Question 202628
In an effort to reduce the number of bottles that contain less than 1.90 liters, the bottler sets the filling machine so the mean is 2.02 liters and a standard deviation of 0.05 liters. Under these circumstances, please answer below. 
a. P(Between 1.90 and 2.0 liters)
Find the z-values for 1.9 and 2
z(1.9) = (1.9-2.02)/0.05 = -2.4
z(2) = (2-2.02)/0.05 = -0.4
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P(1.9 < x < 2) = P(-2.4 < z < -0.4) = 0.3364
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b. Between 1.90 and 2.10 liters
Use the same procedure as in part "a" to get
P(1.9 < x < 2.1) = 0.9370
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c. Below 1.90 liters or above 2.10 liters
Use the results of part "b" to get:

P(x < 1.9 or x > 2.1) = 1 - 0.9370 = 0.0630
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d. 99% of the bottles contain at least how much soft drink?
Find the z-value of 0.99 by using InVNorm(0.99) = 2.3263
Find the corresponding "x" value using x = z*sigma + u
x = 2.3263*0.05 + 2.02 = 2.1363 liters
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e. 99% of the bottles contain an amount that is between which two values (symmetrically distributed) around the mean
Comment: If 99% is distributed around the mean, each tail
has 0.005 or 0.5%
Find the z-value of 0.005  and use x = z*sigma + u to find the x-values.
InVNorm(0.005) = -2.5758
x = -2.5758*0.05+2.02 = 1.8912 liters
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InVNorm(.995) = =2.5758
x = 2.5758*0.05+2.02 = 2.1488 liters
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Cheers,
Stan H.