Shelia's measured glucose level one hour after a sugary drink varies according to the normal distribution with μ = 132 mg/dL and σ = 11.3 mg/dL. what is the level L such that there is probability only 0.01 that the mean glucose level of 5 test results falls above L?

134.6

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula. This is X when Z has a p-value of 1-0.01 = 0.99. So it is X when Z = 2.325.

The level is L = 134.6

The Central Limit Theorem estabilishes that, for a random variable X, with mean  and standard deviation, large sample size can be approximated to a normal distribution with a mean and standard deviation

the equation for this situation is y=122+45. to solve the equation you would add 45 and 122 to get a total of 167lbs. in conclusion, grace's weight eight weeks ago was 167lbs.  

answer: 94

step-by-step explanation: 79+91+92+94+94 = 450 divided by 5 (since there are 5 number that are being added) = 90