Mathematics, 13.02.2020 19:58 1462264

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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Answer from: dprajapati832000

There is a probability of 0.01 that the mean glucose level of 6 test results falls above 141.7 mg/dl

Z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:

$z=\frac{x-\mu}{\sigma} \\\\x=raw\ score,\mu=mean,\sigma=standard\ deviation$

Given that mean(μ) = 117 mg/dl and standard deviation (σ) = 10.6 mg/dl

The z score that corresponds to an area to the right of 0.01 is 2.33.

Hence:

$\frac{L-117}{10.6} =2.33\\\\L-117=24.7\\\\L=141.7$

Hence There is a probability of 0.01 that the mean glucose level of 6 test results falls above 141.7 mg/dl

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Answer from: brianna5626

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.

Answer from: Quest

29 mil

step-by-step explanation:

Answer from: Quest

There is no graph to go along with this

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