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 128 mg/dl and standard deviation 8 mg/dl. Let L denote a patient's glucose level.

(a) If measurements are made on four different days, find the level L such that there is probability only 0.01 that the mean glucose level of four test results falls above L for Shelia's glucose level distribution. What is the value of L?


(b) If the mean result from the four tests is compared to the criterion 140 mg/dl, what is the probability that Shelia is diagnosed as having gestational diabetes?

Answer 1

The level L with a probability of only 0.01 that the mean glucose level of four tests falls above it is 137.32 mg/dl for Shelia's glucose level distribution. The chance of Shelia being diagnosed with gestational diabetes based on a mean value of 140 mg/dl from four tests is 0.13%.

Step-by-step explanation:

Shelia's concern about potentially suffering from gestational diabetes involves interpreting her blood glucose levels that vary according to a normal distribution. With a known mean and standard deviation, we can calculate probabilities related to diabetes and blood-sugar tests.

Part (a)

To find the level L such that there is only a 0.01 probability the mean glucose level of four test results falls above L, we assume the distribution of the mean of four test results also follows a normal distribution. The standard deviation of the mean of four tests is the standard deviation of a single test divided by the square root of four, which is 8 / √4 = 4 mg/dl.

Using a standard normal table or a Z-score calculator, we find the Z-score that corresponds to a cumulative probability of 0.99, which is approximately 2.33. The value of L is then:

L = mean + (Z × standard deviation of the mean)
L = 128 + (2.33 × 4)
L = 128 + 9.32
L = 137.32 mg/dl

Therefore, there is only a 0.01 probability that the mean of four test results for Shelia's glucose level falls above 137.32 mg/dl.

Part (b)

To calculate the probability that Shelia is diagnosed with gestational diabetes if the mean result from the four tests is compared to the criterion of 140 mg/dl, we need to find the Z-score for 140 using the standard deviation of the mean:

Z = (X - mean) / standard deviation of the mean
Z = (140 - 128) / 4
Z = 3

Looking up the Z-score of 3 in standard normal tables or using a calculator, we find the area to the left of Z is 0.9987. Hence, the probability to the right of Z (being diagnosed with gestational diabetes) is 1 - 0.9987, which equals 0.0013 or 0.13%.

So, there is a 0.13% chance that Shelia will be diagnosed with gestational diabetes based on the mean of the four tests.