S1 June 2005 Q2
2. The following table summarises the distances, to the nearest km, that 134 examiners travelled to attend a meeting in London.
| Distance (km) | Number of examiners |
|---|---|
| 41–45 | 4 |
| 46–50 | 19 |
| 51–60 | 53 |
| 61–70 | 37 |
| 71–90 | 15 |
| 91–150 | 6 |
(a) Give a reason to justify the use of a histogram to represent these data. (1)
(b) Calculate the frequency densities needed to draw a histogram for these data.
(DO NOT DRAW THE HISTOGRAM) (2)
(DO NOT DRAW THE HISTOGRAM) (2)
(c) Use interpolation to estimate the median \(Q_2\), the lower quartile \(Q_1\), and the upper quartile \(Q_3\) of these data. (4)
The mid-point of each class is represented by \(x\) and the corresponding frequency by \(f\). Calculations then give the following values
\[\Sigma fx = 8379.5 \quad \text{and} \quad \Sigma fx^2 = 557489.75\](d) Calculate an estimate of the mean and an estimate of the standard deviation for these data. (4)
One coefficient of skewness is given by
\[\frac{Q_3 - 2Q_2 + Q_1}{Q_3 - Q_1}.\](e) Evaluate this coefficient and comment on the skewness of these data. (4)
(f) Give another justification of your comment in part (e). (1)
| Scheme | Marks |
|---|---|
| Distance is a continuous. | B1 |
| (1) |
Notes
B1 continuous
| Scheme | Marks |
|---|---|
| F.D = freq/class width \(\Rightarrow\) 0.8, 3.8, 5.3, 3.7, 0.75, 0.1 | M1 A1 |
| (2) |
Notes
or the same multiple of
| Scheme | Marks |
|---|---|
| \(Q_2 = 50.5 + \dfrac{(67 - 23)}{53} \times 10 = 58.8\) | M1 A1 |
| \(Q_1 = 52.48;\quad Q_3 = 67.12\) | A1 A1 |
| (4) |
Notes
M1 A1 awrt 58.8/58.9
A1 A1 awrt 52.5/52.6 67.1/67.3
Special case : no working B1 B1 B1 ( \(\equiv\) A’s on the epen)
| Scheme | Marks |
|---|---|
| \(\bar{x} = \dfrac{8379.5}{134} = 62.5335\ldots\) | B1 |
| \(s = \sqrt{\dfrac{557489.75}{134} - \left(\dfrac{8379.5}{134}\right)^2}\) | M1 A1ft |
| \(s = 15.8089\ldots\) ( \(S_{n-1} = 15.86825\ldots\)) | A1 |
| (4) |
Notes
B1 awrt 62.5
A1 awrt 15.8 (15.9)
Special case : answer only B1 B1 ( \(\equiv\) A’s on the epen)
| Scheme | Marks |
|---|---|
| \(\dfrac{Q_3 - 2Q_2 + Q_1}{Q_3 - Q_1} = \dfrac{67.12 - 2 \times 58.8 + 52.48}{67.12 - 52.48}\) | M1 A1ft |
| \(= 0.1366 \Rightarrow\); +ve skew | A1; B1 |
| (4) |
Notes
M1 A1ft subst their \(Q_1\), \(Q_2\) & \(Q_3\); need to show working for A1ft and have reasonable values for quartiles
A1 awrt 0.14
| Scheme | Marks |
|---|---|
| For +ve skew Mean > Median & 62.53 > 58.80 or \(Q_3 - Q_2\ (8.32) \gt Q_2 - Q_1\ (6.32)\) Therefore +ve skew | B1 |
| (1) | |
| (16 marks) |