S1 Edexcel june 2016 paper help neededWatch
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...
For 5b: 25 is through the third interval (as the first two intervals total to ). Hence
to three significant figures, as the width is .
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 ?
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:
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.
Good luck with your exam!