Help with probability and statistic tasks

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TheGood7

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If someone could help me solve these I would really appreciate. I am currently working on them but needs to be sure my calculations are correct.
 
TheGood7 said:
View attachment 16511



If someone could help me solve these I would really appreciate. I am currently working on them but needs to be sure my calculations are correct.
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You have posted four rather involved questions. However, you have shown none of your own work, moreover you have not told what sort of help you need.
This is not a homework service. If you need answers, then please look elsewhere.
 
TheGood7 said:
View attachment 16511



If someone could help me solve these I would really appreciate. I am currently working on them but needs to be sure my calculations are correct.
Click to expand...
I solved them. But how does that help you? Can you show your calculations/work so a helper can actually help you if needed?
 
Number 1 and 2, I am sure I have it correct. But 3 and 4 looked very difficult to solve..
 
And I am sorry for totally forgetting! I have been studying :) Really appreciate you taking your time looking at my post!
 
TheGood7 said:
Number 1 and 2, I am sure I have it correct. But 3 and 4 looked very difficult to solve..
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In that case, please share your work for those problems. We can confirm your work

For 3 and 4, please show us:

how did you start and

exactly where you are stuck.

Those will give us an idea as to where we should start to help you. Additionally,

please post only one problem per thread.

You have posted four problems here - should have been four separate threads.
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Here is for 1 and 2.

To be honest for 3 and 4, i dont know where to start correctly..
 

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TheGood7 said:
To be honest for 3 and 4, i dont know where to start correctly..
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About #3. The vocabulary in the question is problematic.
The term distribution function is not in standard use in probability courses taught in North America.
The function's definition given is that of a CDF, cumulative distribution function rather than a pdf, probability distribution function.
The latter must return to almost zero, where as the CDF reaches a max of 1. If this is the case then \({\bf{a}}=1\)
That seems odd for this sort of question. Nevertheless this means that:
\(\mathscr{P}(-0.5\le X\le 0.5)=\mathscr{P}(0\le X\le 0.5) =(0.5)^2\).
 
First you say that you did them all and want someone here to give you their answers so you can compare them to your own. Then you go on to say in another post that you actually did not do #3 and #4.
So you are trying to cheat by coming here. You are not even asking how to do the problem--you just want the answer.
At this site the helpers frown on cheater.
My advice to you is to go elsewhere.
 
Jomo said:
First you say that you did them all and want someone here to give you their answers so you can compare them to your own. Then you go on to say in another post that you actually did not do #3 and #4.
So you are trying to cheat by coming here. You are not even asking how to do the problem--you just want the answer.
At this site the helpers frown on cheater.
My advice to you is to go elsewhere.
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Please dont help. I tried to do 3 and 4 but it was incorrect so you want me to show something that is totally out of this world? And i cant even understand why you call this a cheat, such a disrespect. It is not like im in a exam right now waiting for you to give me the answers. These are only pre-exam preparation questions.

If you are going to start discussions rather than helping then you can leave this thread. Tired of this childish thing. I am not asking you to rob a bank with me.
 
pka said:
About #3. The vocabulary in the question is problematic.
The term distribution function is not in standard use in probability courses taught in North America.
The function's definition given is that of a CDF, cumulative distribution function rather than a pdf, probability distribution function.
The latter must return to almost zero, where as the CDF reaches a max of 1. If this is the case then \({\bf{a}}=1\)
That seems odd for this sort of question. Nevertheless this means that:
\(\mathscr{P}(-0.5\le X\le 0.5)=\mathscr{P}(0\le X\le 0.5) =(0.5)^2\).
Click to expand...
Thanks alot will check out!
 
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