Question

Shelia's measured glucose level one hour after a sugary drink varies according to the normal distribution with μ = 116 μ=116 mg/dl and σ = 11.3 σ=11.3 mg/dl. What is the level L L such that there is probability only 0.01 that the mean glucose level of 2 test results falls above L L ? Give your answer precise to one decimal place.

Answer 1

Answer:

The level is L = 134.6

Step-by-step explanation:

To solve this question, 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 normally distributed random 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:

$\mu = 116, \sigma = 11.3, n = 2, s = \frac{11.3}{\sqrt{2}} = 7.99$

What is the level L L such that there is probability only 0.01 that the mean glucose level of 2 test results falls above L?

This is X when Z has a pvalue of 1-0.01 = 0.99. So it is X when Z = 2.325.

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

By the Central limit theorem

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

$2.325 = \frac{X - 116}{7.99}$

$X - 116 = 2.325*7.99$

$X = 134.6$

The level is L = 134.6