Shelia’s measured glucose level one hour after a sugary drink varies according to the normal distribution with μ = 117 mg/dl and σ = 10.6 mg

Question

Shelia’s measured glucose level one hour after a sugary drink varies according to the normal distribution with μ = 117 mg/dl and σ = 10.6 mg/dl. What is the level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L ? Give your answer precise to one decimal place.

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Kennedy 3 weeks 2021-12-31T02:25:21+00:00 1 Answer 0 views 0

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    2021-12-31T02:26:44+00:00

    Answer:

    The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

    Step-by-step explanation:

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

    Normal probability distribution

    Problems of normally distributed samples are 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 variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}

    In this problem, we have that:

    \mu = 117, \sigma = 10.6, n = 6, s = \frac{10.6}{\sqrt{6}} = 4.33

    What is the level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L ?

    This is the value of X when Z has a pvalue of 1-0.01 = 0.99. So X when Z = 2.33.

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

    By the Central Limit Theorem

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

    2.33 = \frac{X - 117}{4.33}

    X - 117 = 2.33*4.33

    X = 127.1

    The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

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