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The Edexcel S2 (24/05/12 - AM) Revision Thread

A magazine has a large number of subscribers who each pay a membership fee that is due on
January 1st each year. Not all subscribers pay their fee by the due date. Based on
correspondence from the subscribers, the editor of the magazine believes that 40% of
subscribers wish to change the name of the magazine. Before making this change the editor
decides to carry out a sample survey to obtain the opinions of the subscribers. He uses only
those members who have paid their fee on time.

The full survey is to be carried out using 200 randomly chosen subscribers.
(g) Again assuming the editor’s belief to be correct and using a suitable approximation, find
the probability that in this sample there will be least 71 but fewer than 83 subscribers who
agree to the name being changed.

Anyone else done this? I get 0.5553 but my mark scheme says 0.5533

(edited 11 years ago)

Reply 381

TheUltimateProof

8

Original post by Troll the Trolls

no thats all wrong.

P(X<4) will give >6 on oppisite side of rectangle
P(X>6) will give >6 on current side

1/6 + 1/2 = 2/3 which is the correct answer

Yep I used this method. To get through the hard part with questions like this I just ask myself if I am missing out any other possibilities, which was the 1/6 for the current side.

Reply 382

rsaf

Original post by Quantus Life

113. A magazine has a large number of subscribers who each pay a membership fee that is due on
January 1st each year. Not all subscribers pay their fee by the due date. Based on
correspondence from the subscribers, the editor of the magazine believes that 40% of
subscribers wish to change the name of the magazine. Before making this change the editor
decides to carry out a sample survey to obtain the opinions of the subscribers. He uses only
those members who have paid their fee on time.

The full survey is to be carried out using 200 randomly chosen subscribers.
(g) Again assuming the editor’s belief to be correct and using a suitable approximation, find
the probability that in this sample there will be least 71 but fewer than 83 subscribers who
agree to the name being changed.

Anyone else done this? I get 0.5553 but my mark scheme says 0.5533

I just did the question, and I am also getting 0.5553. May be there is a misprint.

Reply 383

Quantus Life

I guess so, easy enough to do

Reply 384

Lostmoon9

2

Original post by Scorcher

The three independent random variables A, B and C each has a continuous uniform distribution over the interval [0, 5].

(a) Find P(A > 3).

I keep getting these questions wrong. I assumed it would be 2/6 because surely it would be P(A=4) + P(A=5). But the answer is 2/5. Is the 0 not included in continuous uniform distribution? I keep making silly mistakes on this sort of stuff.

I keep assuming there are 6 possibilities (0, 1, 2, 3, 4, 5) but I dunno :/

For this question all you do is
(X-a)/(b-a) - the formula to find F(X) in a continuous uniform distribution.

P(A>3)
= 1 -P(A<3)
= 1- (3-0)/(5-0)
= 1-3/5
= 2/5

Reply 385

Kenan and Kel

19

Could you be asked how to derive the formula for the variance of a continuous random variable from its first principles?

Reply 386

Warfare

Did anyone do this

A rectangle has a perimeter of 20cm. The length X cm of one side of this rectange is uniformly distributed between 1 cm and 7cm. Find the probability that the length of the longer side of the rectange is more than 6 cm long.

Reply 387

Rotuham

I ****ing hate S2.I screwed it up the last time around. :mad:

Reply 388

Lostmoon9

2

Original post by WaNaBe

Does anyone know how to do June 2010 question 3? Have no clue how to do it.

First we find out P(more than 6).

P(X>6)
= 1 - P(≤6)
= 1 - (6-1)/(7-1)
= 1 - 5/6
= 1/6

P(X>6) is just one side. We still need to work the other side. They have told us that the perimeter for the rectangle is 20 cm.

The formula for perimeter of a rectangle: P = 2L + 2W
Lets assume that L = 6
So its
P = 2L + 2W
20=2(6) + 2W
20-12=2W
therefore W= 4.

now we do
P(X<4)
=(4-1)/(7-1)
=3/6.

So we P(X>6) + P(X<4)
1/2 + 1/6
= 2/3

Do you understand this now?

Reply 389

sl96

hey guys if they ask P(x>2) how wud u work it out if u had a pdf, cdf or uniform distribution or cumulative uniform and can somebody explain that sampling example in ch6

Reply 390

XxAmxX

hiya-i just did June 2011 GRRRR. So many questions-if anyone can answer any of them i would be very grateful!

