# Question 6:

## Part (i) - Mean and Standard Deviation

**Mean:**

$\frac{(25\times147)+(45\times109)+(55\times182)+(70\times317)+(140\times175)}{930}$ | M1 | For midpoints (at least 3 correct); allow 25.005, 45.005 etc leading to answer 70.20

$\frac{151250}{80} = £70.19$ or $£70.2$ | A1 | CAO (exact answer 70.19355...); correct answers from calculator statistical functions gain full marks

**Standard Deviation:**

$\Sigma x^2f = (25^2\times147)+(45^2\times109)+(55^2\times182)+(70^2\times317)+(140^2\times175)$ $= 5846450$ | M1 | For attempt at $S_{xx}$; should include sum of at least 3 correct multiples $fx^2 - \Sigma x^2/n$; do not FT incorrect mean

$S_{xx} = 5846450 - \frac{65280^2}{930} = 1264215.161$ | A1 | exact answer 36.88949...; condone answer of £36.90

$s = \sqrt{\frac{1264215}{929}} = \sqrt{1360.83} = 36.89$ or $£36.9$ | [4] | Allow any answer between 36.87 and 36.90 without checking working

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## Part (ii)

$\frac{100}{120}\times175 = 145.83$ | M1* | For 175/120; or $\frac{20}{120}\times175 = 29.166$

$145.83/930 = 0.1568$ | *M1dep |

So 15.7% | A1 [3] | Accept 16% with working

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## Part (iii) - Histogram

| Price | Frequency | Group width | Frequency density |
|-------|-----------|-------------|-------------------|
| $10\leq x\leq40$ | 147 | 30 | 4.90 |
| $40<x\leq50$ | 109 | 10 | 10.90 |
| $50<x\leq60$ | 182 | 10 | 18.20 |
| $60<x\leq80$ | 317 | 20 | 15.85 |
| $80<x\leq200$ | 175 | 120 | 1.46 |

Frequency densities correct | M1 | For fds - at least 3 correct; accept any suitable unit for fd

Allow 15.9 and 1.5; condone 1.45 | A1 |

Linear scales on both axes and label on both axes (allow horizontal axis labelled $x$); vertical scale starting from zero | B1 | Linear scale and label on vertical axis IN RELATION to first M1 mark

Width of bars (within half a square); NO GAPS ALLOWED | B1 |

Height of bars | B1 [5] | FT of heights dep on at least 3 heights correct; all must agree with their fds

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## Part (iv)

Positive skewness | B1 [1] | Allow +ve

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## Part (v)

Area for men from 100 to 200 $= 100\times2 = 200$ | M1 | Or $\frac{100}{120}\times240$

$200/990 = 0.202$, so 20.2% | A1 | 20% with working

Cannot be certain as both figures are estimates | E1 [3] | Independent; allow 'grouped data so cannot be certain' or 'values are not exact so cannot be certain' oe

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## Part (vi)

Men's running shoes have a lower average price than women's (mean is only £68.83 compared to £70.19) or equivalent for women | E1 | FT their mean; do NOT condone lower central tendency or lower mean; allow 'on the whole' or similar in place of 'average'

Men's running shoes have more variation in price than women's (sd is £42.93 compared to £36.89) or equivalent for women | E1 [2] | FT their SD; allow 'more spread' or similar but not 'higher range' or 'higher variance' or 'less distributed'

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