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Series of S2 questions for S2 Exam next week.

1) Image 2, question 6, can someone help me with this question? (one on the right)


2) When defining P.D.F or C.D.F, I don't understand how the limits are done on the right-hand side. Sometimes, they are greater than or equal to and sometimes just greater than, can someone help me out with this? I am really confused. (one on the left)


3) Image 4, the limits I used were 6 and u, but shouldn't it have been x and 4, with x at the top and then shouldn't I have equated F(X) to 0.75. I am meant to find the upper quartile, I did this a year ago and looking at my exercise book now, I don't understand why I did it as I have. Can someone explain this to me? (one in the middle)


Much appreciated, this is all for the S2 exam next week. Thanks!
(edited 12 years ago)

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Reply 1
Bump.
Reply 2
Original post by Deep456
Bump.


2) When defining P.D.F or C.D.F, I don't understand how the limits are done on the right-hand side. Sometimes, they are greater than or equal to and sometimes just greater than, can someone help me out with this? I am really confused. (one on the left)

c.d.f is a continuous function so it doesn't matter if it is the sign is greater than or greater than equal to i.e. 0 <= X < 5 is equal to 0 < X <=5

Whereas p.d.f is a discrete function which means that it only takes in integers like 1,2,3,4,etc so in this case 0 <= X < 5 is equal to 0 <= X <= 4
Reply 3
Original post by cazzy-joe
2) When defining P.D.F or C.D.F, I don't understand how the limits are done on the right-hand side. Sometimes, they are greater than or equal to and sometimes just greater than, can someone help me out with this? I am really confused. (one on the left)

c.d.f is a continuous function so it doesn't matter if it is the sign is greater than or greater than equal to i.e. 0 <= X < 5 is equal to 0 < X <=5

Whereas p.d.f is a discrete function which means that it only takes in integers like 1,2,3,4,etc so in this case 0 <= X < 5 is equal to 0 <= X <= 4



So, when I convert it from one thing to another, how do I do it? Can you give me some examples?
Reply 4
Original post by Deep456
So, when I convert it from one thing to another, how do I do it? Can you give me some examples?


Suppose limits given is -3 < X <= 5 (this range is any number from -3 to 5 BUT EXCLUDING -3)

For c.d.f you substitute upper limit=5 and lower limit = -3
c.d.f is continuous so the sign doesn't matter anyway

but for p.d.f, it is discrete so you have to convert to -2 <= X <= 5 (you have to convert as p.d.f only takes in integers but the range does not include -3 so the next number to be included is -2.
upper limit would be 5 and lower limit is -2.

Now let's take -3 <= X < 5

for c.d.f upper limit would be 5 and lower limit would be -3

but for p.d.f it converts to -3 <= X <= 4 (as the range does not include 5 so last number to be included is 4)
upper limit=4 and lower limit=-3
Reply 5
Original post by cazzy-joe
Suppose limits given is -3 < X <= 5 (this range is any number from -3 to 5 BUT EXCLUDING -3)

For c.d.f you substitute upper limit=5 and lower limit = -3
c.d.f is continuous so the sign doesn't matter anyway

but for p.d.f, it is discrete so you have to convert to -2 <= X <= 5 (you have to convert as p.d.f only takes in integers but the range does not include -3 so the next number to be included is -2.
upper limit would be 5 and lower limit is -2.

Now let's take -3 <= X < 5

for c.d.f upper limit would be 5 and lower limit would be -3

but for p.d.f it converts to -3 <= X <= 4 (as the range does not include 5 so last number to be included is 4)
upper limit=4 and lower limit=-3


So p.d.f s always have =< right?
Reply 6
Original post by Deep456
So p.d.f s always have =< right?


Yepp:biggrin:

Sorry I can't help with other questions....I have forgotten everything:frown:
Reply 7
Original post by cazzy-joe
Yepp:biggrin:

Sorry I can't help with other questions....I have forgotten everything:frown:


You see 4.a from the picture above,

The f(x) been defined as 2<=x<3, 0 otherwise, why?
Reply 8
Original post by Deep456
You see 4.a from the picture above,

The f(x) been defined as 2<=x<3, 0 otherwise, why?


tbh I have no idea why it says so in solutions but I have always done it the way I said:frown:
Reply 9
Original post by cazzy-joe
tbh I have no idea why it says so in solutions but I have always done it the way I said:frown:


I am so confused what to do then, haha.
Original post by Deep456
I am so confused what to do then, haha.


Have you try all other questions??
If you do other questions correctly this one might be a typo. There are lots of typo with edexcel textbooks.
Original post by cazzy-joe
Have you try all other questions??
If you do other questions correctly this one might be a typo. There are lots of typo with edexcel textbooks.


Yeah, they all have different ways. I just don't know what I should do.
Original post by Deep456
Yeah, they all have different ways. I just don't know what I should do.


I have a CD of new textbook and the answer for that question is 2 <= x <=3
(edited 12 years ago)
Original post by Deep456
X


Is that the textbook by James Nicholson? If so, could you tell me if it comes with a CD that has worked solutions (like the official Edexcel series)?
Original post by snow leopard
Is that the textbook by James Nicholson? If so, could you tell me if it comes with a CD that has worked solutions (like the official Edexcel series)?


No, my one is an old one and it doesn't come with a CD.
can someone help me with S2 June 2003 Q4 bi please?

It says :A six- sided die is biased. When the die is thrown the number 5 is twice as likely to appear as any other number. All the other faces are likely to appear. The die is thrown repeatedly.

Find the probabilty that :

The first 5 will occur on the sixth throw.
Original post by hankypanky3
can someone help me with S2 June 2003 Q4 bi please?

It says :A six- sided die is biased. When the die is thrown the number 5 is twice as likely to appear as any other number. All the other faces are likely to appear. The die is thrown repeatedly.

Find the probabilty that :

The first 5 will occur on the sixth throw.


Probability it will be 5= 2/7

Is it (5/7)^5*(2/7)=0.053?
Original post by Deep456
Probability it will be 5= 2/7

Is it (5/7)^5*(2/7)=0.053?


Correct indeed ! :smile:
Thank you. I did (1/7)^5 *(2/7)
I see where I went wrong!
can someone help me with S2 June 2003 Q4 bi please?

It says :A six- sided die is biased. When the die is thrown the number 5 is twice as likely to appear as any other number. All the other faces are likely to appear. The die is thrown repeatedly.

Find the probability that :

The first 5 will occur on the sixth throw.
Probability it will be 5= 2/7

Is it (5/7)^5*(2/7)=0.053?

Argh everything I know is wrong. Isnt the probability of any given dice 1/6. So technically shouldn't the probability of getting a 5 be 2/6 not 2/7 where did you guys get 2/7 from :O

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