Question

Shelia's doctor is concerned that she may suffer from gestational diabetes (high blood glucose levels during pregnancy). There is variation both in the actual glucose level and in the blood test that measures the level. A patient is classified as having gestational diabetes if the glucose level is above 140 milligrams per deciliter one hour after a sugary drink is ingested. Shelia's measured glucose level one hour after ingesting the sugary drink varies according to the Normal distribution with mean 129 mg/dl and standard deviation 9 mg/dl. Let L denote a patient's glucose level.

If measurements are made on three different days, find the level LL such that there is probability only 0.02 that the mean glucose level of three test results falls above L for Shelia's glucose level distribution. What is the value of L?

Answer 1

To determine the glucose level L that falls above the top 2% of the mean glucose level distribution of three tests for Shelia, calculate the standard error of the mean and use the Z-value for the 98th percentile. The resulting L is approximately 139.5 mg/dl.

Step-by-step explanation:

To find the glucose level L that corresponds to the top 2% of the distribution of means for three glucose tests (mean glucose level of three test results), we must first determine the standard error of the mean for Shelia's glucose level distribution, and then find the Z-value that corresponds to the 98th percentile due to our interest in the top 2%.

Since Shelia's measured glucose level varies normally with mean (μ) of 129 mg/dl and a standard deviation (σ) of 9 mg/dl, and given the number of tests (n) is 3, the standard error of the mean (SEM) would be:

SEM = σ / √n

SEM = 9 / √3

SEM = 9 / 1.732

SEM = 5.2 mg/dl (approximately)

To find the correct Z-value for the 98th percentile, we use standard Z-tables or a normal distribution calculator. The Z-value corresponding to the 0.98 probability (or the top 2%) is typically around 2.05.

Thus, we calculate L using the following formula:

L = μ + (Z * SEM)

L = 129 + (2.05 * 5.2)

L ≈ 139.5 mg/dl

Therefore, there is only a 0.02 probability that the mean glucose level of three test results falls above approximately 139.5 mg/dl for Shelia's glucose level distribution.