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S3 - Combinations of random variables question.

Hi, I'm stuck on part c) for this question:

In the vegetable section of a local supermarket, leeks are on sale either loose (and unprepared) or prepared in packs of 6. The weights of unprepared leeks are modelled by the random variable X which has the Normal distribution with mean 260 grams and standard deviation 23 grams. The prepared leeks have had 40% of their weight removed, so that their weights, Y, are modelled by Y=0.6X.
(a) Find the probability that a randomly chosen unprepared leek weighs less than 290 grams.
(b) Find the probability that a randomly chosen prepared leek weighs more than 155 grams.
(c) Find the probability that the total weight of 4 randomly chosen prepared leeks in a pack is less than 913 grams.

For part b) I found the distribution for Y to be Y ~ N (56, 190.44) and so I thought for part C, I would multiply the mean and variance by 4 to give me the new distribution.

But instead, the answer requires me to multiply it by 6 - the number of leeks in a pack and then simply works out P(X < 913). Why do they use 6 instead of 4? And where do they take into account the fact that we're working out the probability of the total weight of 4 prepared leeks?

Thank you for you kind help,

Maulik.
Reply 1
Oops, posted in this in wrong place. :s-smilie:

Can someone move it please?
Reply 2
You doing the OCR paper ths friday??

It seems like a typo in the question wheres the question from??

Also for your distribution of Y is the mean 56 or 156
Reply 3
D90
You doing the OCR paper ths friday??

It seems like a typo in the question wheres the question from??

Also for your distribution of Y is the mean 56 or 156


Aah ok, yep the mean is meant to be 156 sorry.

Edexcel S3 this Wednesday. :frown:
Reply 4
I would say that the answer to part c is basically 1. It's so probable that it's going to happen that it is off our tables. The answers to stats questions are never like that though. Are you sure about this question and the mark scheme actually being correct? and was your answer to part b correct?
Maulik2
Hi, I'm stuck on part c) for this question:

In the vegetable section of a local supermarket, leeks are on sale either loose (and unprepared) or prepared in packs of 6. The weights of unprepared leeks are modelled by the random variable X which has the Normal distribution with mean 260 grams and standard deviation 23 grams. The prepared leeks have had 40% of their weight removed, so that their weights, Y, are modelled by Y=0.6X.
(a) Find the probability that a randomly chosen unprepared leek weighs less than 290 grams.
(b) Find the probability that a randomly chosen prepared leek weighs more than 155 grams.
(c) Find the probability that the total weight of 4 randomly chosen prepared leeks in a pack is less than 913 grams.

For part b) I found the distribution for Y to be Y ~ N (56, 190.44) and so I thought for part C, I would multiply the mean and variance by 4 to give me the new distribution.

But instead, the answer requires me to multiply it by 6 - the number of leeks in a pack and then simply works out P(X < 913). Why do they use 6 instead of 4? And where do they take into account the fact that we're working out the probability of the total weight of 4 prepared leeks?

Thank you for you kind help,

Maulik.

It's a typo. It comes from MathsNet A-level question doesn't it? If you 'change' the question, you see that part c's "total weight of X" doesn't change to match the "number in a pack" number, which does change. It stays stuck as 4. The question thus should be:

(c) Find the probability that the total weight of 6 randomly chosen prepared leeks in a pack is less than 913 grams.


The question is based on the S3 exam from OCR MEI, Unit 4768, January 2008.
Reply 6
Aah thanks for that and yep its from mathsnet, I'm going through all the exam q's on that and the specimen + last paper as thats all the material I have available. =]

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