# S1 Edexcel june 2016 paper help needed

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#1
Q3b) read the examiners report and it said 'The responses to part (b) confirmed that few students are aware of the simple relationship between Syy and standard deviation'

what is the relationship?

Secondly question 5B) I got 3.21 using the line method, The answer is 3.47 but why when median of 50 is 25 so confusing...

http://www.examsolutions.net/exampap...-s1-june-2016/
Last edited by RK; 1 year ago
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4 years ago
#2
For 3b: or , where is the number of data, and is the population variance, of which is an unbiased estimate (sample variance)

For 5b: 25 is through the third interval (as the first two intervals total to ). Hence

to three significant figures, as the width is .
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4 years ago
#3
The variance of y is the sum of ( (y - ybar) ^2)/n

Syy is the sum of (y-ybar)^2
Thus the variance = Syy/n and sd = the square root of Syy/n
In this case, the variance = 9.461^2 and n = 8, so Syy = 8 * 9.461^2 = 716

I'm not sure how you got 3.21. The median value (the 25th) is 16 into a group that is 17 wide. That is it is 16/17 of the way across the group. As the who;le group spans values from 3 - 3.5, it is 0.5 wide. Hence the median is 3 + 16/17 * 0.5 =3.47.

Can you explain how you got to 3.21 ?
Last edited by RK; 1 year ago
1
4 years ago
#4
I think it is probably better to use the method in the mark scheme, as there is no mention of mid points. The mark scheme uses the method or
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4 years ago
#5
For 3b: to three significant figures
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4 years ago
#6
.

I'll try and give a good method for attempting questions like 5b:

- half the total number of data. call this

- find the cumulative frequencies. find the first interval whose cumulative frequency is at least . this is the 'main interval' to consider

- let where is the cumulative frequency of the previous interval

- the width of the 'main interval' is and its frequency (non-cumulative) is called .

- the proportion of the frequency taken by the median is and so the proportion of the interval is

- add this onto the lower value of the interval to approximate the median

Applying this to the question:

and the cumulative frequencies are:

0<=w<2, 1
2<=w<3, 9
3<=w<3.5, 26
...

So we take the interval with cumulative frequency 26 as the 'main interval'. Hence .

Now, and . Therefore .

Adding this onto 3 (the lower value of the interval) gives 3.47 as the approximation to the median.

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