2007 General Maths Paper Q24(iii) (1 Viewer)

[cxlxoxk]

Back to back stem and leaf of age of students at "Barry's Ballroom Dancing Studio":

9...[1]...1, 2 , 3

7...[2]...0, 2, 2, 2, 4, 5

5...[3]...0, 0, 1, 7

5, 2...[4]...6, 7

3, 2, 0...[5]...2

4, 4, 2, 1...[6]...4, 4

the [1-6] is the centre column.

Liam decides to use a grouped frequency distribution table to calculate the mean age of the students at "Barry's Ballroom Dancing Studio".

For age group 30-39, what is the value of the product of the class centre and the frequency?

How do i do this? I've been told the answer is 175. But i dunno how to do it.

 

[PC]

This is a really nasty question. But it does specifically ask for one thing, the product of the class centre of the 30-39 group and its frequency.

There are five scores in this group; 35, 30, 30, 31, 37, so the frequency is 5.
The class centre is simply the average (or median) of the smallest and largest possible scores of the group, so class centre is (30+39)/2 = 69/2 = 34.5.

So the product of the class centre and the frequency is 5 x 34.5 = 172.5.

Putting things in perspective, the question is a good way of making sure we know how to deal with group frequency tables. Most people know that in the normal frequency distribution tables, you do score (x), tally, frequency (f) and the fx column, and the mean is worked out by doing ∑fx / ∑f. For grouped frequency tables, as soon as you make up the groups, you lose the identity of each individual score. Instead of using scores, you use the class centre. So to work out the mean, you do the same thing, ∑fx / ∑f, but this time the fx column is the frequency multiplied by the class centre ... which is what you're asked to find in this question.

Now for a bit of a moan. The DA5 topic goes on about COMPARING data sets and having BACK TO BACK stem and leaf plots for two sets of data, but this question talks about the males and the females being a single set of data. No wonder students get confused!