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Y10 maths homework help please!

(4 Posts)
Tansie Sat 08-Feb-14 12:06:37

'Representing Data'- DS has a histogram, showing the height of bushes in a garden:

y vertical axis is called 'frequency density' marked 0.5; 1; 1.5; 2; 2.5 (no units given)
x horizontal axis is called height (m) 5; 10; 15; 20; 25; 30

The histogram 'blocks' sit between say the 5 and 10, not above say 5 or 10.

Q1 says - How many bushes are there in the garden?

I 'get' you can't just count blocks, especially as they're not whole numbers on the 'frequency density' axis i.e. you can't have '2 and a quarter bushes'! The 'blocks' don't just represent 'a unit/bush'

I see from DS's exercise book that such histograms are formed by drawing a series of straight lines from the middle of the top of each column across to the middle of the top of the adjacent column (a line graph, if you like) but I can't see how to translate those 'points' into a head count (of bushes)!

Q2 asks 'How many bushes have a height of between 5 and 15 cm?

Given that the first 'potential histogram column' is empty, and the heights are given in m not cm, I think that's a trick question, the answer being 'none'...?


nickymanchester Sat 08-Feb-14 13:03:40

In a histogram, the frequency is actually the area - not just the height of the bar. So, it's width x height = frequency

So, if the first block goes from 5 to 10 then it has a width of 5. If that block has a height of 2 then the total frequency is 5 x 2 = 10 bushes

With Q2 it sounds like that might be a typo and they mean meters.

Here is something from BBC tha might help explain it better than I just did:-

Tansie Sat 08-Feb-14 13:35:58

I realise the whole thing is a lot more complicated than I thought!

Bear in mind the y axis isn't frequency, it's frequency density..

This is defined as the frequency over the class width (!!)

As mentioned my histogram blocks don't have an absolute value written beneath the centre of each block on the x axis; they appear to have a range, ie 0-5; 5-10, so in my example, the '0-5' value is empty ie I assume no bushes in that range; but the next column, (5-10 m) has a value of about 2.25 high as read off the FD (x) axis, i.e. has a frequency density of 2.25. The class width of each of these is as follows:

The (empty) 0-5 m column measures bushes that are from 0m high to those >5m high (a class width of 5) given that a bush of exactly 5m would be measured in next column-
The 5-10m column measures the bushes that are > or = to 5m up to those >10m (a class width of 5) etc

So, if I know my frequency densities (as measured on my histogram); and I know my class widths (as calculated above), surely I can extrapolate the formula

FD= freq
class width

so: freq = FD x CW

The only ishoo I have with this is that the maths are a bit random. Admittedly I have to guess the FD using a ruler across the top of each histogram column across to the y axis scale (all on blank, not graph, paper) and I'm seeing sort of 2.25 ish. 0.2 ish so my end result is 26.25... bushes!

Tansie Sat 08-Feb-14 14:35:15

Thanks nicky- I cam now see that my 26.25 happened because I'd done my rough measurement off the y axis, making is .25 instead of .2; done that way, it's 26!

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