Question

Please explain using EXCEL

Suppose an expectant mother’s doctor is concerned that the patient may suffer from gestational diabetes (high blood glucose levels during pregnancy).    A patient is considered to have gestational diabetes if her glucose level is above 140 milligrams per deciliter (mg/dl) one hour after the intake of a sugary drink. Assume the patient’s measured glucose level one hour after taking a sugary drink is distributed normally with μ = 125 mg/dl and σ = 10 mg/dl.

1.            If a single glucose measurement is made to calculate this mean of 125, what is the probability that the patient is diagnosed as having gestational diabetes? (That is, what is the probability that a measurement for this patient would equal or exceed 140 mg/dl?) (Round to four digits.)

2.           If, instead, the attending physician takes measurements on three separate days and computes a mean measurement of 125 mg./dl., which, yes, happens to be the same value found in (a).   Compute the probability that the patient is diagnosed as having gestational diabetes, which, again, means a glucose measurement of 140 mg/dl. (Round to four digits.)

3.            What is the corresponding glucose level that corresponds to a right-tail probability of 5% (.05) for a sample mean of 125 calculated from three observations? (Round to one decimal place.)

Answer 1

Here the patient’s measured glucose level is normally distributed with
XN(125,102)

a)The probability is found using the following EXCEL function.

P(X>140)

=1-NORM.DIST(28000,25000,4000,TRUE) = 0.0668
b)Here the measured glucose level $\overline{X}$ is normally distributed with $\overline{X}\sim N(125,10^2/3)$

The probability $P(\overline{X}\geqslant 140)$ is found using the following EXCEL function.

=1-NORM.DIST(140,125,10/SQRT(3),TRUE) = 0.0047

c) Inverse (Quantile of the distribution is found below). Right probability of 5% means 95% quantile. $\Phi ^{-1}\left ( 0.95,\mu =125,\sigma =10/\sqrt{3} \right )$ is found using the following EXCEL function.

=NORMINV(0.95,125,10/SQRT(3)) =  134.5