Question

Sheila's measured glucose level one hour after a sugary drink varies according to the Normal distribution with μμ = 120 mg/dl and σσ = 20 mg/dl. What is the level L such that there is probability only 0.15 that the mean glucose level of 4 test results falls above L?

Answer 1

solution

Given that,

mean = = 120

standard deviation = = 20

n = 4

= 120

= / n =20 / 4=10

Using standard normal table,

P(Z > z) =0.15

= 1 - P(Z < z) = 0.15

= P(Z < z ) = 1 - 0.15

= P(Z < z ) = 0.85

= P(Z < 1.04 ) = 0.85  

z = 1.04 ( using z table)

Using z-score formula  

= z * +   

= 1.04 *10+120

= 130.4

=130

L==130