Question

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.

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
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