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The filling volume of an automatic filling machine, used to fill cans of carbonated drinks.
is distributed normally, with an average of 12.4 fluid ounces and a standard deviation of 0.1 fluid ounce.
(A) What is the probability of the filling volume being less than 12 fluid ounces?
(B) If all cans smaller than 12.1 or larger than 12.6 ounces are rejected. What proportion of cans will be rejected?
(C) Determine the specifications that are symmetrical around the average, that include 99% of all cans.
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(A) Question (A) is to determine the area under the given normal distribution curve on the left of the raw marks 12.
Use your regular calculator like TI-83/84 with its standard function normalcdf
z1 z2 mean SD <<<---=== formatting pattern.
P(X < 12) = normalcdf(-9999, 12, 12.4, 0.1) = 0.0000
It gives very small number less than positive 0 (zero) with 4 decimal zeroes after the decimal point.
Excel standard function NORM.DIST gives the value of P (X < 12) = 3.16712E-05.
(B) Question (B) is to determine the area under the given normal distribution curve out of the raw marks 12.1 and 12.6
Use your regular calculator like TI-83/84 with its standard function normalcdf and find the complementary probability
z1 z2 mean SD <<<---=== formatting pattern.
P(12.1 < X < 12.6) = normalcdf(12.1, 12.6, 12.4, 0.1) = 0.9759.
So, the probability for question (B) is 1 - 0.9759 = 0.0241.
It means that the proportion of cans rejected is 0.0241. ANSWER
Parts (A) and (B) are solved, explained and answered.