The following are the lengths of stay (in days) for a random sample of 19 patients discharged from a particular hospital:13,9, 5, 11, 6, 3, 12, 10, 11, 7, 3, 9, 5, 2, 2, 10, 10, 10, 12Send data to calculatorDraw the histogram for these data using an initial class boundary of 1.5, an ending class boundary of 13.5, and 6 classes of equal width. Note that you can addor remove classes from the figure. Label each class with its endpoints.Frequency0:00 11:07+6+5+5?L-000anginaDLength of stay in days)ExplikationCheck

The following are the lengths of stay in days for a random sample of 19 patients discharged from a particular hospital139 5 11 6 3 12 10 11 7 3 9 5 2 2 10 10 10 class=

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initial class boundary of 1.5

ending class boundary of 13.5

Number of classes: 6

Class width: Subtract initial class boundary from ending class boundary and divide the result into the number of classes:

[tex]\frac{13.5-1.5}{6}=2[/tex]

Classes:

1.5 - 3.5

3.5 - 5.5

5.5 - 7.5

7.5 - 9.5

9.5 - 11.5

11.5 - 13.5

Order the given data to find easily the frequency of each class:

[tex]\begin{gathered} \\ \\ 2,2,3,3,5,5,6,7,9,9,10,10,10,10,11,11,12,12,13 \end{gathered}[/tex]

Then, you get frequencies of:

1.5 - 3.5: 4

3.5 - 5.5: 2

5.5 - 7.5: 2

7.5 - 9.5: 2

9.5 - 11.5: 6

11.5 - 13.5: 3

And you get the next histogram:

Ver imagen TanylahG670494
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