Question 1189095: A soft drink machine is regulated so that it discharges an average of 200 milliliters per cup. If the
amount of drink is normally distributed with a standard deviation equal to 15 millimeters, a) what fraction of the cups will contain more than 240 milliliters?
b) c) what is the probability that a cup contains between 191 and 209 milliliters? how many cups will likely overflow if 230 milliliters cups are used for the next 1000 di
d)
below what value do we get the smallest 25% of the drinks?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! z(240)>(240-200/15 or +2.67
probability z> 2.67=0.0038
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between 191 and 209 is the same way.
calculator 2ndVARS(191,209,200,15) and that will equal twice the probability of z=9/15 or -.6
=0.4515.
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230 ml cups are two sd over or 0.02275 probability, so 22.75 or 23 cups would be expected to overflow
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z(0.25)=-0.6745=(x-200)/15
-10.12=x-200
x=189.88 or 190 ml
Note: there is an error in the problem. The SD is 15 milliliters. All units are in ml.
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