[cooldudeman]

why doesnt the total area add up to 140?? and generally, how would you calculate the width of a bar?

[fail_in_a_can]

what paper is this ?

[cooldudeman]

(Original post by fail_in_a_can)
what paper is this ?

jan8

[fail_in_a_can]

will have a look now

[ThaChronic]

You have to find the area for each bar and add them together, the total area is equal to 70. So you have to double the area of each bar to make it equal to 140. Therefore, the answer is 12.

[cooldudeman]

(Original post by ThaChronic)
You have to find the area for each bar and add them together, the total area is equal to 70. So you have to double the area of each bar to make it equal to 140. Therefore, the answer is 12.

but why did you have to double the 70 to make it equal the total frequency?
i saw in other histograms, the total area was equal to the total frequency.

[otrivine]

(Original post by ThaChronic)
You have to find the area for each bar and add them together, the total area is equal to 70. So you have to double the area of each bar to make it equal to 140. Therefore, the answer is 12.

can you not find the area from 90.5-78.5 and times by 0.5 = 6 then cant you do 140/6?

[mguindy]

Total area = 70

Next bit = 6

[fail_in_a_can]

Im not very good at explaining stuff but ill give it a stab

Basically, their are 140 runners, This doesnt mean the area is 140.... The area is W x F.... So the answer for area would be calculated by adding all the W x F together


Width 1 1 4 2 3 5 3 12
Freq. Density 6 7 2 6 5.5 2 1.5 0.5

So, you multiply all of them and add the sum to get the area

Should get 70 (6+7+8+12+16.5+10+4.5+6)

[mguindy]

find the area of each column then add them up !

[otrivine]

so what do we do with the 140 ? do you mean to find the frequency but is the start pint 59.5?

[cooldudeman]

(Original post by fail_in_a_can)
Im not very good at explaining stuff but ill give it a stab

Basically, their are 140 runners, This doesnt mean the area is 140.... The area is W x F.... So the answer for area would be calculated by adding all the W x F together


Width 1 1 4 2 3 5 3 12
Freq. Density 6 7 2 6 5.5 2 1.5 0.5

So, you multiply all of them and add the sum to get the area

Should get 70 (6+7+8+12+16.5+10+4.5+6)

but why doesnt it mean the area would be 140?

[Hobo389]

(Original post by cooldudeman)
but why doesnt it mean the area would be 140?

Because it isn't always the same as the frequency because it is frequency density, it is a representation of the frequency. Namely, half of it in this case.


Or twice it? Haven't read properly.

[Mystogan.]

In a histogram the area is proportional to the frequency. Area = K x Frequency. So if the Area is 70 and there are 140 runners k is 1/2.

You then use this when it asks to work out the frequency from the area for individual bars etc. The reason you found the area and frequency was the same question was because k was 1 so you were just kinda lucky

[cooldudeman]

(Original post by Hobo389)
Because it isn't always the same as the frequency because it is frequency density, it is a representation of the frequency. Namely, half of it in this case.


Or twice it? Haven't read properly.

so its always a fraction?

[cooldudeman]

(Original post by Kishan91)
In a histogram the area is proportional to the frequency. Area = K x Frequency. So if the Area is 70 and there are 140 runners k is 1/2.

You then use this when it asks to work out the frequency from the area for individual bars etc. The reason you found the area and frequency was the same question was because k was 1 so you were just kinda lucky

thanks you've answered my question.

[otrivine]

(Original post by Kishan91)
In a histogram the area is proportional to the frequency. Area = K x Frequency. So if the Area is 70 and there are 140 runners k is 1/2.

You then use this when it asks to work out the frequency from the area for individual bars etc. The reason you found the area and frequency was the same question was because k was 1 so you were just kinda lucky

so what do we do with the 140 ? do you mean to find the frequency but is the start pint 59.5?

[Mystogan.]

(Original post by cooldudeman)
so its always a fraction?

It depends on what the area and frequency is. E.g. if the area was 1200 and the frequency was 2, then k would be 2

[Mystogan.]

You multiple the class width and the frequency density of each bar to get the Area E.g. for the first bar it would be 1 * 6 so 6cm.

Altogether the Area should be 70 and then you get a formula Area = 1/2 x Frequency

For (a) you do 90.5 - 78.5 = 12 = Class Width For this bar the Frequency Density is 1 so

12 x 1 = 12cm^2 = Area

12 = 1/2 x Frequency

Frequency = 24

[otrivine]

(Original post by Kishan91)
It depends on what the area and frequency is. E.g. if the area was 1200 and the frequency was 2, then k would be 2

wait but why is it not
6=K x 140