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 her glucose level is above 120 milligrams per deciliter (mg/dl) one hour after a sugary drink. Sheila's measured glucose level one hour after the sugary drink varies according to the Normal distribution with μ = 105 mg/dl and σ = 14 mg/dl.

Let X = Sheila's measured glucose level one hour after a sugary drink

(a) P(X > 120) = (Use 3 decimal places)

Suppose measurements are made on 6 separate days and the mean result is compared with the criterion 120 mg/dl.

(b) P(X > 120) = (Use 3 decimal places)

(c) What sample mean blood glucose level is higher than 95% of all other sample mean blood glucose levels? Hint: this requires a backward Normal calculation. (Use 2 decimal places)

Answer 1

Solution:-

μ = 105 mg/dl and σ = 14 mg/dl

a) P(X > 120) = 0.142

x = 120

By applying normal distribution:-

z = 1.07

P(z > 1.07) = 0.1423

b) P(X > 120) = 0.004.

x = 120, n = 6

By applying normal distribution:-

z = 2.624

P(z > 2.624) = 0.0043

c) The sample mean blood glucose level is higher than 95% of all other sample mean blood glucose levels is 128.03.

p-value for the top 5% = 0.95

z-score for the p-value = 1.645

By applying normal distribution:-

x = 128.03