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23 January, 12:39

The weight of oranges growing in an orchard is normally distributed with a mean weight of 6 oz. and a standard deviation of 0.5 oz. Using the empirical rule, determine what interval would represent weights of the middle 95% of all oranges from this orchard.

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  1. 23 January, 12:59
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    The interval that would represent weights of the middle 95% of all oranges from this orchard is from 5 oz to 7 oz.

    Step-by-step explanation:

    The Empirical Rule states that, for a normally distributed random variable:

    68% of the measures are within 1 standard deviation of the mean.

    95% of the measures are within 2 standard deviation of the mean.

    99.7% of the measures are within 3 standard deviations of the mean.

    In this problem, we have that:

    Mean = 6

    Standard deviation = 0.5

    Middle 95% of weights:

    By the Empirical Rule, within 2 standard deviations of the mean.

    6 - 2*0.5 = 5

    6 + 2*0.5 = 7

    The interval that would represent weights of the middle 95% of all oranges from this orchard is from 5 oz to 7 oz.
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