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s1 cumulative distribution function-Discrete random variables watch

1. If
x 1 2 3
F(x) 3/4 7/8 1

Find F(2.6)

I know how to go about calculating this but..
The answer says that we should use F(2) ..however why is it that we dont round 2.6 up to 3 and calculate F(3)?

And is this the case with all discrete random variables??

confused :/
2. (Original post by maamm)
If
x 1 2 3
F(x) 3/4 7/8 1

Find F(2.6)

I know how to go about calculating this but..
The answer says that we should use F(2) ..however why is it that we dont round 2.6 up to 3 and calculate F(3)?

And is this the case with all discrete random variables??

confused :/
Think about what F(2.6) means. It's the probability that your random variable will be less than or equal to 2.6. There is no rounding involved.

The only values <=2.6 are 1 and 2, so it's F(2).

Another example. Consider a standard 6 sided die. What's the probability that your roll of the die is <=3.5? Clearly it can only be 1,2, or 3, so F(3) which is 1/2.
3. Great explanation! Thank you! ....

So, with continuous variables, would I round?
4. (Original post by maamm)
Great explanation! Thank you! ....

So, with continuous variables, would I round?
Again, no rounding.
5. (Original post by ghostwalker)
Again, no rounding.

i agree with what your saying but for some reason in my textbook it keeps saying that you have to always round it up.

for example in a question on discrete uniform distribution they say
find P(1.792<or equal to X<or equal to 4.981) and then change this to P(2<or equal to X<or equal to 5)
I find it hard to believe that Edexcel could make such a mistake in their books....

dont know what to dooo
6. (Original post by maamm)
dont know what to dooo
Perhaps scan/link to the specific page, with full context, and I might be able to clarify things.
7. (Original post by ghostwalker)
Perhaps scan/link to the specific page, with full context, and I might be able to clarify things.
I cant scan the page..but one of the questions goes like this:
find P(1.792<or equal to X<or eql to 4.981)

The answer that they obtain is by rounding to P(2<or eql to X<or eql to5)
8. (Original post by maamm)
I cant scan the page..but one of the questions goes like this:
find P(1.792<or equal to X<or eql to 4.981)

The answer that they obtain is by rounding to P(2<or eql to X<or eql to5)
If it's a continuous r.v., then I'd use 1.792 and 4.981, or as close as I'd could get if I was reading it off a graph.

If it's a discrete r.v., with integer values, then as per my previous argument, I'd use 1 and 4.

What's the book, and I'll try and find it online?

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Updated: January 15, 2013
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