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 127 mg/dl and standard deviation 9 mg/dl. Let LL denote a patient's glucose level.

(a) If measurements are made on four different days, find the level LL such that there is probability only 0.05 that the mean glucose level of four test results falls above LL for Shelia's glucose level distribution. What is the value of LL?

(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

Given,

= 127, = 9

The central limit theorem is

P( < x) = P( Z < x - / ( / sqrt(n) )

a)

We have to calculate L such that P( > L) = 0.05

Therefore,

P( > L) = 0.05

P( Z > L - / ( / sqrt(n) ) = 0.05

P( Z < L - / ( / sqrt(n) ) = 0.95

From the Z table, z-score for the probability of 0.95 is 1.645

Therefore,

L - / ( / sqrt(n) = 1.645

Put the value of , and n in above expression and solve for L.

L - 127 / (9 / sqrt(4 ) ) = 1.645

L - 127 = 1.645 * (9 / sqrt(4 ) )

= 134.4025

Value of L = 134.40

b)

Using central limit theorem,

P( < x) = P( Z < x - / ( / sqrt(n) )

P( < 140) = P( Z < 140 - 127 / (9 / sqrt(4) ) )

= P( Z < 2.89)

= 0.9981