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# Statistics 1 Histograms! (Estimate median) Watch

1. Hey as the title says I was wondering how would I calculate the estimate median for a histogram?
Can I use the formula (n+1)/2 and then look at the frequency?
If I can but then what do i put as the median would i put for example 25-30 due to it being continuous data.
Thanks
2. Find the total number of samples (eg 60).

Then halve that number (30) and then look for the bar which contains teh 30th highest value.

For example:
0<x<5 - 10
5<x<10 - 15
10<x<15 - 22
15<x<20 - 13

the cumulative total from 0<x<10 is 25. But our value is the 30th value. This lies in 10<x<15, hence the answer is the interpolated value of this.

At x=10, y=25
At x=15, y=47

So we need to find the value of x when y=30.

This can be done by similar triangles, or by finding the equation of the line [y-y1=m(x-x1)] and substituting the value of y=30.
3. (Original post by TheTallOne)
Find the total number of samples (eg 60).

Then halve that number (30) and then look for the bar which contains teh 30th highest value.

For example:
0<x<5 - 10
5<x<10 - 15
10<x<15 - 22
15<x<20 - 13

the cumulative total from 0<x<10 is 25. But our value is the 30th value. This lies in 10<x<15, hence the answer is the interpolated value of this.

At x=10, y=25
At x=15, y=47

So we need to find the value of x when y=30.

This can be done by similar triangles, or by finding the equation of the line [y-y1=m(x-x1)] and substituting the value of y=30.

Sorry I don't get the part in bold I dont understand how you worked out y given that the original values are unknown. Could you explain please?
4. (Original post by JBKProductions)
[/B]
Sorry I don't get the part in bold I dont understand how you worked out y given that the original values are unknown. Could you explain please?
Ok, assume we had the data above.
0<x<5 - 10
5<x<10 - 15
10<x<15 - 22
15<x<20 - 13

The total number of samples is 60. Thus the median value is at 30.

If you take the cumulative frequencies:
0<x<5 = 10
0<x<10 = 25
0<x<15=47
0<x<20 = 60

Since the median is 30, it must lie between 25 and 47.

Let x be the measured value (height, etc) and y be the cumulative frequency.

Look at this diagram of cumulative frequency.

As you can see, at the value of x=10, the CF is 25. and at the value of x=15, CF is 47.

Between the value of 25 and 47 is the median, 30. You see the black line drawn from the value of 30 (going horizontal). This meets the diagonal line drawn from the lowest value (x=10 y=25) to the highest value (x=15 y=47). You are interpolating this part of the bar as if it is a straight line between x=10 and x=15.

So you can find an equation of this diagonal line.

And find m:

So,

Then sub in the value y=30 and find x. (Obviously x is somewhere between 10 and 15)
Spoiler:
Show

x=11.13
5. (Original post by TheTallOne)
Ok, assume we had the data above.
0<x<5 - 10
5<x<10 - 15
10<x<15 - 22
15<x<20 - 13

The total number of samples is 60. Thus the median value is at 30.

If you take the cumulative frequencies:
0<x<5 = 10
0<x<10 = 25
0<x<15=47
0<x<20 = 60

Since the median is 30, it must lie between 25 and 47.

Let x be the measured value (height, etc) and y be the cumulative frequency.

Look at this diagram of cumulative frequency.

As you can see, at the value of x=10, the CF is 25. and at the value of x=15, CF is 47.

Between the value of 25 and 47 is the median, 30. You see the black line drawn from the value of 30 (going horizontal). This meets the diagonal line drawn from the lowest value (x=10 y=25) to the highest value (x=15 y=47). You are interpolating this part of the bar as if it is a straight line between x=10 and x=15.

So you can find an equation of this diagonal line.

And find m:

So,

Then sub in the value y=30 and find x. (Obviously x is somewhere between 10 and 15)
Spoiler:
Show

x=11.13
oh ok thanks
6. (Original post by JBKProductions)
oh ok thanks
No problem.
7. Hi, The example quoted on the forum is of a standard distribution shape for the histogrm. How do you go about finding the median for something that is not quite so 'standard distribution' in shape or am I perceiving it wrong?

Updated: October 29, 2008
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