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Linear interpolation help!!!

I don't get how to do linear interpolation to work out part d it's just not working:frown:interpolation q.png
(edited 4 years ago)

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Reply 1
Original post by Sidd1
I don't get how to do linear interpolation to work out part d it's just not working:frown:interpolation q.png

Is this question from one of the Edexcel textbooks?
Reply 2
Original post by Sidd1
I don't get how to do linear interpolation to work out part d it's just not working:frown:interpolation q.png

What did you get for the mean and std dev and how did you try to do the linear interpolation?
Reply 3
Original post by OL12345
Is this question from one of the Edexcel textbooks?

Yeah from the AS Stats and Mechanics book
Reply 4
Original post by mqb2766
What did you get for the mean and std dev and how did you try to do the linear interpolation?

For the mean I got 23.4 and std dev: 7.32
I then added the mean and standard deviation and got 30.72 and then used linear interpolation but got 22.43?

I know it would be easier to show my working out but I can't at the moment:frown:
Er okay; calculate the maximum gust of wind within 1 std deviation and figure out which category it is in; once you have done that add another column to the table with cumulative frequency: plot a cumulative frequency graph and look for the segment line which represents your category: calculate the gradient of this line using the upper and lower bounds with the upper and lower frequencies; and then form an equation with your maximum gust of wind .within one standard deviation and x to denote its frequency: this will again be a gradient equation with 2 known x,y values and since you worked out the gradient of the line earlier you can solve the equation
Reply 6
Original post by Levi.-
Er okay; calculate the maximum gust of wind within 1 std deviation and figure out which category it is in; once you have done that add another column to the table with cumulative frequency: plot a cumulative frequency graph and look for the segment line which represents your category: calculate the gradient of this line using the upper and lower bounds with the upper and lower frequencies; and then form an equation with your maximum gust of wind .within one standard deviation and x to denote its frequency: this will again be a gradient equation with 2 known x,y values and since you worked out the gradient of the line earlier you can solve the equation

Ummmm this seems so longgg is there no other way without working out the gradient etc?
Reply 7
Original post by Sidd1
For the mean I got 23.4 and std dev: 7.32
I then added the mean and standard deviation and got 30.72 and then used linear interpolation but got 22.43?

I know it would be easier to show my working out but I can't at the moment:frown:


Obviously you want the mean +/- std dev so approximately 16 <-> 31. 16 lies in the 2nd interval and 31 in the last interval.
Just do the linear interpolation on each of these intervals. Not much working ...
(edited 4 years ago)
Reply 9

Yeah I had a look at that but I don't get it still:frown:
Original post by Sidd1
Ummmm this seems so longgg is there no other way without working out the gradient etc?

should only take about 3 mins; you only need to do the gradient of the specific category's line and you needn't plot the graph; there are other methods tho
Original post by Sidd1
Yeah I had a look at that but I don't get it still:frown:


How do you think you do linear interpolation for mean-stddev?
Reply 12
Original post by mqb2766
Obviously you want the mean +/- std dev so approximately 16 <-> 31. 16 lies in the 2nd interval and 31 in the last interval.
Just do the linear interpolation on each of these intervals. Not much working ...

Hang on, so the 16 and 31 represent the intervals and not the frequency? I thought they were the frequencies that's how I ended up with the wrong intervals. Right, okay makes sense thanks:smile:
(edited 4 years ago)
Original post by Sidd1
Hang on, so the 16 and 31 represent the intervals and not the frequency? I thought they were the frequencies that's how I ended up with the wrong intervals. Right, okay makes sense thanks:smile:

mean and stddev must be the original varaible, not the frequency or probability.
Reply 14
Original post by Levi.-
should only take about 3 mins; you only need to do the gradient of the specific category's line and you needn't plot the graph; there are other methods tho

Oh okayyyy I prefer another method but thank you anywayy:smile:
Original post by Sidd1
Oh okayyyy I prefer another method but thank you anywayy:smile:

Yeah i was just doing this **** today and forgot all of my stats stuff so this was the method i found most useful
Reply 16
Original post by mqb2766
Obviously you want the mean +/- std dev so approximately 16 <-> 31. 16 lies in the 2nd interval and 31 in the last interval.
Just do the linear interpolation on each of these intervals. Not much working ...


Okay I worked it out I got 50.252 and then 6.3
But they subtracted them? Why is that?
Original post by Sidd1
Okay I worked it out I got 50.252 and then 6.3
But they subtracted them? Why is that?

They want the number of days its within the range, so the "width" of the frequencies.
Reply 18
Original post by mqb2766
They want the number of days its within the range, so the "width" of the frequencies.

oh okay thanks:smile: You're so good at Maths I wish I was someone who naturally kinda got it:frown:
Original post by Sidd1
I don't get how to do linear interpolation to work out part d it's just not working:frown:interpolation q.png

you find your upper boundry and lower boundry. upper = mean + sd and lower = mean -sd . you use linear interpolation to find the value of x for your upper and lower boundry ( remembering the x goes inbetween the class values ) then take them away

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