Below are the times (in days) it takes for a sample of 17 customers from Andrew's computer store to pay their invoices. 19, 15, 43, 39, 35, 31, 27, 34, 34, 30, 30, 26, 26, 26, 21, 21, 17
Draw the histogram for these data using an initial class boundary of 14.5, an ending class boundary of 49.5, and 5 classes of equal width. Note that you can add
or remove classes from the figure. Label each class with its endpoints.
Frequency
:lolo:0
x
?
7-
6+
5+
4+
3+
2+
0
0
0
Time (in days)

Step-by-step explanation:

Hello!

The variable of interest is X: time it takes a customer from Andrew's computer store to pay his invoices.

You have the information of a sample of n= 17 customers

19, 15, 43, 39, 35, 31, 27, 34, 34, 30, 30, 26, 26, 26, 21, 21, 17

To determine the class width of the intervals for the divide the difference between the ending and initial class boundaries by the number of intervals that you want to determine:

Class width: (49.5-14.5)/5= 7

Then, starting from the initial class boundary, you have to add the class width to determine the next boundary, and so on until the ending class boundary:

Initial class boundary: 14.5

14.5 + 5.6= 20.1

1st interval: [14.5; 21.5]

and so on:

[21.5; 28.5]

[28.5; 35.5]

[35.5; 42.5]

[42.5; 49.5]

Once you determined all class intervals, you have to order the values of the data set from least to greatest and then count how many observations correspond to each interval and arrange it in a frequency table.

15, 17, 19, 21, 21, 26, 26, 26, 27, 30, 30, 31, 34, 34, 35, 39, 43

[14.5; 21.5] ⇒ 5

[21.5; 28.5] ⇒ 4

[28.5; 35.5] ⇒ 6

[35.5; 42.5] ⇒ 1

[42.5; 49.5] ⇒ 1

Once you have the data set organized in the table, you can proceed to draw the histogram.

(See attachment)

I hope this helps!

answer: d. $20

step-by-step explanation:

just took the test

1. 3. can also be expressed as (22/7)

2. -4.5

3. 3.6

happy to

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