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3. The variable x was measured to the nearest whole number. Forty observations are given in
the table below.
x 10 15 16 18 19
Frequency 15 9 16
A histogram was drawn and the bar representing the 10 15 class has a width of 2 cm and a height of 5 cm. For the 16 18 class find
(a) the width,

I'm really stuck even though this question is only worth one mark!

frequency = fdx class width
15=fdx5
fd=3

if a class width of 5 has a width on the histogram of 2

5k=2
k=0.4

so a class width of 2 should be
2x0.4= 0.8

but the answer is 1cm! and it's only worth one mark so i must be missing something incredibly obvious but i can't see what! :frown:
Original post by YLY
...


The class width is the difference between the class boundaries.

So, what are the class boundaries for the 10-15 interval? They're not 10 and 15.
If you're not sure, reread the first sentence of your post.
(edited 12 years ago)
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YLY
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Original post by ghostwalker
The class width is the difference between the class boundaries.

So, what are the class boundaries for the 10-15 interval? They're not 10 and 15.
If you're not sure, reread the first sentence of your post.


I see now!
15.5-9.5=6
6/3=2

18.5-15.5=3
3/3=1

thank you!
Original post by YLY
S1

3. The variable x was measured to the nearest whole number. Forty observations are given in
the table below.
x 10 15 16 18 19
Frequency 15 9 16
A histogram was drawn and the bar representing the 10 15 class has a width of 2 cm and a height of 5 cm. For the 16 18 class find
(a) the width,

I'm really stuck even though this question is only worth one mark!

frequency = fdx class width
15=fdx5
fd=3

if a class width of 5 has a width on the histogram of 2

5k=2
k=0.4

so a class width of 2 should be
2x0.4= 0.8

but the answer is 1cm! and it's only worth one mark so i must be missing something incredibly obvious but i can't see what! :frown:

Fequency density= Frequency/width
15.5-9.5 = 6
And we are given the width of 2cm and we have to find the Fd and the frequency is now 6
6/2 = 3
Then
We are asked to find the width of 16—18
18.5-15.5 = 3 that the frequency
Then we have to use the frequency 3 and Frequency density to find the width
Which is 3/3 = 1cm

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