In: Math
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 = 12 mg/dl.
Let X = Sheila's measured glucose level one hour after a sugary drink
(a) P(X > 130) =
Suppose measurements are made on 3 separate days and the mean result is compared with the criterion 130 mg/dl. (b) P(X > 130) =
(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)
a)
µ = 115
σ = 12.0
right tailed
X ≥ 130
Z = (X - µ ) / σ = 1.25
P(X ≥ 130 ) = P(Z ≥
1.25 ) = P ( Z <
-1.25 ) =
0.1056(answer)
excel formula : =NORMSDIST(-1.25)
b)
µ = 115
σ = 12
n= 3
right tailed
X ≥ 130
Z = (X - µ )/(σ/√n) = 2.165
P(X ≥ 130 ) = P(Z ≥
2.165 ) = P ( Z <
-2.165 ) = 0.0152
c)
µ = 115
σ = 12
proportion= 0.95
Z value at 0.95 =
1.645 (excel formula =NORMSINV(α))
z=(x-µ)/σ/√n
so, X=zσ/√n+µ= 1.645 * 12/√3
+ 115
X = 126.3958802
mean blood glucose level is higher than 95% of all other sample mean blood glucose levels = 126.40 mg/dl