S1 June 2018 Q2
2. The following grouped frequency distribution summarises the number of minutes, to the nearest minute, that a random sample of 100 motorists were delayed by roadworks on a stretch of motorway one Monday.
| Delay (minutes) | Number of motorists (f) | Delay midpoint (\(x\)) |
|---|---|---|
| 3–6 | 38 | 4.5 |
| 7–8 | 25 | 7.5 |
| 9–10 | 18 | 9.5 |
| 11–15 | 12 | 13 |
| 16–20 | 7 | 18 |
(You may use \(\sum \mathrm{f}x^2 = 8096.25\))
A histogram has been drawn to represent these data.
The bar representing a delay of (3–6) minutes has a width of 2 cm and a height of 9.5 cm.
One coefficient of skewness is given by \(\dfrac{3(\text{mean} - \text{median})}{\text{standard deviation}}\)
On the following Friday, the coefficient of skewness for the delays on this stretch of motorway was – 0.22
| Scheme | Marks |
|---|---|
| (3 – 6 ) mins has width 4 and is 2cm, (11 – 15) mins has width 5 so is 2.5(cm) | B1 |
| (3 – 6) mins has frequency of 38 and area of 19 cm\(^2\) so 2 people(per cm\(^2\))(o.e.) or frequency density \(= \dfrac{38}{4} = 9.5 =\) height | M1 |
| (11 – 15) mins has area of \(2.5 \times h\) cm\(^2\) so \(h = \dfrac{12}{2 \times 2.5} = \)2.4 (cm) allow \(\frac{12}{5}\) | A1 |
| (3) |
Notes
B1 for width of 2.5 (cm) allow \(\frac{5}{2}\)
M1 for 2 people per cm\(^2\) or a correct numerical equ’n for \(h\) or their width\(\times\)height = 6
A1 for height of 2.4 (cm) [If just see 2.4 and 2.5 it must be clear which is \(h\) and which \(w\)]
| Scheme | Marks |
|---|---|
| \(Q_2 = (6.5) + \dfrac{12}{25} \times 2\) or \((8.5) - \dfrac{13}{25} \times 2\) | M1 |
| = awrt 7.46 | A1 |
| (2) |
Notes
M1 for a correct expr’n with sign (ignoring end point). Condone 12.5 for use of (\(n\) + 1)
A1 for awrt 7.46 (or 7.5 if using (\(n\) + 1) but must see evidence of (\(n\) + 1) approach)
| Scheme | Marks |
|---|---|
| \(\sum \mathrm{f}x = 38 \times 4.5 + \ldots + 7 \times 18 = 811.5\) and \(\bar{x} = \dfrac{811.5}{100}\), = awrt 8.12 | M1, A1 |
| (2) |
Notes
M1 for an attempt at \(\Sigma \mathrm{f}x\) ( i.e. full expression or \(650 \lt \Sigma \mathrm{f}x \lt 950\)) and division by 100
\(\Sigma \mathrm{f}x\) may be in the table.
A1 for 8.115 or awrt 8.12 (allow 8.11) [May be in (d) but must be labelled e.g. \(\bar{x} = \ldots\)]
| Scheme | Marks |
|---|---|
| \(\sigma = \sqrt{\dfrac{8096.25}{100} - \bar{x}^2} = \sqrt{80.9625 - \text{"}65.85\ldots\text{"}} = \sqrt{15.1(0)\ldots}\), = awrt 3.89 | M1, A1 |
| (2) |
Notes
M1 for a correct expression (ft their mean) including \(\sqrt{\ }\). Allow \(s\) leading to \(\sqrt{15.26\ldots}\)
A1 for awrt 3.89 Allow use of \(s\) = awrt 3.91 [Correct ans. only to (c) or (d) full marks]
| Scheme | Marks |
|---|---|
| \(\text{Skewness} = \dfrac{3(\text{"}8.12\text{"} - \text{"}7.46\text{"})}{\text{"}3.89\text{"}} = 0.5055\ldots =\) awrt 0.47 ~ 0.51 | B1 |
| (1) |
Notes
B1 for a correct expression seen using their values ( \(\sigma\) must be > 0) or awrt 0.47 ~ 0.51
| Scheme | Marks |
|---|---|
| Skewness for Monday and Friday are different (o.e.) | B1 |
| Suggests more longer delays on Friday (o.e.) [look for diagrams to support this.] | B1 |
| (2) | |
| (12 marks) |
Notes
1st B1 for a comment that skewness is different (only commenting on“correlation” is B0)
If ans. to (e) > 0 allow B1 for e.g. “skewness on Fri is < 0”[“on Fri” may be implied]
2nd B1 for a comment about length of delay e.g. “more long ones (on Fri.)
or “longer delays on Fri.”