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Edexcel S2 Revision Thread

can anyone shed some light on why (in the jan 08 s2 paper)

in part 8d
m^2 – 4m + 4 = 0.5
where did the +4 come from??

(the question is to find the median of:
f(x)=2(x-2) 2<x<3 )
Thanks guys

Reply 41

Liverpool F.C.™

11

error502

can anyone shed some light on why (in the jan 08 s2 paper)

in part 8d
m^2 – 4m + 4 = 0.5
where did the +4 come from??

(the question is to find the median of:
f(x)=2(x-2) 2<x<3 )
Thanks guys

\displaystyle\int^m_2 2(x-2)\ dx = 0.5

\left[ \frac{2x^2}{2} - 4x \right]_2^m = 0.5

[(m^2 - 4m) - (2^2 - 4(2))] = 0.5

(m^2 - 4m) - (4 - 8) = 0.5

(m^2 - 4m) - (- 4) = 0.5

m^2 - 4m + 4 = 0.5

Reply 42

samir12

OP

16

Liverpool F.C.™

Yes, you can. How do you know which one is better ? (Without working out the actual Binomial distribution). My conditions satisfy both Poisson and Normal when choosing if it is a good approximation.

If you look in the S2 book (new one) on page 89, there is a paragraph which explains in which case to either go to normal or poisson, it does say in the paragraph that if you are not sure of which approximation to use, then calculate np, and if it is less than or equal to 10 then the poisson approximation should be used, but if it is greater then 10, then normal approximation should be used.

Reply 43

himegan

when finding the median do you always have to intergrate with limits of m and the lower limit?

Reply 44

samir12

OP

16

himegan

when finding the median do you always have to intergrate with limits of m and the lower limit?

Well the whole point is to find the + C after you integrate, you can do it Liverpool F.C's way, or just integrate and you know F(x) = 0 when x = lower limit, so jus sub them in and find the +C, then make F(x) = 0.5, and then find your median. But Liverpool F.C's way works fine, as you know the area between the lower limit (2) and the median (m) is 0.5, so if you integrate f(x) using limits m and 2, it will equal 0.5, also this works for upper quartile and lower quartile only you equate it to 0.75 or 0.25 depending on what quartile you are finding.

thank you!

sorry to be a pain...

i understand Liverpool FC's method. But, when i try to do this again, with the 'normal' method of trying to find the integral of f(x), then equate that to 0.5, I CANT FIND C!?!? when i substitute the upperbound i get c=3, and when i substitute the lowerbound i get c=4.

f(x)= 2x-4 2<x<3

what am i doing wrong??
thankyou!

Reply 47

Liverpool F.C.™

11

himegan

when finding the median do you always have to intergrate with limits of m and the lower limit?

Not always, when you have the c.d.f, F(x) = 0.5 since the c.d.f. is the integrated function of the p.d.f.

If you're given a p.d.f. you must integrate first.

Think of it this way, essentially the median is when the probability less then or equal to a certain value is equal to 0.5, P(X ≤ m) = 0.5. Since we are finding an area under a function we must integrate, the lower limit must the lowest value and since we don't know the upper limit to make the area 0.5, we just label it m.

Reply 48

Liverpool F.C.™

11

error502

sorry to be a pain...

i understand Liverpool FC's method. But, when i try to do this again, with the 'normal' method of trying to find the integral of f(x), then equate that to 0.5, I CANT FIND C!?!? when i substitute the upperbound i get c=3, and when i substitute the lowerbound i get c=4.

f(x)= 2x-4 2<x<3

what am i doing wrong??
thankyou!

Can you show me exactly what you are doing ? I'm not getting what you're trying to say.

Reply 49

samir12

OP

16

error502

sorry to be a pain...

i understand Liverpool FC's method. But, when i try to do this again, with the 'normal' method of trying to find the integral of f(x), then equate that to 0.5, I CANT FIND C!?!? when i substitute the upperbound i get c=3, and when i substitute the lowerbound i get c=4.

f(x)= 2x-4 2<x<3

what am i doing wrong??
thankyou!

If I were you, I would stick to Liverpool FC's way if you are confused about the way I told you, look at the post above he fully explains it. Otherwise here is the way I usually do it:

First lets find F(x), so integrate f(x) to get x^2 - 4x + C

so F(x) = x^2 - 4x + C

we know F(x) = 0, when x = 2, so (2)^2 - 4(2) + C = 0, C = 4 (this is where you went wrong, you equated it to 0.5 to find +C, sorry if i wasn't clear on it,)

so F(x) = x^2 - 4x + 4, now we equate it to 0.5 to find the median

x^2 - 4x + 4 = 0.5, and then just find x

Reply 50

error502

My bad! forgot to put F(x)=1, when using the upperbound! left it as F(x)=0!!

okay, so now iv got c= 4

shouldn't i be able to equate F(0.5)= median??

so, x^2 -4x + 4 =median where x=0.5

which comes out as median=2.25
the mark scheme says it's 2.71 though

Reply 51

error502

okay! GOT IT! so equate F(x) to 0.5, not x=0.5!!!!!!!!!!!!!!!!!! THANK YOUUU!

Reply 52

Liverpool F.C.™

11

error502

okay! GOT IT! so equate F(x) to 0.5, not x=0.5!!!!!!!!!!!!!!!!!! THANK YOUUU!

Be careful not to confuse the p.d.f and c.d.f. in the exam. Very important, and could lead you to get the whole question wrong.

Reply 53

error502

thank you, help much appretiated!

Reply 54

J.J.M

error502

My bad! forgot to put F(x)=1, when using the upperbound! left it as F(x)=0!!

okay, so now iv got c= 4

shouldn't i be able to equate F(0.5)= median??

so, x^2 -4x + 4 =median where x=0.5

which comes out as median=2.25
the mark scheme says it's 2.71 though

Its F(0.5) not x=0.5!!

Reply 55

J.J.M

I've gone through all the edexcel papers. I was just wondering if you think it would be helpful to go through OCR and AQA papers as well... opinions anyone??

Reply 56

TwilightKnight

11

Anyone else finding it hard to remember all of the Populations and Sample definitions? Ugh.

That might just be because im revising for 4 Maths exams at the same time, but it's still a bitch.

Reply 57

laurenlodge

19

TwilightKnight

Anyone else finding it hard to remember all of the Populations and Sample definitions? Ugh.

That might just be because im revising for 4 Maths exams at the same time, but it's still a bitch.

Snap on both accounts!

All the words are the same/similar and none of it's sticking :frown:

As for the insane amount of maths exams.... :sigh:

Reply 58

felt_monkey

12

Oh bugger I've forgotten to learn those definitions! MUST remember to tomorrow...

Reply 59

rajiv17

Liverpool F.C.™

You can approximate using a Normal. (Since np>5 and nq>5)

So a binomial distribution X~B[n,p] ≈ X~N[μ,σ²]

Where μ = np; σ² = np(1-p)

So in your case that would be X ~ N[10,25/3]

P(X≥16) Using Continuity correction would be P(X>15.5)

Standardising it:

P(Z>\frac{15.5-10}{\sqrt {\frac{25}{3}}})

So P(Z>1.91) (Answer is to 3 sig. fig.)

1 - P(Z<1.91)

Using tables P(Z<1.91) = 0.9719

∴ 1 - 0.9719 = 0.0281

right, but can i find it throught the binomial method?

Edexcel S2 Revision Thread

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