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#76 Re: Help Me ! » PDF to CDF » 2011-06-15 14:14:38

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

#77 Help Me ! » Joint Density » 2011-06-15 14:13:15

lindah
Replies: 10

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:


May I ask if my bounds correct in calculating this?

Q3) I had:

Q4) I feel it is related to Q3, and wanted to see if I have Q3 correct first

Thank you

#78 Re: Help Me ! » PDF to CDF » 2011-06-15 13:27:21

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!

#79 Re: Help Me ! » PDF to CDF » 2011-06-15 12:10:37

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. sad

#80 Re: Help Me ! » PDF to CDF » 2011-06-08 13:23:08

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

#81 Re: Help Me ! » PDF to CDF » 2011-06-08 12:06:58

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?

#82 Help Me ! » PDF to CDF » 2011-06-08 08:48:29

lindah
Replies: 31

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

for -1<x<0
fpr 0<x<1
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

#83 Re: Help Me ! » Defective parts » 2011-06-08 00:00:38

Hi bobby,

Yes, that's very clear! I knew I'd learn a better way off you.
Thanks very much!!!

#84 Re: Help Me ! » Defective parts » 2011-06-07 23:21:52

bobbym,

Unless its too much work, could you possibly share with me how you would calculate it?

#85 Re: Help Me ! » Defective parts » 2011-06-07 23:02:26

Should the combination be from 5 defective parts rather than 3?

#86 Re: Help Me ! » Moment Generating Moment » 2011-06-07 22:59:17

Thank you very much  for clarifying!!

#87 Re: Help Me ! » Defective parts » 2011-06-07 22:35:04

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:

#88 Re: Help Me ! » Moment Generating Moment » 2011-06-07 22:15:41

gAr wrote:

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?

#89 Re: Help Me ! » Moment Generating Moment » 2011-06-07 21:15:20

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.

#90 Help Me ! » Defective parts » 2011-06-07 21:13:16

lindah
Replies: 9

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

#91 Help Me ! » Moment Generating Moment » 2011-06-07 20:00:42

lindah
Replies: 13

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

#92 Re: Help Me ! » Tossing coins » 2011-06-07 19:44:39

Thank you very much!!!

#93 Help Me ! » Tossing coins » 2011-06-07 13:20:53

lindah
Replies: 3

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

#94 Help Me ! » Regression: Transformed variables » 2011-06-02 11:25:25

lindah
Replies: 0

Hi guys,

I have a dilemma about a log-transformed regression


And so the transformation is

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:
where t - is the critical value

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

#95 Re: Help Me ! » Parameter Estimation: Statistics » 2011-06-01 21:13:39

Hi bobby,

Great, thanks for the assistance!! I'll remember to post the answer, when I get it

#96 Re: Help Me ! » Parameter Estimation: Statistics » 2011-06-01 20:09:50

Hi bobby,

Yes for the first and second moments, that is what I got exactly.

#97 Re: Help Me ! » Parameter Estimation: Statistics » 2011-06-01 18:50:56

I used:

For

is where I get a bit confused as other examples e.g. normal and gamma distributions seem to incorporate
.
I used:

Would this b correct?

After this I rearranged the first equation by taking the log of both sides, while for the second I rewrote it as:


and rearranged this in terms of

#98 Re: Help Me ! » Parameter Estimation: Statistics » 2011-06-01 18:18:27

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
?

#99 Re: Help Me ! » Parameter Estimation: Statistics » 2011-06-01 17:58:44

Sorry that was a typo, I did use

for the second part.
Am I on the right track at all?

#100 Re: Help Me ! » Parameter Estimation: Statistics » 2011-06-01 17:51:08

Hi bobby,

I first looked for E[X] by calculating:


Then I equated that to
or
and solved for a.

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 sad

EDIT: Typo

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