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Math Help - Finding Sample size? (Normal Distribution)

  1. #1
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    Finding Sample size? (Normal Distribution)

    I'm not really sure what to name this, and im pretty sure it belongs here, im sorry if it doesnt.

    Question

    The number of loaves of white bread demanded daily at a bakery is normally distributed with mean 6800 loaves and variance 84000. the company decides to produce a sufficient number of loaves so that it will fully supply demand on 95% of the days.

    (a) How many loaves of bread should the compan produce?

    I know i need to find n = CI * standard dev. / E .. i can figure out CI and SD. but i cant find E..

    (b) Based on (a), on what percentage of days will the company be left with more than 500 loaves of unsold bread?

    I'm not sure on how to do this one!


    Any amount of help would be amazing!!
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  2. #2
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    For part (a)
    You want to find P(Y<y_1)=.95
    so by standardising,
    you want \frac{y_1-\mu}{\sigma}=z_{.95}
    So y_1=1.6449*\sigma+\mu
    y_1=1.6449\cdot 289.83+6800=7276.737
    So they must make 7277 loaves

    (b)..
    since they bake 7277 loaves, you want the probability that they sell Y less than 7277-500=6777
    so P(Y<6777)...
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  3. #3
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    Quote Originally Posted by superjen View Post
    I'm not really sure what to name this, and im pretty sure it belongs here, im sorry if it doesnt.

    Question

    The number of loaves of white bread demanded daily at a bakery is normally distributed with mean 6800 loaves and variance 84000. the company decides to produce a sufficient number of loaves so that it will fully supply demand on 95% of the days.

    (a) How many loaves of bread should the compan produce?

    I know i need to find n = CI * standard dev. / E .. i can figure out CI and SD. but i cant find E..

    (b) Based on (a), on what percentage of days will the company be left with more than 500 loaves of unsold bread?

    I'm not sure on how to do this one!


    Any amount of help would be amazing!!
    (a) Let Pr(Z > z*) = 0.05 and Pr(N > n*) = 0.05. Then z^* = \frac{n^* - 6800}{\sqrt{84000}}. Your job is to get z* and then solve for n*.

    (b) Calculate Pr(N > n* + 500) and multiply the answer by 100.
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