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I haven't ate out in a long time, and after this I ever more reluctant.
I knew there was something funny about the octopus
http://imageshack.us/photo/my-images/690/61270887.png/
Hi guys,
May I again confirm my answers/thoughts with you
Q1) I evaluated the double integral and it was equal to 1. I also substituted x=10, 20 into the function and confirmed it is >0
Q2) I drew the graph http://imageshack.us/photo/my-images/4/23301964.png/ for limits of integration
For the marginal density of x I have:
For the marginal density of y I have:
Q3) I had:
Q4) I feel it is related to Q3, and wanted to see if I have Q3 correct first
Thank you
Hi bobbym,
I am feeling a bit better now. However looking at some questions have me nauseous again.
I was at a seafood buffet. Never again!
Hi guys,
Thank you for your help and clarifying the expected value - so it is -5/26.
I actually had food poisoning before the exam and been sick afterwards so will have to sit a supplementary in a few weeks.
Hi bobbym,
Isn't that simply the integral of the function?
I'm working of this:
http://en.wikipedia.org/wiki/Expected_v … m_variable
Argh! So far I've only seen cdf-> pdf questions in my course
I'm not sure about the 0 for x>1 for the cdf, do you have any hints? Edit: Should this be 1, since a cdf ranges from 0 to 1?
c) I haven't dealt with a piecewise version before - can you give me a hint how I would set this up?
http://imageshack.us/photo/my-images/3/19637915.png/
Hi guys,
My last question before crunch time, if you guys may. This one is a bit long (sorry)
a) I have found k = 6/13 after using the property that a density function should equal 1
b) In finding the CDF, I integrate each term in the piecewise function and solved for the constant as follows:
I solved for them and for A= 9/13 and B = 9/13. Is this correct?
For the CDF I'll obtain
0 for x<-1
c) In trying to obtain the mean I integrate
But I am obtaining a negative mean?
d) I can complete this myself if I get the mean correct
e) To find the probability X>0 Ive used
Hi bobby,
Yes, that's very clear! I knew I'd learn a better way off you.
Thanks very much!!!
bobbym,
Unless its too much work, could you possibly share with me how you would calculate it?
Should the combination be from 5 defective parts rather than 3?
Thank you very much for clarifying!!
Hi bobbym,
Thanks for the tip, I've updated all my posts.
I initially did it per your answer, but it seems with probability questions I over think it.
So is the rationale behind this:
p=1 out of 5 chance of obtaining a defective part from the sample
For the 3 parts chosen, the possible combinations with at least one defective part is:
But which algebra was messy?
Sorry, it wasn't a reference your workings!!
The question on the paper says its worth 2 marks, for all the working (I used integration by parts and quotient rule for the derivative)
May I ask how you took the limit? Is it L'Hopitals rule?
I've got my exam tomorrow, so thank you for taking the time to give feedback on this one!!!
The algebra/working was very messy.
http://imageshack.us/photo/my-images/688/68986489.png/
Hi guys,
Please ignore part a)
Sorry for the stream of questions, I'm completing a past paper that does not have answers - so trying the bets I can
With part b) may I confirm if my work is correct:
I've decided to take the complement (1 - sample of 3 having no defective items)
I calculated this as
If this is correct, what would be the method using binomial probability?
Thanks
Lin
http://img819.imageshack.us/i/57381179.png/
Hi guys,
May I confirm if the MGF I have derived is correct?
My answer after:
Thanks
Lin
Thank you very much!!!
http://imageshack.us/photo/my-images/685/52395533.png/
Hi guys,
Please ignore the second bit about the normdist.
May I confirm if my workings are correct?
1) State which distro it is from - Geometric
2) P(more than five tosses required) = 1 - (P[X=1] + P[X=2] + p[X=3] + P[X=4] + P[X=5])
X= getting TT
p= 0.5 x 0.5 = 0.25
I get
= 1 - (0.25 + 0.1875 + 0.141 + 0.1054 + 0.079)
= 0.2371
Hi guys,
I have a dilemma about a log-transformed regression
Assuming I have found the coefficient parameters for
and , I am required to test the hypothesis that by building the 95% confidence interval using and .I am unsure about transforming back or keeping the interval. Here is my thinking
1) I use the standard errors of the coefficients of
and from the regression output and build a confidence interval:2) Take the exponential to get back to normal scale and if the confidence interval contains 1, then accept the null hypothesis
Would this be the correct procedure?
Some people have said to me that this distorts the standard errors of the coefficient.
I would also like to note that the original data (pre-transformed) has no scale and was just raw numbers
Thanks in advance for any feedback,
Linda
Hi bobby,
Great, thanks for the assistance!! I'll remember to post the answer, when I get it
Hi bobby,
Yes for the first and second moments, that is what I got exactly.
For
is where I get a bit confused as other examples e.g. normal and gamma distributions seem to incorporate .After this I rearranged the first equation by taking the log of both sides, while for the second I rewrote it as:
This is how the function was originally defined so I am assuming it is like an exp function (continuous), I think the
you mention is actually ?Sorry that was a typo, I did use
for the second part.Hi bobby,
I first looked for E[X] by calculating:
For E[X^2] I did the same:
I am not completely sure how to go about method of moments, since for most examples - at this step onwards it jumps to the answer very rapidly and I can't follow
EDIT: Typo