For the question attached and solution.
find Pr(X > 9 |X > 8.5).
What did they do to the denominator?
They just split it up - it's a bit more clear if we include both bounds:
=\mathbb{P}(8.5<X<10))
So, splitting this up, we get:
=\mathbb{P}(8.5<X<9)+\mathbb{P}(9<X<10)=\mathbb{P}(8.5<X<9)+\mathbb{P}(X>9))
There's not really too much reason to do this - just that they had already calculated the probability that X was greater than 9, so this might've made the calculation slightly simpler?
Are we required to know the limit formulas attached?
You should know how to use them, yes. Consider the following probability density functions:
=e^{-x},x>0)
=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}, x\in\mathbb{R})
These pdfs go off to infinity - so to actually integrate them and ensure that the area under both curves is 1, you need to know how to integrate up to infinity - which, initially, will mean being able to use those limit formula. As an experiment (/for some practice using them), you should check that the following is a valid pdf:
=\lambda e^{-\lambda x},x>0)
For the image attached.
isn't there meant to be an open circle at x=0 ?
There is an open circle at x=0? Remember, pdfs should be defined *everywhere*, so it's about where the open/closed circle appears as a coordinate, not just the x-value.
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