Question

Sheila's measured glucose level one hour after a sugary drink varies according to the Normal distribution with μ = 130 mg/dl and σ = 15 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 ,

= 130

= 15

A sample of size n = 4 is taken from this population.

Let be the mean of sample.

The sampling distribution of the is approximately normal with

Mean = = 130

SD =      = 15/​4 = 7.5

For level L , P( > L ) = 0.15

For z , P(Z > z) = 0.15

P(Z < z) = 1 - 0.15

P(Z < z) = 0.85

From z table , P(Z < 1.036) = 0.95

z = 1.036

Using z score formula

L = + (z * ) = 130 + (1.036 * 7.5) = 137.77

Answer : Required L = 137.77