1, *B1) |
| | **7** | **[14]** |A continuous random variable $X$ is believed to have the probability density function f given by
$$f ( x ) = \begin{cases} \frac { 3 } { 10 } \left( 5 x - x ^ { 2 } - 4 \right) & 2 \leqslant x < 4 \\ 0 & \text { otherwise } \end{cases}$$
A random sample of 60 observations was taken and these values are summarised in the following grouped frequency table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
Interval & $2 \leqslant x < 2.4$ & $2.4 \leqslant x < 2.8$ & $2.8 \leqslant x < 3.2$ & $3.2 \leqslant x < 3.6$ & $3.6 \leqslant x < 4$ \\
\hline
Observed frequency & 19 & 17 & 16 & 8 & 0 \\
\hline
\end{tabular}
\end{center}
The estimated mean, based on the grouped data in the table above, is 2.69 , correct to 2 decimal places. It is decided that a goodness of fit test will only be conducted if the mean predicted from the probability density function is within $10 \%$ of the estimated mean. Show that this condition is satisfied.
The relevant expected frequencies are as follows.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
Interval & $2 \leqslant x < 2.4$ & $2.4 \leqslant x < 2.8$ & $2.8 \leqslant x < 3.2$ & $3.2 \leqslant x < 3.6$ & $3.6 \leqslant x < 4$ \\
\hline
Expected frequency & 15.456 & 16.032 & 14.304 & 10.272 & 3.936 \\
\hline
\end{tabular}
\end{center}
Show how the expected frequency for the interval $3.2 \leqslant x < 3.6$ is obtained.
Carry out the goodness of fit test at the 10\% significance level.
\hfill \mbox{\textit{CAIE FP2 2012 Q10 OR}}