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Edexcel A level S2 June 2009

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Waterstorm

The grade boundries being so high brought my hopes down, I was getting around 90%, but I saw the boundries being nearly that high was crazy :sad:

but could the UMS be any lower than percentage.. i.e. if you got 90% in raw marks.. could that be less than 90 UMS?

Reply 101

struz

For 5 c):

Why in the answer do they do the following...

...rather than

P (X\leq 4) - P (X\leq 1)

Reply 102

sharp357

I'm having trouble drawing a cumulative distribution function when given the probability density function (i keep seeing random numbers being added after one of the lines being integrated). Say for example question 7b on here: http://www.carlgauss.co.uk/PDFs/A-Level%20Papers/S2/S2%20-%202004%20June%20Paper.pdf

Could someone talk me through what they do and why?

Reply 103

username233985

struz

For 5 c):

Why in the answer do they do the following...

...rather than

P (X\leq 4) - P (X\leq 1)

because thats the answer..... you have to get the numbers 1,2,3,4 in, so you have to do less than 5 - less than 0

make sense?

Reply 104

Rocafella89

thegreatest

What do you mean by sample frames and distrubtions things? like the definations or what?

yeah the defenitions.. and sometimes they do quastions on them.. like q6c jan 2006

Reply 105

*Rachie*

struz

For 5 c):

Why in the answer do they do the following...

...rather than

P (X\leq 4) - P (X\leq 1)

It's the probability that P( 1<=1 X <=4)
So P(1<=X) is the same as 1-P(x=0)
So it's P(X<=4) - P(X=0).

Reply 106

Waterstorm

Ali M

but could the UMS be any lower than percentage.. i.e. if you got 90% in raw marks.. could that be less than 90 UMS?

80% UMS = A, but if the raw marks were say 85% for an A, then you need to get that 5% more from raw marks to get the A, if you see what I'm saying. So if you got 90% raw marks, you woudn't get as much UMS as would if the grade boundries were lower :frown:

Reply 107

username233985

sharp357

the cdf is the "probability up to this point", so you have to add on the value of the area underneath the graph of the first part of the pdf, then take away the value of the area that would be under the second part, but below the range of the function ( if that makes sense )

i answered someones question about this earlier on in this thread if you want to have a look.

Reply 108

Rocafella89

sharp357

your doing the integeration correct, but then you need to put +C on the end.

to calculate the value of c, subsititute EITHER:

the lower boundary and make = 0 and then u will get the value for c

or

the upper boundary and make = 1 then u will get the value for c

the way you decide if you use the upper or lower is like this:

"always use the upper boundary, except if you have two seperate parts to integrate, in which case you would use the lower boundary in the first one and make equal to zero and the upper boundary in the second one which you would make equal to zero"

hope that makes it a bit clearer?

Reply 109

lilangel890

Am hoping S2 will be a lot easier than S1, thankfully it's more predictable.

Just hope we don't get a sampling dist'n. Ack

Reply 110

Waterstorm

struz

For 5 c):

Why in the answer do they do the following...

...rather than

P (X\leq 4) - P (X\leq 1)

The question is asking to include the 1, you don't wanna cut it out. If you draw it, you'll see.

Reply 111

struz

*Rachie*

It's the probability that P( 1<=1 X <=4)
So P(1<=X) is the same as 1-P(x=0)
So it's P(X<=4) - P(X=0).

mmm, that's what i thought but why is it not then P(X<=4) - (1 - P(X=0))?

Reply 112

struz

ahhh i think i see it now! thanks for all the help peoples! :smile:

Reply 113

Rocafella89

Waterstorm

that sucks.. really! guess all we can do is aim for 100% then we wont be affected (i think) :P

i hate bloody UMS

Reply 114

Waterstorm

Ali M

that sucks.. really! guess all we can do is aim for 100% then we wont be affected (i think) :P

i hate bloody UMS

Yeah, something like my Biology the grade boundries is like 70% for an A, so we got a whole 10% below the 80%.. which is awesome, so if we do get about 80%, in UMS it's loads! :yep:

Mind you, there're that low cause it's pretty difficult to get 80% though, but still :p:

Reply 115

*Rachie*

hparsons4

u want 2 find the values 6,7,8,9,10,11,12,13,14.
Therefore u want 2 take values less than 14, away from all values less than 5, leavin u with the values wanted. Hope tht makes sense.

Can anyone help me, i have no clue wot 2 do wen ur asked 4 a sampling distribution?

If you're talking about the mean sampling distribution, here's what you do.
Say you have a bag with counters in. 60% (3/5) of the counters have the number 0 on them and the rest, 40% (2/5) , have 1 on them.
If we randomly pick out three counters at a time (and replace them afterward), the possible sample combinations are:

000

100
010
001

110
101
011

111

In the first line, the mean is 1 [(1+1+1)/3]
In the second group, the mean is 1/3 (as all samples are the same).
In the third group, the mean is 2/3.
In the last group, the mean is also 1.

So the mean could be 0, 1/3, 2/3, or 1.

The probabilities of obtaining each of the above results are as follows:
For 000, the probability is (3/5)^3 = 27/125
For 100, or any combination, the probability is (3/5)^2 x (2/5) = 18/125
For 110, or any combination, the probability is (2/5)^2 x (3/5) = 12/125
For 111, the probability is (2/5)^3 = 8/125.

Then you put all this information in a table (see attached image).
And that's your distribution. Hope that helps!

prob1.bmp270.6KB

Reply 116

username233985

ok i have a question, with sampling distributions say for example,

3 coins are chosen out of a bag, they have numbers 0 (25%), 1(50%) and 2(25%) on
find sampling distribution of the mode.

so i wrote all the different combinations, then find the modes, i dont understand how you find the mode of the combo where you have one of each coin i.e. 0,1,2. what is the mode of this??

any help??

Reply 117

sharp357

Ali M

Ok so if there is just 1 line to integrate i can solve it with the lower or upper class boundary but if there is more than 1 line to integrate i should use the rule you put above? (also what if there are 3 or 4 lines to integrate?) I understand it all apart from my last question so thanks

Reply 118

thenewromance

What are you meant to do when asked for a 'sampling distribution' of something. There was a question about 75% of coins being 5p and 25% being 10p. 3 were selected and you had to state the sampling distribution. For 7 marks. Any ideas?