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    Sorry for the loads of questions but I'm desperately trying to learn S2 before a mock this week

    I can't seem to understand this question:
    A hotel has 10 rooms, which are always occupied, and in each room there is a complimentary packet of biscuits. If on any one night the packet is opened, it is replaced by a new packet the next day. Probability of a packet being opened is 0.4. Manager wants to be at least 95% sure that he has enough packets in stock to replace the opened ones. Find the smallest number of packets he needs

    Thank you
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    (Original post by olivia30)
    Sorry for the loads of questions but I'm desperately trying to learn S2 before a mock this week
    Don't fret it! This is what the forum is for. Also I might be getting a bit annoying, sorry...

    I can't seem to understand this question:
    A hotel has 10 rooms, which are always occupied, and in each room there is a complimentary packet of biscuits. If on any one night the packet is opened, it is replaced by a new packet the next day. Probability of a packet being opened is 0.4. Manager wants to be at least 95% sure that he has enough packets in stock to replace the opened ones. Find the smallest number of packets he needs

    Thank you
    So let's have X \sim B(10, 0.4), you want to find P(X \leq n) \geq 0.95. Make sense?
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    (Original post by Zacken)
    Don't fret it! This is what the forum is for. Also I might be getting a bit annoying, sorry...



    So let's have X \sim B(10, 0.4), you want to find P(X \leq n) \geq 0.95. Make sense?
    I was trying to do >= n for some reason :facepalm: thank you so much, again!
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    For some clarification:

    (Original post by olivia30)
    A hotel has 10 rooms
    Should be making you think binomial.

    If on any one night the packet is opened, it is replaced by a new packet the next day.
    Keeps the number of trials constant.

    Probability of a packet being opened
    Two outcomes, success/failure: packet being opened/packet not being opened/


    is 0.4.
    Probability is constant.

    Manager wants to be at least 95% sure that he has enough packets in stock to replace the opened ones. Find the smallest number of packets he needs
    Cumulative binomial.
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    Only 4 out of 10 will eat the biscuits? OCR, you crazy?
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    (Original post by olivia30)
    I was trying to do >= n for some reason :facepalm: thank you so much, again!
    Ah, yes. The whole "manager and stocks" thing is overdone, so it's good that you learnt from this one because you'll encounter quite a few more problems using a very similar scenario! Glad I helped.

    P.S: Don't worry about how many questions you're posting on the forum, if you are at all, it's completely fine.
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    (Original post by tinkerbella~)
    Only 4 out of 10 will eat the biscuits? OCR, you crazy?
    What's the probability that after eating all ten biscuits tinkerbella will complain about being too full?
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    (Original post by Zacken)
    For some clarification:



    Should be making you think binomial.



    Keeps the number of trials constant.



    Two outcomes, success/failure: packet being opened/packet not being opened/




    Probability is constant.



    Cumulative binomial.
    This is great, thank you!
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    (Original post by olivia30)
    This is great, thank you!
    Sorry, I posted that thinking you weren't sure how to formulate an expressions, turns out you'd already gotten the binomial part of it, and I was just rehashing what you already knew. Apologies.
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    I have another question that I am a bit confused over, can I send it to one of you? It's from a past paper
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    (Original post by olivia30)
    I have another question that I am a bit confused over, can I send it to one of you? It's from a past paper
    I don't mind having a look, which paper?
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    (Original post by olivia30)
    I have another question that I am a bit confused over, can I send it to one of you? It's from a past paper
    Sure thing.
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    Can one of you help me with the last part of Q9 from Jan 13? I should know how to do this but I don't http://www.ocr.org.uk/Images/144610-...atistics-2.pdf

    tinkerbella~ Zacken @other clever maths people
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    (Original post by olivia30)
    Can one of you help me with the last part of Q9 from Jan 13? I should know how to do this but I don't http://www.ocr.org.uk/Images/144610-...atistics-2.pdf

    tinkerbella~ Zacken @other clever maths people
    Does it help if I say that for the only way that p=0.5 occurs for the second test is if there is a Type II error on the first test?
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    (Original post by olivia30)
    Can one of you help me with the last part of Q9 from Jan 13? I should know how to do this but I don't http://www.ocr.org.uk/Images/144610-...atistics-2.pdf

    tinkerbella~ Zacken @other clever maths people
    I think I should probably go into some more detail, actually:

    Draw a tree diagram, the first branch is a split into p = 0.5 (with probability 0.2) or p = 0.6 (with probability 0.8).

    The second split is Reject H0/Do not reject H0 (with probabilities that you should have found previously).

    The third split is again Reject H0/Do not reject H0, but remember that p is adjusted to 0.6 if the null hypothesis was rejected the first time around in the 0.5 branch.

    Then just multiply along the tree and add the relevant probabilities.
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    (Original post by Zacken)
    I think I should probably go into some more detail, actually:

    Draw a tree diagram, the first branch is a split into p = 0.5 (with probability 0.2) or p = 0.6 (with probability 0.8).

    The second split is Reject H0/Do not reject H0 (with probabilities that you should have found previously).

    The third split is again Reject H0/Do not reject H0, but remember that p is adjusted to 0.6 if the null hypothesis was rejected the first time around in the 0.5 branch.

    Then just multiply along the tree and add the relevant probabilities.
    That's really helpful, thank you so much! I need to start using tree diagrams more often
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    (Original post by olivia30)
    That's really helpful, thank you so much! I need to start using tree diagrams more often
    Does it work?
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    (Original post by Zacken)
    Does it work?
    I think so, I got 0.083 or something but I haven't checked the mark scheme yet
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    (Original post by olivia30)
    I think so, I got 0.083 or something but I haven't checked the mark scheme yet
    Let me know when you do! :lol:
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    (Original post by Zacken)
    Let me know when you do! :lol:
    Well the answer is 0.0827 but I was using rounded values and my phones calculator so that's close enough thanks for all your help!
 
 
 
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