# Question 6:

## Part (a)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\times4.2, 4\times4, 4\times3.5, 10\times1$ $(= 8.4+16+14+10 = 48.4)$ | M1 | Attempt to find number between 10 and 30; 2 correct products or 48 or 48.4 seen |
| $\left[\frac{48.4}{90}\right] = \frac{121}{225} = 0.53777...$ **0.53~0.54** (2sf OK) | A1 | For 2sf answer in $[0.53 \sim 0.54]$; NB use of 48 gives 0.5333... |

## Part (b)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Data is not symmetric or is skew (normal is symmetric); "Even" distribution or diagram alone not sufficient | B1 | Comment suggesting not suitable based on lack of symmetry or "not bell shaped" |

## Part (c)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int xe^{-x}\,dx = \int x\,d(-e^{-x})$ | M1 | Attempting integration by parts in right direction; must have $u=x$ and $v=\pm e^{-x}$ |
| $= \left[-xe^{-x}\right] - \int(-e^{-x})\,dx \quad (+c)$ | A1 | Correct first step and expression for second integral |
| $\int_0^n xe^{-x}\,dx = \left[-xe^{-x}-e^{-x}\right]_0^n = \left(-ne^{-n}-e^{-n}\right)-\left[-(0)-1\right]$ | dM1 | Dep on 1st M1; all integration attempted and some use of at least one limit |
| $= 1-(n+1)e^{-n}$ (*) | A1cso* | cso with no incorrect working; minimum is correct integration and use of limits |

## Part (d)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Require area $= 90$, i.e. $k\int_0^n xe^{-x}\,dx = 90$ (ignore limits) | M1 | Realising need area under curve (implied by integral) $= 90$ |
| Using result in (c) with $n=4$: $k\left[1-5e^{-4}\right]=90$ | M1 | Use of (c) with $n=4$ and set $=90$; may be implied by 99.07... or better |
| $(k=)$ **99**(.0729...) (*) | A1cso* | For $k=99$ or awrt 99.1; allow $k=99$ with area $=$ awrt 89.9 as conclusion for 3/3 |

## Part (e)(i)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $[P(10 < T < 30) =]\ 0.64863...$ awrt **0.649** | B1 | For awrt 0.649 |

## Part (e)(ii)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $[\text{No. of patients}=]\ (99)\left[\left(1-4e^{-3}\right)-\left(1-2e^{-1}\right)\right]\ (=53.1..)$ | M1 | Use of (c) with $n=1$ and $n=3$; don't need the 99; implied by awrt 0.54 |
| $\text{Prob} = \dfrac{0.5366...\times99}{90} = 0.59027...$ awrt **0.590 or 0.591** | A1 | Allow 0.59 from correct working |

## Part (f)

| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. Patients might stay longer than 40 hours | B1 | Comment in context about upper limit for time ($t$ or $x$); time/hour may be implied |