Question

What is the value of L?

Answer 1

Each sample of Shelia's glucose comes from a normal distribution with mean 128 and standard deviation 10. The average of three samples is normal with the same mean and a standard deviation of
`(10)/(\sqrt 3)`.

By the 68-95-99.7 rule a one sided probability of 2% is a bit more than two standard deviations above the mean (2.5%). We look up 1-.02=.98 in the normal table that's the integral of the standard normal from 0 to z and get z=2.06.

For the observed average to have probability 2% it needs to be 2.06 standard deviations above the mean, so

`L = 128 + 2.05 \cdot (10)/(\sqrt 3) \approx 139.8`

Part a: 139.8

For the sample average compared to 140, we compute how many standard deviations 140 is above our mean of 128.

`z = (140 - 128)/( 10 / √(3)) =\approx 2.08`

That's virtually the same as the last result, but we look it up and get .98124, so p=1-98124=.019.

Part b: .019