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    Good bit of last minute signing up in a desperate attempt at an A... :roll:

    anyway, trying to do this Q, just not entirely sure about methods for it:


    "6) A hotel has 10 rooms, which are always occupied, and in each room there is a complimentary packet of biscuits, if on any night the packet is opened, it is replaced be a new packet the next day. The hotel manager knows from experience that, fore each room and for each night, the probability that the packet is opened is 0.4, independantly of other rooms and other nights." (i mean, what hotel manager wouldn't know that), "He keeps spare packets in stock

    i) The manager needs to be at least 95% sure that he has enough packet in stock each day to be able to replace those that are opened. Find the smallest number of packets that he needs to have in stock each day.

    ii) At the start of the week (7 days), there are 35 packets in stock, Use a suitable approximation to calculate the probability that this is enough to meet the demands in the coming week"

    The answers I got were 7 packets and 0.992 respectively, I'm just not sure about the methods I'm using, really I just guessed at methods, which were using X~B(10,0.4) with P(X<=n) >= 0.95, then used tables,
    and for the second part trying X~Po(4*7), using an approx. of X~N(28,28), then finding P(X<=35) with a continuity correction of +0.5 = P(Z<=1.417) = 0.912

    It doesn't help that I've lost my textbook, so I can't look anything like this up. Also the tables for the normal / poisson distribution seem to be nowhere to be found on the internet.



    edit: also on this paper:

    4) X~N(10,sigma^2);
    P(X<7) = p
    P(X<13) = 2p

    i) show that p is 1/3
    ii) find the value of sigma
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    Could you scan and post that paper on hear please, as i havnt done Jan 04 and feel i need the practice.

    Right method for the first part

    Second part
    X~B(70,04)
    V~N(28,16.8)

    Packets are discrete variables

    Therefore
    P(V>34.5)=1-p(v<34.5)

    Can you do the rest?
    If not ill put the answer up, ill check answer to first part as well.
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    the answer to the first part is correct,

    My explaination to the second part works out the probabilty he dosnt meet demand.

    Should be

    P(V<35.5)

    I think only your Binomial to Normal approximation was wrong.
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    (Original post by PhilC)
    the answer to the first part is correct,

    My explaination to the second part works out the probabilty he dosnt meet demand.

    Should be

    P(V<35.5)

    I think only your Binomial to Normal approximation was wrong.

    yeh, thanks, the approximation was probably wrong because I used the poisson, I wasn't sure if you could just multiply n by 7, it seems like a complicated situation with them being replaced every night... I could work out the rest, if I had the table for the normal distribution. lol...

    And I got the paper on this forum, I can't find the thread, and it's too big to upload alone so I've just uploaded it to :

    http://zogger.blamethepixel.com/s2%20jan%2004.JPG

    (bit under 1MB in size) would have compressed it mroe but the quality was already a bit low..
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    Cheers,
    I think when you are given that a number of events occur and a probability its best to use Binomial, then use either Normal of Poission approximations.
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    oh and learn the rules about converting ie np>5 etc. It makes your life a lot easier (ie you don't have to guess which approx to use)
 
 
 
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