Solutions HWB04

2.2:

For the ungrouped data as given in the table on page 25, the frequency polygon is the broken line joining the points

(1,0), (2,1), (3,1), (4,2), (5,2), (6,1), (7,0), (8,0), (9,0), (10,1), (11,0).

For the grouped data as given in the table on page 29, the frequency polygon is the broken line joining the points

(-1.5,0), (1,1), (3.5,3), (5.5,3), (9,1), (14,0).

2.12:

This phenomena is explained in the article handed out with the material of
the first class.

If you ask each student the size of his/her classes, the perceived
average size is larger than the actual average. Suppose there were just
two classes, one of size 2 and one of size 12. The average class size is
(2 + 12)/2 = 7.

Yet 12 of the students report their class size as 12 and two students
report their class sizes as 2. So if you average out the reported class
size of the 14 students, you get
{(12 + 12 + 12 + 12 + ... + 12) + (2 + 2)}/14 = {144 + 4}/14 = 10.57
So the actual mean is 7, the perceived mean is 10.57.

2.26: Various answers are possible depending on the grouping.

Suppose we group as follows

(each class includes left endpoint, but not right endpoint)

Class freq

[2,4) 2

[4,6) 1

[6,8) 8

[8,10) 3

[10,12) 2
Sum: 16.

Then the ogive is the broken line joining the points

(2,0), (4,2), (6,3), (8,11), (10,14), (12,16).

Converting the vertical axis to percentages (where 16 = 100%),

The ogive is the broken line joining the points

(2,0%), (4,12.5%), (6,18.75%), (8,68.75%), (10,87.5%), (12,100%)

So the 50^th percentile is 7.25, because the line joining (6,18.75%) and (6,68.75%) goes thru the point (7.25,50%).

And the 90^th percentile is 11.6, because the line joining (10,87.75%) and (12,100%) goes thru the point (11.6,90%).