1. ## frequency diagram problem

Question:

Part c , estimate the total number of marks scored by all students in the group?

Is the answer frequency X class width for each group?

Regarding the class width, do we have to find the mid point of the class?

Many thanks

2. Originally Posted by nazz
Question:

Part c , estimate the total number of marks scored by all students in the group?

Is the answer frequency X class width for each group?

Regarding the class width, do we have to find the mid point of the class?

Many thanks
Yes, you'll need to find the mid-point of each class. For example, you've got 8 students who scored an average of 4.5 (the total of their marks is 8x4.5 or 36) and another 20 students who scored an average of 14.5 (the total of these students' marks is 20x14.5 or 290). Simply total up ALL the students' marks.

3. ## frequency diagram problem

Thanks for the reply, one more question, as the bar charts are not joined together , regarding the class width do we have to link them by doing the following:

0-9.5
9.5- 19.5
19.5-29.5
29.5-39.5
39.5-49.5

Then take the mid point of each class and multiply it by number of students.

Or just take the classes as the way they are shown in the diagram and then take the mid points.

Both methods will give different mid points for the classes.

4. Originally Posted by nazz
Thanks for the reply, one more question, as the bar charts are not joined together , regarding the class width do we have to link them by doing the following:

0-9.5
9.5- 19.5
19.5-29.5
29.5-39.5
39.5-49.5

Then take the mid point of each class and multiply it by number of students.

Or just take the classes as the way they are shown in the diagram and then take the mid points.

Both methods will give different mid points for the classes.

Great question!! The frequency histogram you provide is actually a poorly constructed one. By definition (all the definitions I've seen anyway), the classes should be touching. And usually you would just see the center of each class listed instead of the entire range. You are correct about having to connect them prior to finding the midpoint.