S1 June 2014 Q6
6. The times, in seconds, spent in a queue at a supermarket by 85 randomly selected customers, are summarised in the table below.
| Time (seconds) | Number of customers, \(f\) |
|---|---|
| 0 – 30 | 2 |
| 30 – 60 | 10 |
| 60 – 70 | 17 |
| 70 – 80 | 25 |
| 80 – 100 | 25 |
| 100 – 150 | 6 |
A histogram was drawn to represent these data. The 30 – 60 group was represented by a bar of width 1.5 cm and height 1 cm.
Given that \(x\) denotes the midpoint of each group in the table and
\[\sum fx = 6460 \qquad \sum fx^2 = 529\,400\]for the above data. (3)
One measure of skewness is given by
\[\text{coefficient of skewness} = \frac{3(\text{mean} - \text{median})}{\text{standard deviation}}\]| Scheme | Marks |
|---|---|
| 70 – 80 group - width 0.5 (cm) | B1 |
| 1.5 cm\(^2\) is 10 customers or 3.75cm\(^2\) is 25 customers or \(0.5c = 3.75\) or \(\dfrac{2.5}{\frac{1}{3}}\) | M1 |
| 70 – 80 group - height 7.5 (cm) | A1 |
| (3) |
Notes
B1 for 0.5
M1 for one of the given statements or any method where “their width” \(\times\) “their height” = 3.75. Correct height scores M1A1 independent of width so B0M1A1 is possible.
| Scheme | Marks |
|---|---|
| Median = (70) + \(\dfrac{13.5}{25}\times 10\) allow \((n + 1) = (70) + \dfrac{14}{25}\times 10\) | M1 |
| \(=\) 75.4 ( or if using \((n + 1)\) allow 75.6) | A1 |
| (2) |
Notes
M1 for a correct fraction: \(+\dfrac{k}{25}\times 10\) where \(k = 13.5\) or 14 for \((n + 1)\) case.
NB may work down so look out for (80) \(-\dfrac{11.5}{25}\times 10\) etc Beware: \(69.5 + \dfrac{13.5}{25}\times 11 = 75.44\) (but M0)
| Scheme | Marks |
|---|---|
| \(\left[\text{Mean} = \dfrac{6460}{85}\right] =\) 76 | B1 |
| \(\sigma = \sqrt{\dfrac{529400}{85} - 76^2}\) | M1 |
| \(= 21.2658\ldots\) (\(s = 21.3920\)) awrt 21.3 | A1 |
| (3) |
Notes
M1 for a correct expression with square root, ft their mean
A1 for awrt 21.3 or, if clearly using \(s\) allow awrt 21.4. Must be evaluated...no surds.
| Scheme | Marks |
|---|---|
| Coeff’ of skewness \(= \dfrac{3(76 - 75.4)}{21.2658\ldots} = 0.08464\ldots\) awrt 0.08 (awrt 0.06 for 75.6) | M1 A1 |
| There is (very slight) positive skew or the data is almost symmetrical (or both) Any mention of “correlation” is B0 | B1ft |
| (3) | |
| (11 marks) |
Notes
M1 sub. their values into formula allow use of \(s\) but their \(\sigma\) or \(s\) must be > 0
A1 for awrt 0.08 but accept 0.085 No fraction
B1ft for a correct comment compatible with their coefficient. Allow “symmetrical” for |coeff’| < 0.25. They may say it is “slightly skew” so omit “positive” but do not allow “negative” if coef’ +ve. Condone “strongly” positive skew.