Question 4b-i kinda understand, but i did 1-(5/6)^2 Why is it cubed? Would that not mean that they are all the wrong length?

Question 5b-I did 6(0.2650)(0.735^5) + 6(0.735^6)...But i dont need the 6 times on the second part and dont understand why not..

Question 5c-Where do they get 30 from to approximate with !?

Question 6c-Why do we use significance of 2.5% and not 5%?

Thanks. x

Reply 391

XxAmxX

Original post by sl96

hey guys if they ask P(x>2) how wud u work it out if u had a pdf, cdf or uniform distribution or cumulative uniform and can somebody explain that sampling example in ch6

Wouldnt you just integrate between the upper limit and 2 PDF/CDF...

For the uniform dist. you can use the top limit-2 X (PDF)

...Hoping thats right

(edited 11 years ago)

Reply 392

Lostmoon9

2

Original post by XxAmxX

hiya-i just did June 2011 GRRRR. So many questions-if anyone can answer any of them i would be very grateful!

Question 4b-i kinda understand, but i did 1-(5/6)^2 Why is it cubed? Would that not mean that they are all the wrong length?

Question 5b-I did 6(0.2650)(0.735^5) + 6(0.735^6)...But i dont need the 6 times on the second part and dont understand why not..

Question 5c-Where do they get 30 from to approximate with !?

Question 6c-Why do we use significance of 2.5% and not 5%?

Thanks. x

Question 4B. They cube it because player must select 3 sticks.

Question 5B.
X~B(6,0.2650)
P(X<2)
=P(X≤1)
=P(X=0) + P(X=1)
=(0.735)^6 + 6(0.2650)^1(0.735)^5
= 0.4987

Question 5C.
They got 30 because 60 times 0.5.

There is no 6C.
For 6B. We use 10% level of significance.

Reply 393

XxAmxX

Original post by sl96

hey guys if they ask P(x>2) how wud u work it out if u had a pdf, cdf or uniform distribution or cumulative uniform and can somebody explain that sampling example in ch6

I lied.....you have to do 1-F(2) :biggrin:

Reply 394

Mathematix

2

Original post by sl96

hey guys if they ask P(x>2) how wud u work it out if u had a pdf, cdf or uniform distribution or cumulative uniform and can somebody explain that sampling example in ch6

Since it's already been answered...

Good luck in the exam :eek:

Reply 395

Braniac101

when doing a 2 tailed test, you half the significance level.
so if you are testng p(x>16) in a 2 tailed test becuase question says different form at 5% level,
do you reject if probability is less than 0.025, or do you use 0.5 a syou ar eonly considering the one tail at the moment even though it is 2 tailed?

Reply 396

masterhr1

5

An airline knows that overall 3% of passengers do not turn up for flights. The
airline decides to adopt a policy of selling more tickets than there are seats on a flight. For an aircraft with 196 seats, the airline sold 200 tickets for a particular flight.

By using a suitable approximation, find the probability that:

(c) there is at least one empty seat on this flight.

How would you do this question? I done P(X>=1) so 1-P(X=0) but according to the mark scheme its wrong :frown:

Can someone help?

Reply 397

rfuzzy

Can someone please tell me how to solve Q4 of June 2007?Please help!
Thanks=)

Reply 398

JordanS94

16

Original post by rfuzzy

Can someone please tell me how to solve Q4 of June 2007?Please help!
Thanks=)

If you have the edexcel S2 book look at page 100 :wink:

It explains it better than what I could :smile:

It pretty simple tbh just a lot of work

Reply 399

Quantus Life

Original post by masterhr1

An airline knows that overall 3% of passengers do not turn up for flights. The
airline decides to adopt a policy of selling more tickets than there are seats on a flight. For an aircraft with 196 seats, the airline sold 200 tickets for a particular flight.

By using a suitable approximation, find the probability that:

(c) there is at least one empty seat on this flight.

How would you do this question? I done P(X>=1) so 1-P(X=0) but according to the mark scheme its wrong :frown:

Can someone help?

Approximate to Poisson so Lambda = 6 (200 x 0.03)
This is the Mean number of seats vacant on each flight. For at least one seat to be free there would have to be at least 5 passengers not turn up as there are 196 seats and 200 passengers booked. 200 - 5 = 195 so one free seat. Therefore P(X > 4) = 1 - P(X <= 4).

Hope that helps.

The Edexcel S2 (24/05/12 - AM) Revision Thread

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