Question

Shelia's measured glucose level one hour after a sugary drink varies according to the normal distribution with ?=117μ=117 mg/dl and ?=10.6σ=10.6 mg/dl.

What is the level ?L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above ?L? Give your answer precise to one decimal place.

?=

Answer 1

We have to calculate L such that P( > L) = 0.01

Using central limit theorem,

P( < x) = P( Z < x - / / sqrt(n) )

So,

P( > L) = 0.01

P( Z > L - 117 / 10.6 / sqrt(6) ) = 0.01

P( Z < L - 117 / 10.6 / sqrt(6) ) = 0.99

From the Z table, z-score for the probability of 0.99 is 2.3263

L - 117 / ( 10.6 / sqrt(6) ) = 2.3263

Solve for L

L = 127.1