Yes friend, your definition of W is wrong. Note that W is not sum of random variables X/3 and 2Y/3. It is mixture distribution with mixture component X, Y and mixture weights 1/3 and 2/3 respectively. hope it will help
I have done the following question:
In a hospital, 1/3 of the people who use a particular elevator are adults and 2/3 are children. The weight of the adults follows a normal distribution with mean 70 kg and sd 18 kg. The weight of the children follows a normal distribution with mean 18 kg and sd 8kg. Calculate the probability that a person entering the elevator weighs more than 45 kg.
I did the following: 1/3 Pr(X > 45) + 2/3 Pr(Y > 45) = 0.311705
and got the right answer.
However, I also tried doing it by defining a new random variable W = X/3 + 2Y/3:
I found that W is a normal random variable with mean 40.66667 and sd 8.02773. Then I calculated Pr(W > 45) and got 0.294669 which is wrong. I don't know why this method is wrong. If someone could explain the error I'd be grateful. Is there some different way that the normal random variables get combined?
Thanks.
The reason I tried this second method is that another part of the question asks to calculate the probability that the weight of three people entering the lift will exceed 110 kg. I figured that I could just use the normal distribution corresponding to V = W + W + W and calculate Pr(V > 110). But this will be wrong if my definition of W is wrong ....
Any help will be appreciated.
Last edited by Dason; 11-22-2010 at 09:16 AM.
Yes friend, your definition of W is wrong. Note that W is not sum of random variables X/3 and 2Y/3. It is mixture distribution with mixture component X, Y and mixture weights 1/3 and 2/3 respectively. hope it will help
Thankyou for your reply. Can you explain a bit more about mixture distributions, how would it apply to my question? Do the 'mixture weights' of 1/3 and 2/3 mean that W = aX + bY where a and b depend on 1/3 and 2/3 in some special way? I will try to research this more in the meantime.
Thanks again for your reply.
With Excel you can simulate distributions with an even number of values. 10 or 20 for each distribution may give you a pretty good approximation to reality. Then you can use +/-/x or divide on them and see how it affects the outcome distribution.
It is labor intensive but may give you insight into what's going on at the nuts and bolts level.
OK. So there's no way of coming up with a 'mixture' pdf that's based on knowing the pdf's of X and Y and knowing the probabilities of 1/3 and 2/3 .....? Is a numerical simulation the only way of dealing with the situation?
I have used Google to try and research the idea of 'mixture distribution' - although a lot of what I read was over my head I got the feeling that the question is in fact a lot more difficult than I thought.
Any further information is warmly accepted.
Friend, as told by Dason, your final random variable isn't really the sum of two different ones. Let me explain it to you. Let X denote N(70, 18^2) distributed R.V. of adult weight and Y be N(18,8^2) distributed R.V. of children weight.
Now Lets come to your first question, What is the probability that a random person entering elevator weighs more than 45.
i.e. probability that an adult entered elevator AND his weight was greater than 45 OR an child entered elevator AND his weight was greater than 45.
i.e. p = P[X>45]/3 + 2*P[Y>45]/3
which is a mixture distribution of Normal R.V. X and Y (also known as mixture components) with mixture weights 1/3 and 2/3 respectively. And you have calculated it in right manner.
But P[W>45] = P[(X/3+ 2*Y/3)>45] is not same as P[X>45]/3 + 2*P[Y>45]/3
what you are trying to do by defining new variable W is, you are making adults to loose their weight 1/3 and children to loose 2/3 respectively. and then putting one adult and one child in elevator (randomly) , computing the probability that the "total load" in elevator is greater than 45. I guess now you understand why your second method is wrong.
That makes a lot of sense, especially the part I have highlighted in red. Thankyou!
My only remaining question now is:
I think I understand how to answer this using a simulation (as suggested in an earlier reply) to model the question. But can it be done by constructing a pdf (using some clever formula?) and then using the pdf to calculate a probability? Is there some way of constructing a 'mixture' pdf to answer the question? Or is more information needed?another part of the question asks to calculate the probability that the weight of three people entering the lift will exceed 110 kg.
Thanks to all who have read this thread so far.
Yes, it it quite straightforward. The probability you need =My only remaining question now is:
Quote:
another part of the question asks to calculate the probability that the weight of three people entering the lift will exceed 110 kg.
I think I understand how to answer this using a simulation (as suggested in an earlier reply) to model the question. But can it be done by constructing a pdf (using some clever formula?) and then using the pdf to calculate a probability? Is there some way of constructing a 'mixture' pdf to answer the question?
P[3X>110] * P[all three people are adult]
+ P[2X+Y > 110] *P[two adult and one child]
+ P[X+2Y > 110] * P[one adult and two child]
+ P[3Y > 110] * P[all three are child]
i.e. you mixture components will be 3X, 2X+Y, X+2Y and 3Y with mixture weight being a RV ~ Bin(3, 2/3) taking values 0, 1, 2, 3 respectively. Let me know if it matches with your answer.
Last edited by PRADIPMASKE30; 11-23-2010 at 11:49 AM. Reason: correction
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