# S1 InterpolationWatch

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#1

In part d) I don't understand the method in the mark scheme at all and my own method didn't work. Could someone please explain it?

In part e) I used a very long method to get the answer. Is there an easier way to solve it?

Last edited by Presto; 4 weeks ago
0
4 weeks ago
#2
Hey, Presto.

( ! )

I've just read 3 posts from you. All to do with stats. All asking for explanations of basic concepts.
Linear Interpolation
Venn Diagrams
Probability
Normal Distribution

Please go and try to learn some of these topics, and then come back and post your working for someone here to help try and explain what you might then still be getting wrong, and help you get it right.

Thanks.
0
#3
(Original post by begbie68)
Hey, Presto.

( ! )

I've just read 3 posts from you. All to do with stats. All asking for explanations of basic concepts.
Linear Interpolation
Venn Diagrams
Probability
Normal Distribution

Please go and try to learn some of these topics, and then come back and post your working for someone here to help try and explain what you might then still be getting wrong, and help you get it right.

Thanks.
I have gone through them actually. I just suck at S1 .-.
I made a really silly mistake in part d); I took the wrong frequencies ><

In part e) first I calculated the frequency when the weight was 1.6 using interpolation

Got 34.8 which I rounded off to 35
Then 35 - 32 = 3
Then (1 + 12 + 11 + 3)/48 = 27/48

This took a good 4-5 minutes and by the end I was unsure about whether it was correct or not.
Can you please tell me of a better method to approach this q?
0
4 weeks ago
#4
between 1.1g and 1.5g, there are 12 + 11 = 23 beans

in the next class, 1.5 to 1.7, the mass we need (1.6) is bang in the middle, so half of the beans in that class will be less then 1.6
half of 8(frequency for this class) = 4

hence, 23 + 4 = 27 beans are btwn 1.1g and 1.6g
(Original post by Presto)
I have gone through them actually. I just suck at S1 .-.
I made a really silly mistake in part d); I took the wrong frequencies ><

In part e) first I calculated the frequency when the weight was 1.6 using interpolation

Got 34.8 which I rounded off to 35
Then 35 - 32 = 3
Then (1 + 12 + 11 + 3)/48 = 27/48

This took a good 4-5 minutes and by the end I was unsure about whether it was correct or not.
Can you please tell me of a better method to approach this q?
1
#5
(Original post by begbie68)
between 1.1g and 1.5g, there are 12 + 11 = 23 beans

in the next class, 1.5 to 1.7, the mass we need (1.6) is bang in the middle, so half of the beans in that class will be less then 1.6
half of 8(frequency for this class) = 4

hence, 23 + 4 = 27 beans are btwn 1.1g and 1.6g
Can't believe I didn't think of that >>
Thank you!
0
4 weeks ago
#6
in part d, we have a 'choice' about which bean to take as the median bean.

Usually, when n (the number of items in our sample) is greater than 50, then the median item is the (n/2)th item.
When n is less than 50, the median item is the (n+1)/2 th item.

Hence the MS allowing both

if we choose the 24th bean to be the median, we need to know the class interval in which that lies, and further, what is the position of that item within the interval.
In the question, there are 9 + 12 = 21 beans in the first two classes. So the 24th bean will be the third item in the next class.
Now we use linear interpolation to find its mass.

We make an assumption : that the beans in this class are spread evenly. the spread of masses is 0.2g, and there are 11 of them, so the difference in mass btwn successive beans in this class will be 0.2 / 11.
we want the third bean in this class, so this will be 3 * 0.2/11 more than the lower limit of the class. eg 1.3 + 0.6/11

You should now see how the alternate response in the MS has been calculated (for using the 24.5th bean as our median bean)

Can you see how much explanation is needed here to explain interpolation fully & properly?
Now you understand the reason for my first post to you.

Go and learn the basics. Look up youtube videos of how to use these types of calculations BEFORE you start asking such open questions on here.
0
#7
(Original post by begbie68)
in part d, we have a 'choice' about which bean to take as the median bean.

Usually, when n (the number of items in our sample) is greater than 50, then the median item is the (n/2)th item.
When n is less than 50, the median item is the (n+1)/2 th item.

Hence the MS allowing both

if we choose the 24th bean to be the median, we need to know the class interval in which that lies, and further, what is the position of that item within the interval.
In the question, there are 9 + 12 = 21 beans in the first two classes. So the 24th bean will be the third item in the next class.
Now we use linear interpolation to find its mass.

We make an assumption : that the beans in this class are spread evenly. the spread of masses is 0.2g, and there are 11 of them, so the difference in mass btwn successive beans in this class will be 0.2 / 11.
we want the third bean in this class, so this will be 3 * 0.2/11 more than the lower limit of the class. eg 1.3 + 0.6/11

You should now see how the alternate response in the MS has been calculated (for using the 24.5th bean as our median bean)

Can you see how much explanation is needed here to explain interpolation fully & properly?
Now you understand the reason for my first post to you.

Go and learn the basics. Look up youtube videos of how to use these types of calculations BEFORE you start asking such open questions on here.
I thought the general rule was that we took (n+1)/2 for an even set of data (like 48 here) and n/2 if odd

Thank you so much for your help!
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