Question

Sheila'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 (mg/dl) one hour after a sugary drink is ingested. Sheila's measured glucose level one hour after ingesting the sugary drink varies according to the Normal distribution with μ = 123 mg/dl and σ = 10 mg/dl. What is the level L such that there is probability only 0.05 that the mean glucose level of 2 test results falls above L for Sheila's glucose level distribution? (Round your answer to one decimal place.)

          = mg/dl

Answer 1

Using central limit theorem,

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

So.

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

L - / / sqrt(n) = 1.645

L - 123 / 10 / sqrt(2) = 1.645

Solve for L

L = 134.6 mg / dL