Question

ch 15. #10 (15.31) Shelia's measured glucose level one hour after a sugary drink varies according to the Normal distribution with μ = 116 mg/dl and σ = 10.4 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?

L = ????

Answer 1

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

We have to find L such that

P( > L ) = 0.01

That is

P( Z > L - 116 / 10.4 / sqrt(4) ) = 0.01

Find z score in z table for probability 0.01 (Right tailed), we get z score = 2.3263

Therefore,

L - 116 / 10.4 / sqrt(4) = 2.3263

L - 116 / 5.2 = 2.3263

L - 116 = 12.0968

L = 128.0968