Question

Shelia's measured glucose level one hour after a sugary drink varies according to the normal distribution with ?=132μ=132 mg/dl and ?=10.5σ=10.5 mg/dl.

What is the level L such that there is probability only 0.05 that the mean glucose level of 4 test results falls above ?L? Give your answer precise to one decimal place.

Answer 1

We have mean = 132 mg/dl and standard deviation = 10.5 mg/dl

we have to find the level L such that there is only 0.05 probability that the mean glucose level of 4 test results fall above L

so, sample size is n = 4

first, we will find the z score corresponding to the probability value of 0.05

So,

using z distribution for z critical value corresponding to the p value of 0.05, we get

z score = 1.645

Now, using the z formula, we can write setting the given values, we get

multiplying both sides by 5.25

we get

1.645*5.25 = (L-132)*(5.25/5.25)...................................................(5.25/5.25 becomes 1)

8.6363 = (L-132)

adding 132 on each side, we get

8.6363 + 132 = L-132+132...............................(-132+132 becomes 0)

140.6363 = L

rounding it to one decimal, we get L = 140.6 mg/dl

So, required level is L = 140.6 mg/dl