M = mean = 102.1
mu = 101.5
sigma = 1.6
n = 36
z = (102.1 - 101.5)/(1.6/sqrt(36)) = 2.25
P(M < 102.1) = P(z < 2.25) = 0.9878
Aiko L.
asked 08/15/218. Suppose the amount of a popular sport drink in bottles leaving the filling machine has a normal distribution with mean 101.5 milliliters (mL) and standard deviation 1.6. If 36 bottles are randomly selected, find the probability that the mean content is less than 102.1 mL
M = mean = 102.1
mu = 101.5
sigma = 1.6
n = 36
z = (102.1 - 101.5)/(1.6/sqrt(36)) = 2.25
P(M < 102.1) = P(z < 2.25) = 0.9878
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