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 130 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 = 115 mg/dl and σ = 14 mg/dl. *I would really appreciate a step by step to solve these problems*

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

(a) P(X > 130) = (Use 3 decimal places)

Suppose measurements are made on 3 separate days and the mean result is compared with the criterion 130 mg/dl.

(b) P(X > 130) = (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

It is given that Sheila's Glucose level aftet an hour of sugary drink is Normal with mg/dl and mg/dl.

We know that if X~Normal, has a standard normal variate.

(a). P(X>130)=?

When X=130, we have

  

(b). The measurements are made with three sepearte days. The mean will have a distribution with . Now will have a std normal variate.

When

Therefore

  

  

(c). What sample mean blood glucose level is higher than 95% of all other sample mean blood glucose levels?

We need to find that value of a such that P(Z>a)=0.05

We get this value as 1.6449 (using NORM.S.INV(.95) function of EXCEL).

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