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  1. #1
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    lost:(

    A soft-drink machine can be regulated so that it discharges an average of x ounces per cup. If the ounces of fill are normally distributed with standard deviation 0.3 ounce, give the setting for x so that 8-ounce cups will overflow only 1% of the time.


    from the normal distribution table, the probability of X where X is greater than x is always greater than 0.5. how do i find a value x such that the probability of it overflowing is only 1%?
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  2. #2
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    Quote Originally Posted by alexandrabel90 View Post
    A soft-drink machine can be regulated so that it discharges an average of x ounces per cup. If the ounces of fill are normally distributed with standard deviation 0.3 ounce, give the setting for x so that 8-ounce cups will overflow only 1% of the time.


    from the normal distribution table, the probability of X where X is greater than x is always greater than 0.5. how do i find a value x such that the probability of it overflowing is only 1%?
    Hi alexandrabel90,

    The part in blue is the probability that Z\le X when X\ge x

    for Z=\frac{X-x}{0.3}

    In this case you need to work with the region in the tails of graph of the normal curve.

    The 0.5% to the right of 99.5% added to the first 0.5%
    constitutes the 1% region in a 2-tailed situation,
    or the first 1% or last 1% (9% to 100%) constitutes the 1% region
    in a 1-tail situation.
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