| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Calculate y on x from raw data table |
| Difficulty | Moderate -0.3 This is a straightforward application of standard linear regression formulas with all summations provided. Students need to substitute into memorized formulas for regression lines and correlation coefficient—routine A-level Further Maths statistics with no conceptual challenges or problem-solving required, making it slightly easier than average. |
| Spec | 5.08a Pearson correlation: calculate pmcc5.09b Least squares regression: concepts5.09c Calculate regression line |
| Delegate | \(A\) | \(B\) | \(C\) | \(D\) | \(E\) | \(F\) | \(G\) | \(H\) |
| \(x\) | 90 | 46 | 72 | 98 | 52 | 65 | 105 | 82 |
| \(y\) | 90 | 55 | 69 | 85 | 45 | 50 | 110 | 74 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(b = (47136 - 610 \times 578/8)/(49682 - 610^2/8) = 3063 \cdot 5/3169 \cdot 5 = 0 \cdot 966[6]\) | B1 | Calculate gradient \(b\) |
| \(y = 578/8 + 0 \cdot 967(x - 610/8) = 72 \cdot 2[5] + 0 \cdot 967(x - 76 \cdot 2[5])\) or \(-1 \cdot 45 + 0 \cdot 967x\) | M1 A1 | Regression line of \(y\) on \(x\) (A.E.F.) |
| \(b' = (47136 - 610 \times 578/8)/(45212 - 578^2/8) = 3063 \cdot 5/3451 \cdot 5 = 0 \cdot 887[6]\) | B1 | Calculate gradient \(b'\) |
| \(x = 610/8 + 0 \cdot 888(y - 578/8) = 76 \cdot 2[5] + 0 \cdot 888(y - 72 \cdot 2[5])\) or \(12 \cdot 1 + 0 \cdot 888y\) | M1 A1 | Regression line of \(x\) on \(y\) (A.E.F.) |
| Use regression line \(x\) on \(y\) at \(y = 100\): \(x = 101\) [mins] | M1 A1 | S.R. Using \(y\) on \(x\) at \(y=100\): \(x = 105\) [mins] gives (B1) |
| \(r^2 = bb' = 0 \cdot 8580\), \(r = 0 \cdot 926\) | M1 A1 | *EITHER* method |
| \(r = (47136 - 610 \times 578/8)/\sqrt{\{(49682 - 610^2/8)(45212 - 578^2/8)\}} = 3063 \cdot 5/\sqrt{(3169 \cdot 5 \times 3451 \cdot 5)} = 0 \cdot 926\) | (M1 A1) | *OR* method |
# Question 10:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $b = (47136 - 610 \times 578/8)/(49682 - 610^2/8) = 3063 \cdot 5/3169 \cdot 5 = 0 \cdot 966[6]$ | B1 | Calculate gradient $b$ |
| $y = 578/8 + 0 \cdot 967(x - 610/8) = 72 \cdot 2[5] + 0 \cdot 967(x - 76 \cdot 2[5])$ or $-1 \cdot 45 + 0 \cdot 967x$ | M1 A1 | Regression line of $y$ on $x$ (A.E.F.) |
| $b' = (47136 - 610 \times 578/8)/(45212 - 578^2/8) = 3063 \cdot 5/3451 \cdot 5 = 0 \cdot 887[6]$ | B1 | Calculate gradient $b'$ |
| $x = 610/8 + 0 \cdot 888(y - 578/8) = 76 \cdot 2[5] + 0 \cdot 888(y - 72 \cdot 2[5])$ or $12 \cdot 1 + 0 \cdot 888y$ | M1 A1 | Regression line of $x$ on $y$ (A.E.F.) |
| Use regression line $x$ on $y$ at $y = 100$: $x = 101$ [mins] | M1 A1 | **S.R.** Using $y$ on $x$ at $y=100$: $x = 105$ [mins] gives (B1) |
| $r^2 = bb' = 0 \cdot 8580$, $r = 0 \cdot 926$ | M1 A1 | *EITHER* method |
| $r = (47136 - 610 \times 578/8)/\sqrt{\{(49682 - 610^2/8)(45212 - 578^2/8)\}} = 3063 \cdot 5/\sqrt{(3169 \cdot 5 \times 3451 \cdot 5)} = 0 \cdot 926$ | (M1 A1) | *OR* method |
---10 Delegates who travelled to a conference were asked to report the distance, $y \mathrm {~km}$, that they had travelled and the time taken, $x$ minutes. The values reported by a random sample of 8 delegates are given in the following table.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | }
\hline
Delegate & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ & $G$ & $H$ \\
\hline
$x$ & 90 & 46 & 72 & 98 & 52 & 65 & 105 & 82 \\
\hline
$y$ & 90 & 55 & 69 & 85 & 45 & 50 & 110 & 74 \\
\hline
\end{tabular}
\end{center}
$$\left[ \Sigma x = 610 , \Sigma x ^ { 2 } = 49682 , \Sigma y = 578 , \Sigma y ^ { 2 } = 45212 , \Sigma x y = 47136 . \right]$$
Find the equations of the regression lines of $y$ on $x$ and of $x$ on $y$.
Estimate the time taken by a delegate who travelled 100 km to the conference.
Calculate the product moment correlation coefficient for this sample.
\hfill \mbox{\textit{CAIE FP2 2012 Q10 [10]}}