# My brothers stats homeworkWatch

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#1
Does anyone have any idea how to answer this?

4) You are catering for 16 people and provide 4 vegetarian and 12 non-vegetarian meals. (Hint: you are assuming that the probability someone is a vegetarian is 4/16 ie 0.25) You can answer this with Excel using the function ‘Statistical’ and then ‘BINOMDIST’ option. Or, you can answer it using statistical tables.
a) What is the probability that you will have at least one disappointed guest? (You know that the meat-eaters do not like the vegetarian option.)
b) What is the probability that all the disappointed guests will be vegetarians?
c) Suppose you were to provide 17 meals for 16 guests. Would it be more sensible to provide an extra vegetarian meal or non-vegetarian? Show your reasoning.
d) How does your decision in part c.) affect the probability of disappointment?

Please don't just give me the anwsers I just don't know where to start.

I'm studying economics at southampton.

Thank you
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6 years ago
#2
(Original post by Liamnut)
Does anyone have any idea how to answer this?

4) You are catering for 16 people and provide 4 vegetarian and 12 non-vegetarian meals. (Hint: you are assuming that the probability someone is a vegetarian is 4/16 ie 0.25) You can answer this with Excel using the function ‘Statistical’ and then ‘BINOMDIST’ option. Or, you can answer it using statistical tables.
a) What is the probability that you will have at least one disappointed guest? (You know that the meat-eaters do not like the vegetarian option.)
b) What is the probability that all the disappointed guests will be vegetarians?
c) Suppose you were to provide 17 meals for 16 guests. Would it be more sensible to provide an extra vegetarian meal or non-vegetarian? Show your reasoning.
d) How does your decision in part c.) affect the probability of disappointment?

Please don't just give me the anwsers I just don't know where to start.

I'm studying economics at southampton.

Thank you
Well I suspect for part(a) that you'll get "at least one disappointment" if the actual breakdown of the 16 people is anything other than the assumed distribution i.e. anything other than 12 "successes" and "4 failures" (or vice versa if you want to call "vegetarianism" a "success" ).

So you're looking for 1 - P(12 nonV and 4 V) from a binomial distribution with 16 trials and using the probabilities given.
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#3
(Original post by davros)
Well I suspect for part(a) that you'll get "at least one disappointment" if the actual breakdown of the 16 people is anything other than the assumed distribution i.e. anything other than 12 "successes" and "4 failures" (or vice versa if you want to call "vegetarianism" a "success" ).

So you're looking for 1 - P(12 nonV and 4 V) from a binomial distribution with 16 trials and using the probabilities given.
So for a,

P(s)+p(f)=1, using the table to find p(s), I got p(f)=0.772

For b,

I have no clue where to start. Is it p(v)*p(f)? Where p(f)=disappointed

Thanks
0
6 years ago
#4
(Original post by Liamnut)
So for a,

P(s)+p(f)=1, using the table I got p(f)=0.772

For b,

I have no clue where to start. Is it p(v)*p(f)? Where p(f)=disappointed

Thanks
Well for (b) you can cope with 4 vegetarians without disappointment. So if all the disapponinted guests are vegetarians you need to add the probs of 5 6, 7 etc vegetarians up to the max possible, Alternatively, again work it out as 1 - Something where the Something covers the cases of no vegetarian disappointments.
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