# Question 8:

## Part (a)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Graph has maximum height of 2, labelled, and passes through (2,0) | B1 | Max height of 2 must be labelled |
| Shape must be between $x=2$ and $x=3$ with no other lines drawn (patios accepted) | B1 | Line must be between 2 and 3; can get mark even if patio cannot be seen |
| Correct straight line shape | B1 | Line must be straight and correct shape |
| Mode = 3 | B1 | Only accept 3 |

## Part (b)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_2^3 2x(x-2)\,dx = \left[\dfrac{2x^3}{3} - 2x^2\right]_2^3$ | M1 A1 | Attempt to find $\int xf(x)\,dx$; need to see $x^n \to x^{n+1}$, ignore limits; A1 for correct integration ignoring limits |

## Part (c)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $= 2\dfrac{2}{3}$ | A1 | Accept $2\dfrac{2}{3}$ or awrt 2.67 or $2.\dot{6}$ |

## Part (d)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_2^m 2(x-2)\,dx = 0.5$ | M1 | Using $\int f(x)\,dx = 0.5$ |
| $\left[x^2 - 4x\right]_2^m = 0.5$ leading to $m^2 - 4m + 4 = 0.5$ | A1 | $m^2 - 4m + 4 = 0.5$ oe |
| $m^2 - 4m + 3.5 = 0$ | M1 | Attempting to solve quadratic |
| $m = \dfrac{4 \pm \sqrt{2}}{2}$, so $m = 2.71$ | A1 | awrt 2.71 or $\dfrac{4+\sqrt{2}}{2}$ or $2+\dfrac{\sqrt{2}}{2}$ oe |

## Part (e)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Negative skew | B1 | First B1 for "negative" |
| mean $<$ median $<$ mode | B1dep | Second B1 for mean $<$ median $<$ mode; need all 3 or may explain using diagram |