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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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 117, \sigma = 10.6, n = 6, s = \frac{10.6}{\sqrt{6}} = 4.33[/tex]
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.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]2.33 = \frac{X - 117}{4.33}[/tex]
[tex]X - 117 = 2.33*4.33[/tex]
[tex]X = 127.1[/tex]
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.
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:
[tex]z=\frac{x-\mu}{\sigma} \\\\x=raw\ score,\mu=mean,\sigma=standard\ deviation[/tex]
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:
[tex]\frac{L-117}{10.6} =2.33\\\\L-117=24.7\\\\L=141.7[/tex]
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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