Question

the following are the lengths of stay (in days) for a random sample of 22 patients discharged from a particular hospital: 13, 9, 5, 11, 6, 3, 12, 10, 11, 7, 3, 9, 10, 10, 12, 12, 8, 8, 4, 4, 11, 11 send data to calculator draw the histogram for these data using an initial class boundary of 2.5 and a class width of 2. note that you can add or remove classes from the figure. label each class with its endpoints.

Explanation:

Step1: Determine class intervals

The initial class - boundary is 2.5 and class width is 2. The first class interval is 2.5 - 4.5, the second is 4.5 - 6.5, the third is 6.5 - 8.5, the fourth is 8.5 - 10.5, the fifth is 10.5 - 12.5, the sixth is 12.5 - 14.5.

Step2: Count frequencies

For the class 2.5 - 4.5, the values 3, 3, 4, 4 fall in it, so the frequency is 4.
For the class 4.5 - 6.5, the values 5, 6 fall in it, so the frequency is 2.
For the class 6.5 - 8.5, the values 7, 8, 8 fall in it, so the frequency is 3.
For the class 8.5 - 10.5, the values 9, 9, 10, 10 fall in it, so the frequency is 4.
For the class 10.5 - 12.5, the values 11, 11, 11, 11, 12, 12, 12 fall in it, so the frequency is 7.
For the class 12.5 - 14.5, the value 13 falls in it, so the frequency is 1.

Step3: Draw histogram

On the x - axis, label the class intervals (2.5 - 4.5, 4.5 - 6.5, 6.5 - 8.5, 8.5 - 10.5, 10.5 - 12.5, 12.5 - 14.5). On the y - axis, label the frequencies from 0 to 8. Draw bars with heights corresponding to the frequencies for each class interval.

Answer:

The histogram has class intervals 2.5 - 4.5 with frequency 4, 4.5 - 6.5 with frequency 2, 6.5 - 8.5 with frequency 3, 8.5 - 10.5 with frequency 4, 10.5 - 12.5 with frequency 7, 12.5 - 14.5 with frequency 1.