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Math Help - water carton problem

  1. #1
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    dfsdfdf
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    Question water carton problem

    Any help offered in solving the below question will be much appreciated.
    Thanks
    Kingman

    The volume of water in milliliters in cartons is normally distributed with mean 't' and the standard deviation 8. Measurements were taken of the volume in 900 of these cartons and it is found that 225 of them contained more then 1002 milliliters.
    1) Calculate the value of 't' (0.674)
    2) Three of them 900 cartons are chosen at random. Calculate the probability that exactly 2 of them contain more than 1002 milliliters. (0.140)
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  2. #2
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    1. Okay, to find \mu:

    Let X= the number of mL of water in a carton
    pr(X\geq 1,002)=\frac{225}{900}=0.25

    This is given in the problem.

    Take note that pr(X\geq 1,002)=0.25 is the same as pr(X\leq 1,002)=0.75

    Now standardize the distribution.

    pr(X\leq 1,002)=0.75= pr(Z\leq \frac{1,002-\mu}{8})=0.675 (Taken from z-table)

    Now, solve for \mu and you should get \mu = 996.6

    2. This is a simple binomial distribution:

    X~binomial(n=3,p=0.25)

    X= the number of cartons containing more than 1,002 mL of water.

    {n\choose x}(p)^{x}(1-p)^{(n-x)}

    pr(X=2)= {3\choose 2}(0.25)^{2}(0.75)\approx 0.1406

    Hmm.. I now see that you gave the answers in parentheses. I don't understand why (1) is 0.674. It's approximately the same number I took from the z-table but the problem states that we're trying to find \mu not Z. Pretty weird. Maybe someone can correct me on that.
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  3. #3
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    Thanks

    Yes there is a typo error in the question.you are right .
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