Sheila's measured glucose level one hour after a sugary drink varies according to the normal distribution with μμ = 120 mg/dl and σσ = 20 mg/dl. what is the level l such that there is probability only 0.15 that the mean glucose level of 5 test results falls above l?

Since the sample size is below 30, in this case we use the t statistic. The formula for t score is:

t = (x – u) / (σ / sqrt n)

where,

x = the level l = unknown

u = sample mean = 120 mg / dl

σ = standard deviation = 20 mg / dl

n = sample size or number of results = 5

Using the standard distribution tables for t, we can find the value of t given the probability (P = 0.15) and degrees of freedom (DOF).

t  = 1.036

Going back to the formula for t score:

1.036 = (x – 120) / (20 / sqrt 5)

x = 129.27 mg / dl = l

50.23 i believe is correct but don't quote me on that.