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    A firm of caterers wishes to buy wine for a wedding reception of 200 guests. They estimate that, on average, each guest will drink 45 cl of wine. The volume of wine in the bottles they buy may be assumed to have a distribution with mean 70.5 cl and standard deviation 1.2 cl. Show that if they buy 128 bottles then the caterers can be more than 95% certain that their requirements can be met.

    Please answer!!
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    (Original post by sharryborgan;[url="tel:66451844")
    66451844[/url]]A firm of caterers wishes to buy wine for a wedding reception of 200 guests. They estimate that, on average, each guest will drink 45 cl of wine. The volume of wine in the bottles they buy may be assumed to have a distribution with mean 70.5 cl and standard deviation 1.2 cl. Show that if they buy 128 bottles then the caterers can be more than 95% certain that their requirements can be met.

    Please answer!!
    I'm assuming distribution is normal. In which case I'm unsure whether the question is wrong or I am because my z value is too negative...
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    Not sure if the way I did it is correct, but...

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    \displaystyle \mathrm{V\sim N}\left ( 70.5,1.2^{2} \right )

    \displaystyle \mathrm{X= V_{1} + \cdots +V_{128} \Rightarrow X\sim N}\left (9024 ,184.32\right )

    \displaystyle \mathbb{P}\left (\mathrm{X}> 200\times 45 \right ) \approx \Phi\left ( 1.77 \right ) =0.9616> 0.95
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    Here's how I did it, assuming it was a normal distribution.
    z = (x - 70.5)/1.2
    P(Z<z) = 0.95, therefore P(Z>z) = 0.05
    From tables, z = 1.6449
    Rearranged the equation above, x = 72.47
    There are 128 bottles, so 72.47 * 128 = 9276.16
    There were 200 guests, each want 45cl,
    so 200 * 45 = 9000
    9000 / 9276.16 * 100 = 97.02
    97.02% > 95%
    QED
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    (Original post by sharryborgan)
    Please answer!!
    Er, no. This isn't a "do-my-work-for-me-pls!" site, please add your own thoughts and attempts to the question before we nudge you along the right path.
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    (Original post by Ayman!)
    Not sure if the way I did it is correct, but...
    Spoiler:
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    \displaystyle \mathrm{V\sim N}\left ( 70.5,1.2^{2} \right )

    \displaystyle \mathrm{X= V_{1} + \cdots +V_{128} \Rightarrow X\sim N}\left (9024 ,184.32\right )

    \displaystyle \mathbb{P}\left (\mathrm{X}&gt; 200\times 45 \right ) \approx \Phi\left ( 1.77 \right ) =0.9616&gt; 0.95
    Thanks that's much appreciated
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    (Original post by asinghj)
    Here's how I did it, assuming it was a normal distribution.
    z = (x - 70.5)/1.2
    P(Z<z) = 0.95, therefore P(Z>z) = 0.05
    From tables, z = 1.6449
    Rearranged the equation above, x = 72.47
    There are 128 bottles, so 72.47 * 128 = 9276.16
    There were 200 guests, each want 45cl,
    so 200 * 45 = 9000
    9000 / 9276.16 * 100 = 97.02
    97.02% > 95%
    QED
    Thanks a lot
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    (Original post by Zacken)
    Er, no. This isn't a "do-my-work-for-me-pls!" site, please add your own thoughts and attempts to the question before we nudge you along the right path.
    You dnt even knw the answer anyway. Dumbo.
    This is propa maths

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    (Original post by physicsmaths)
    You dnt even knw the answer anyway. Dumbo.
    This is propa maths

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    At least I do write proper English, as for being a dumbo, I do have a PhD.
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    (Original post by sharryborgan)
    At least I do write proper English, as for being a dumbo, I do have a PhD.
    Well I wasn't talking to you. But now that you have tried to assert your better then me, your struggling on easy maths and I wrote like that on purpose because well, this is the internet and I know Zacken so it was a joke. Get a grip on your life with your PhD, how is that working out for you?


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    (Original post by physicsmaths)
    Well I wasn't talking to you. But now that you have tried to assert your better then me, your struggling on easy maths and I wrote like that on purpose because well, this is the internet and I know Zacken so it was a joke. Get a grip on your life with your PhD, how is that working out for you?


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    I thought you were talking to me. I do not assert that I am better than anyone, My PhD is in Quantum Physics and I do not have much knowledge about stats, hence my question. Let's put this all down to a misunderstanding.
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    I have a question as well from S2!
    When using the poission distribution and I end up with a parameter that does not end with 0.5 or an integer such as 3.75, do I round it up to 4 or is it the average?? 😭😭
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    (Original post by 17lina)
    I have a question as well from S2!
    When using the poission distribution and I end up with a parameter that does not end with 0.5 or an integer such as 3.75, do I round it up to 4 or is it the average?? 😭😭
    I, think that if >3.75 then use 4, if <3.75 then use 3
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    (Original post by 17lina)
    I have a question as well from S2!
    When using the poission distribution and I end up with a parameter that does not end with 0.5 or an integer such as 3.75, do I round it up to 4 or is it the average?? 😭😭
    What is the exact question? Youc an sum exact probabilities


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