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

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 $\geqslant$ for continuous distributions.

I thought so thank you.

What about this question: Find P(-sigma)<X<sigma)

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?

Nope! $>$ is the same as $\geqslant$ for continuous distributions.

Is there any reason why this is btw, just feel like I am following it without really understanding it :/

Is there any reason why this is btw, just feel like I am following it without really understanding it :/

$P(X \geq x) = P(X=x) + P(X > x)$, but for continuous distributions: $P(X=x) = 0$ since, well, intuitively you're asking for one single possibility in a (continuous, hence massive/infinite/uncountably infinite) realm of possibilities.

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.

I thought so thank you.

What about this question: Find P(-sigma)<X<sigma)

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

Thanks but isnt it P(x<sigma) - P(x<sigma) = 0?

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!

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.