Packages of sugar bags for Sweeter Sugar Inc. have an average weight of 16 ounces and a standard deviation of 0.3 ounces. The weights of the

Question

Packages of sugar bags for Sweeter Sugar Inc. have an average weight of 16 ounces and a standard deviation of 0.3 ounces. The weights of the sugar packages are normally distributed. What is the probability that 9 randomly selected packages will have an average weight in excess of 16.025 ounces

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Adalynn 2 weeks 2021-09-15T11:23:20+00:00 1 Answer 0

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    2021-09-15T11:24:33+00:00

    Answer:

    40.13% probability that 9 randomly selected packages will have an average weight in excess of 16.025 ounces

    Step-by-step explanation:

    To solve this question, we have to understand the normal probability distribution and the central limit theorem.

    Normal probability distribution:

    Problems of normally distributed samples can be solved using the z-score formula.

    In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

    Z = \frac{X - \mu}{\sigma}

    The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

    Central limit theorem:

    The Central Limit Theorem estabilishes that, for a random normally distributed variable X, with mean \mu and standard deviation \sigma, the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

    In this problem, we have that:

    \mu = 16, \sigma = 0.3, n = 9, s = \frac{0.3}{\sqrt{9}} = 0.1

    What is the probability that 9 randomly selected packages will have an average weight in excess of 16.025 ounces

    This is 1 subtracted by the pvalue of Z when X = 16.025. So

    Z = \frac{X - \mu}{\sigma}

    By the Central Limit Theorem

    Z = \frac{X - \mu}{s}

    Z = \frac{16.025 - 16}{0.1}

    Z = 0.25

    Z = 0.25 has a pvalue of 0.5987

    1 – 0.5987 = 0.4013

    40.13% probability that 9 randomly selected packages will have an average weight in excess of 16.025 ounces

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