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# Continous Random Variable Questions P(X>0.2) different to P(X>=0.2)? watch

1. Is there any difference so I get that you integrate from 0.2 to being the lower limit, and the upper limit being infinity/highest value?
2. (Original post by Mihael_Keehl)
Is there any difference so I get that you integrate from 0.2 to being the lower limit, and the upper limit being infinity/highest value?
Nope! is the same as for continuous distributions.
3. (Original post by aymanzayedmannan)
Nope! is the same as for continuous distributions.
I thought so thank you.

Where sigma is the s.d.

How do I solve this lmao

I have worked out P(X<sigma) but I never know what to do next: It is page 52 Q6d of the edexcel book if you have it
4. (Original post by Mihael_Keehl)
I thought so thank you.

Where sigma is the s.d.

How do I solve this lmao

I have worked out P(X<sigma) but I never know what to do next: It is page 52 Q6d of the edexcel book if you have it
Which module?
5. (Original post by aymanzayedmannan)
Which module?
lmao sorry S2. I don't know if you are writing it though :P
6. (Original post by aymanzayedmannan)
Nope! is the same as for continuous distributions.
Is there any reason why this is btw, just feel like I am following it without really understanding it :/
7. (Original post by Mihael_Keehl)
Is there any reason why this is btw, just feel like I am following it without really understanding it :/
, but for continuous distributions: since, well, intuitively you're asking for one single possibility in a (continuous, hence massive/infinite/uncountably infinite) realm of possibilities.
8. (Original post by Mihael_Keehl)
Is there any reason why this is btw, just feel like I am following it without really understanding it :/
Think about it - time is a continuous variable and you can have 3.5 seconds but if you were selecting beads from a bowl you couldn't have 3.5 beads. Maybe look over S1 Chapter 8 and 9 if you're still a little confused.
9. (Original post by Zacken)
, but for continuous distributions: since, well, intuitively you're asking for one single possibility in a (continuous, hence massive/infinite/uncountably infinite) realm of possibilities.
(Original post by aymanzayedmannan)
Think about it - time is a continuous variable and you can have 3.5 seconds but if you were selecting beads from a bowl you couldn't have 3.5 beads. Maybe look over S1 Chapter 8 and 9 if you're still a little confused.
ofc ffs this is continuous and not discrete omg.

Did you look at that question aymanzayedmannan
10. (Original post by Mihael_Keehl)
I thought so thank you.

Where sigma is the s.d.

How do I solve this lmao

I have worked out P(X<sigma) but I never know what to do next: It is page 52 Q6d of the edexcel book if you have it
You can do this by thinking about the area under the graph. Is X normally distributed? If so you can just look up the relevant values in the table of the cumulative normal function
11. (Original post by shamika)
You can do this by thinking about the area under the graph. Is X normally distributed? If so you can just look up the relevant values in the table of the cumulative normal function
Thanks but isnt it P(x<sigma) - P(x<sigma) = 0?
12. (Original post by Mihael_Keehl)
Thanks but isnt it P(x<sigma) - P(x<sigma) = 0?
Obviously. But how is that relevant here?
13. (Original post by Mihael_Keehl)
Thanks but isnt it P(x<sigma) - P(x<sigma) = 0?
Yep. But P(X<-sigma) isn't the same as P(X<sigma); note the minus sign!
14. (Original post by shamika)
Yep. But P(X<-sigma) isn't the same as P(X<sigma); note the minus sign!
Thanks yes, so how do you account for the negative sign?

P(X< - sigma) = 1 - P(X<sigma)?
15. (Original post by Mihael_Keehl)
Thanks yes, so how do you account for the negative sign?

P(X< - sigma) = 1 - P(X<sigma)?
Yes. so with some nice re-arranging, you should get: P(x < sigma) = 1/2. Which makes sense if you look at a normal distribution graph.
16. (Original post by Zacken)
Yes. so with some nice re-arranging, you should get: P(x < sigma) = 1/2. Which makes sense if you look at a normal distribution graph.
thnx

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