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Math Help - Help please midterm tomorrow!

  1. #1
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    Help please midterm tomorrow!

    I need help answering this question please:

    A machine fills containers with a mean weight per container of 16.0 oz. If no more than 5% of the containers are to weigh less than 15.8 oz, what must the standard deviations of the weight equal?

    Any help will be much appreciated!
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  2. #2
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    What is the total distance between 15.8 and 16? On a normal curve, what is the score that corresponds to a chance of 0.05? That should help you get started, or jumpstart whatever work you have done so far. Have you tried anything - your other problem mentions "normal" curve, so I would imagine this is a "normally distributed" machine.
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  3. #3
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    This is all the information he gave for the problem, he wasnt very clear :/ . Yes, I think this problem has a normal distribution.
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  4. #4
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    Since the question asks about placing a bound on the variance ( stdev^2) given a mean and a probability, I immediately thought of CHEBYSHEV. (Note usually, Chebyshev is used via a given mean and variance, and one places bounds on the probability.)

    While he provides bounds on the probabilities of random variables of any distribution given a mean and a variance, since we know the probability and the mean we can effectively place bounds on the variance.

    So first of all here is the statement of his inequality:

    <br />
If \; E[X] = \mu, \; Var(X) = \sigma^2, \;then\;for\;any\;a>0,
    P({X\geq\mu+a})\leq\frac{\sigma^2}{\sigma^2+a^2}
    P({X\geq\mu-a})\leq\frac{\sigma^2}{\sigma^2+a^2}


    So in our case, define the random variable X to be difference between 16oz and the realized weight of the container. Clearly, given the wording of the problem, we may take E[X]=0
    In our case we want
    <br />
P({X>(0+(16-15.8)})<.05

    Solving\;for\;\sigma\; \Rightarrow

    (.05)=\frac{\sigma^2}{\sigma^2+.2^2}

    \Rightarrow

    \sigma^2=.010526\;or\;\sigma = .102598

    Note that if you use the Normal distribution as an approximation of the distribution of your random variable, you get \sigma<.12159


    Chebyshev delivers the tighter bounds. Intuitively this makes sense, since it makes no assumption about the distribution of the random variable, other than its having finite moments.
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