The following back-to-back stem-and-leaf diagram shows the times to load an application on 61 smartphones of type A and 43 smartphones of type B.
| Type A |
|
|
|
Type B |
| (7) | 9 7 6 6 4 3 3 |
2 | 1 3 5 8 | (4) |
| (7) | 5 5 4 4 2 2 2 |
3 | 0 4 4 5 6 6 6 7 8 9 | (12) |
| (13) | 9 9 8 8 8 7 6 4 3 2 2 0 |
4 | 0 1 1 2 3 6 8 8 9 9 | (10) |
| (9) | 6 5 5 4 3 2 1 1 0 |
5 | 2 5 6 6 9 | (5) |
| (4) | 9 7 3 0 |
6 | 1 3 8 9 | (4) |
| (6) | 8 7 4 4 1 0 |
7 | 5 7 | (2) |
| (10) | 7 6 6 6 5 3 2 1 0 |
8 | 1 2 4 4 | (4) |
| (5) | 8 6 5 5 5 |
9 | 0 6 | (2) |
Key: 3 | 2 | 1 means 0.23 seconds for type A and 0.21 seconds for type B.
- Find the median and quartiles for smartphones of type A.
- Represent the data by drawing a pair of box-and-whisker plots in a single diagram on graph paper.
- Compare the loading times for these two types of smartphone.
Solution
(i) To find the median and quartiles for type A:
- Arrange the data in ascending order.
- Median position = \(\frac{61 + 1}{2} = 31\), so the median is the 31st value, which is 0.52 seconds.
- Lower quartile (Q1) position = \(\frac{61 + 1}{4} = 15.5\), so Q1 is the average of the 15th and 16th values, which is 0.41 seconds.
- Upper quartile (Q3) position = \(\frac{3(61 + 1)}{4} = 46.5\), so Q3 is the average of the 46th and 47th values, which is 0.79 seconds.
(ii) Draw box-and-whisker plots for both types A and B using the given quartiles and medians.
(iii) Compare the loading times:
- Smartphone B is quicker as its median is lower.
- Smartphone B is slightly less variable as its interquartile range is smaller.
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