# Question 6:

## Part (i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Frequency density calculations: $500\leq x\leq 1000$: fd = 0.014; $1000 < x\leq 1500$: fd = 0.044; $1500 < x\leq 2000$: fd = 0.052; $2000 < x\leq 3000$: fd = 0.018; $3000 < x\leq 5000$: fd = 0.0035 | M1 | At least 4 fds correct. M1 can also be gained from freq per 1000: 14, 44, 52, 18, 3.5 (at least 4 correct) |
| All fds correct | A1 | Accept any suitable unit for fd, e.g. freq per 1000, BUT NOT FD per 1000. Allow fds correct to at least 3dp |
| Linear scales on both axes, label on vertical axis | G1(L1) | Label required on vert axis IN RELATION to first M1 mark. Accept f/w or f/cw. Ignore horizontal label |
| Width of bars drawn at 500, 1000 etc. NO GAPS ALLOWED. Linear scale. | G1(W1) | Must be drawn at 500, 1000 etc NOT 499.5 or 500.5 etc |
| Height of bars correct, dep on at least 3 heights correct | G1(H1) | Visual check only (within one square). If fds not given and one height wrong then max M1A0G1G1G0 |

**[5]**

## Part (ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Do not know exact highest and lowest values so cannot tell what the midrange is. OR counterexample to show it may not be 2750. OR $(500+5000)/2 = 2750$ but very unlikely to be absolutely correct, probably close to true value — element of doubt needed. | E1 | Allow comment such as 'Highest value could be 5000 and lowest could be 500 therefore midrange could be 2750'. NO mark if incorrect calculation. Sight of 1750 AND 3000 (min and max of midrange) scores E1 |

**[1]**

## Part (iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Mean} = \frac{(750\times7)+(1250\times22)+(1750\times26)+(2500\times18)+(4000\times7)}{80}$ | M1 | For midpoints (at least 3 correct). No marks for mean or sd unless using midpoints |
| $= \frac{151250}{80} = 1891$ | A1 | CAO. Accept correct answers for mean (1890 or 1891) and sd (850 or 846 or 845.5) from calculator |
| $\Sigma x^2 f = (750^2\times7)+(1250^2\times22)+(1750^2\times26)+(2500^2\times18)+(4000^2\times7) = 342437500$ | M1 | For sum of at least 3 correct multiples $fx^2$. Allow M1 for anything which rounds to 342400000 |
| $S_{xx} = 342437500 - \frac{151250^2}{80} = 56480469$ | | |
| $s = \sqrt{\frac{56480469}{79}} = \sqrt{714943} = 846$ | A1 | Allow SC1 for RMSD 840.2 or 840 from calculator |
| Only an estimate since the data are grouped. | E1 (indep) | Or any mention of midpoints or 'don't have actual data' or 'data are not exact' |

**[5]**

## Part (iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\bar{x} - 2s = 1891 - (2\times846) = 199$; Allow 200 | M1 | For either. FT any positive mean and their positive sd/rmsd. No marks unless using $\bar{x}+2s$ or $\bar{x}-2s$ |
| $\bar{x} + 2s = 1891 + (2\times846) = 3583$; Allow 3580 or 3600 | A1 | For both (FT). Do NOT penalise over specification here |
| So there are probably some outliers | E1 | Must include element of doubt. Dep on upper limit in range 3000–5000 |

**[3]**

## Part (v):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Number of cars over $2000\text{ cm}^3 = \frac{25}{80}\times 2.5\text{ million} = 781250$ | M1 | For $\frac{25}{80}\times 2.5$ million or $\frac{(18+7)}{80}\times 2.5$ million |
| Duty raised $= 781250\times £1000 = £781\text{ million}$ | M1 (indep) | For something $\times$ £1000 even if this is the first step |
| | A1 | CAO. NB £781250000 is over specified so only 2/3 |

**[3]**

## Part (vi):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Because the numbers of cars sold with engine size greater than $2000\text{ cm}^3$ might be reduced due to the additional duty. | E1 | Allow any other reasonable suggestion. Condone 'sample may not be representative'. Allow 'sample is not of NEW cars' |

**[1]**

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