the following are the lengths of stay (in days) for a random sample of 18 patients discharged from a…
the following are the lengths of stay (in days) for a random sample of 18 patients discharged from a particular hospital: 13, 9, 5, 11, 6, 5, 12, 10, 11, 7, 8, 9, 4, 4, 2, 3, 2. draw the histogram for these data using an initial class boundary of 1.5 and a class width of 2. note that you can add or remove classes. enter the midpoints for each class with its frequency.
Answer
Explanation:
Step1: Determine class intervals
Starting with lower - limit 1.5 and class - width 2, the class intervals are: 1.5 - 3.5, 3.5 - 5.5, 5.5 - 7.5, 7.5 - 9.5, 9.5 - 11.5, 11.5 - 13.5.
Step2: Tally the data
For the data set 13, 9, 5, 11, 6, 5, 12, 10, 11, 7, 8, 9, 4, 4, 2, 3, 2:
- In the interval 1.5 - 3.5, there are 3 values (2, 2, 3).
- In the interval 3.5 - 5.5, there are 4 values (4, 4, 5, 5).
- In the interval 5.5 - 7.5, there are 3 values (6, 7).
- In the interval 7.5 - 9.5, there are 3 values (8, 9, 9).
- In the interval 9.5 - 11.5, there are 3 values (10, 11, 11).
- In the interval 11.5 - 13.5, there are 2 values (12, 13).
Step3: Find mid - points
For the interval 1.5 - 3.5, mid - point $=\frac{1.5 + 3.5}{2}=2.5$. For the interval 3.5 - 5.5, mid - point $=\frac{3.5+5.5}{2}=4.5$. For the interval 5.5 - 7.5, mid - point $=\frac{5.5 + 7.5}{2}=6.5$. For the interval 7.5 - 9.5, mid - point $=\frac{7.5+9.5}{2}=8.5$. For the interval 9.5 - 11.5, mid - point $=\frac{9.5 + 11.5}{2}=10.5$. For the interval 11.5 - 13.5, mid - point $=\frac{11.5+13.5}{2}=12.5$.
Answer:
| Class Interval | Mid - point | Frequency |
|---|---|---|
| 1.5 - 3.5 | 2.5 | 3 |
| 3.5 - 5.5 | 4.5 | 4 |
| 5.5 - 7.5 | 6.5 | 3 |
| 7.5 - 9.5 | 8.5 | 3 |
| 9.5 - 11.5 | 10.5 | 3 |
| 11.5 - 13.5 | 12.5 | 2 |