Question

Shelia's measured glucose level one hour after a sugary drink varies according to the Normal distribution with µ = 118 mg/dl and s = 10 mg/dl.

What is the level L (±0.1) such that there is probability only 0.01 that the mean glucose level of 4 test results falls above L?   

Answer 1

Solution,

Given that,

mean = = 118 mg/ dl

standard deviation = = 10 mg/ dl

n = 4

= 118 mg/ dl

= / n = 10 / 4 = 5 mg /dl

Using standard normal table,

P(Z > z) = 0.01

= 1 - P(Z < z) = 0.01  

= P(Z < z ) = 1 - 0.01

= P(Z < z ) = 0.99

= P(Z < 2.33 ) = 0.99  

z = 2.33

Using z-score formula  

= z * +   

= 2.33 * 5 + 118

= 129.65 mg /dl