Question

(1 point) Sheila's measured glucose level one hour after a sugary drink varies according to the Normal distribution with μμ = 130 mg/dl and σσ = 20 mg/dl. What is the level L such that there is probability only 0.1 that the mean glucose level of 4 test results falls above L?

L=

Answer 1

solution

Given that,

mean = = 130

standard deviation = = 20

n = 4

= 130

= / n =20 / 4=10

Using standard normal table,

P(Z > z) =0.1

= 1 - P(Z < z) = 0.1  

= P(Z < z ) = 1 - 0.1

= P(Z < z ) = 0.9

= P(Z < 1.28 ) = 0.9  

z = 1.28 ( using z table)

Using z-score formula  

= z * +   

= 1.28 *10+130

= 142.8

=143

L==143