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 125 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 = 110 mg/dl and σ = 14 mg/dl.

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

(a) P(X > 125) =  (Use 3 decimal places)

Suppose measurements are made on 3 separate days and the mean result is compared with the criterion 125 mg/dl.

(b) P(X > 125) =  (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

Answer :

given data :-

he Normal distribution with

µ = 110
σ = 14
right tailed

(a). now we need to find out the P(X > 125)

we know that,

X = (X > 125)
Z = (X - µ ) / σ

= (125-110)/14

= 15/14

= 1.071

P(X ≥ 125 ) = P(Z ≥ 1.071 )  

= P ( Z < - 1.071 ) = 0.142

(b). now we need to find out the  P(X > 125)
n= 3
right tailed
P(X > 125)
Z = (X - µ )/(σ/√n)

= (125-110)/14/√3

= 15/(14/1.732)

= 15/8.082

= 1.855

P(X ≥ 125) = P(Z ≥ 1.855)

= P ( Z < -1.855 ) = 0.032

(c). now we need to find out the sample mean blood glucose level is higher than 95% of all other sample mean blood glucose levels...?
proportion= 95%

= 95/100

= 0.95

Z value at 0.95 = 1.645 ( from z table )

we know that,
z=(x-µ)/σ/√n
so,

X=zσ/√n+µ

= 1.645 * 14/√3 + 110

= 1.645 * 14/1.732+110

= 1.645 * 8.0882+110

= 13.29+110

= 123.296
X = 123.29 mean blood glucose level is higher than 95% of all other

sample mean blood glucose levels = 123.29