# Question 4:

## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| Correct trapezoid shape starting and ending on $x$-axis | B1 | Correct shape. Must start and end on the $x$ axis |
| Labels: 1, 3, 6, 10 on $x$-axis and $k$ on $y$-axis | B1 | Fully correct including 1, 3, 6, 10 and $k$. Allow $\frac{1}{6}$ for $k$. Ignore labels for $x$ and $f(x)$ and any extras e.g. $\frac{k}{2}$ |

## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{2}(3+9) \times k = 1$ or $\frac{1}{2}(3-1)k + (6-3)k + \frac{1}{2}(10-6)k = 1$ or $\frac{1}{2}k\left[\frac{x^2}{2}-x\right]_1^3 + k[x]_3^6 + \frac{1}{4}k\left[10x - \frac{x^2}{2}\right]_6^{10} = 1$ | M1 | Setting up area of trapezium $= 1$ or 2 triangles $+$ a rectangle $= 1$ or a fully correct integration including limits $= 1$ |
| $k = \frac{1}{6}$* | A1* cso | Answer given. Correct solution only with no incorrect working |

## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int_1^x \frac{1}{12}(x-1)\,dx$ or $\int \frac{1}{12}(x-1)\,dx$ and using $F(1)=0$ | M1 | For a correct method to find the 2nd line. Allow in terms of $k$ |
| $\int_3^x \frac{1}{6}\,dx + \text{"F(3)"}$ or $\int \frac{1}{6}\,dx$ and using $\text{"F(3)"} = \frac{1}{6}$ | M1 | For a correct method to find the 3rd line, ft their F(3). If using $+c$ method then ft their $F(3) = \frac{1}{6}$. Allow in terms of $k$ |
| $\int_6^x \left(\frac{5}{12} - \frac{1}{24}x\right)dx + \text{"F(6)"}$ or $\int\left(\frac{5}{12}-\frac{1}{24}x\right)dx$ and using $\text{"F(6)"} = \frac{2}{3}$ or $F(10)=1$ | M1 | For a correct method to find the 4th line, ft their F(6). If using $+c$ method then ft their $F(6) = \frac{2}{3}$. Allow in terms of $k$ |
| $F(x) = \begin{cases} 0 & x < 1 \\ \frac{1}{24}(x^2-2x+1) & 1 \leqslant x \leqslant 3 \\ \frac{1}{6}(x-2) & 3 < x \leqslant 6 \\ \frac{1}{48}(20x-x^2-52) \text{ or } 1-\frac{(10-x)^2}{48} & 6 < x \leqslant 10 \\ 1 & x > 10 \end{cases}$ | A1oe, A1oe, A1oe, B1 | Each line correct including inequality (allow $<$ instead of $\leqslant$). B1: 1st and 5th lines correct. All 5 lines must be in terms of the same letter |

## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X > E(X)) = 1 - F\!\left(\frac{61}{12}\right) = 1 - 0.51388\ldots = 0.4861\ldots$ awrt 0.486 | M1 A1 | M1: Use of $1 - F\!\left(\frac{61}{12}\right)$ using their line of $F(x)$ for $3 < x \leqslant 6$. May use integration/area methods. A1: awrt 0.486, allow $\frac{35}{72}$ |

## Part (e)
| Answer | Mark | Guidance |
|--------|------|----------|
| Since (d) $< 0.5$ [the mean is greater than the median] therefore positive (skew), or follow through their sketch in part (a) | M1 A1ft | M1: For correctly comparing part (d) with 0.5 (may be implied by correct comparison of mean and median). Do not allow "mean is greater than the median" on its own. A1ft: For positive skew or ft their answer. Accept "no (or negligible) skew" following reason that "mean $\approx$ median". Allow argument based on sketch in part (a) |

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