Asked by Shiro
The glucose concentration of adults from Municipal X is normally distributed with mean of 75 mg/dL and standard deviation of 18 mg/dL. What is the proportion of the 30 households from Barangay Y with glucose concentration more than 105 mg/dL?
Identify what is given: μ = _ σ = _ x = _
Using the same data from number 1, construct a 90% confidence interval for x.
Identify what is given:
n = __ σ = __ x = __ z-deviate = __
LL = __
UL = __
Identify what is given: μ = _ σ = _ x = _
Using the same data from number 1, construct a 90% confidence interval for x.
Identify what is given:
n = __ σ = __ x = __ z-deviate = __
LL = __
UL = __
Answers
Answered by
R_scott
105 is 30 above the mean ... or 30/18 standard deviations
this is a z-score of 5/3 , or 1.667
use a z-score table to find the portion of the population above this score
this is a z-score of 5/3 , or 1.667
use a z-score table to find the portion of the population above this score
Answered by
Shiro
how about the second one?
Answered by
boyet
n = 30 σ = 18 x = 75 z-deviate = 1.6
LL = 69.5
UL = 80.4
LL = 69.5
UL = 80.4
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