Question

A machine that lls 20oz bottles of pop doesn't actually put exactly 20oz of pop into each bottle, but rather,
the amount of pop in each bottle is actually random, with a Normal Distribution. Operators of the machine
can choose the mean of the Normal Distribution but the variance is xed at 0.09oz2.
(a) If the mean is set to 20oz, what fraction of all bottles will contain less than 20oz of pop?
(b) If the mean is set to 20oz, what is the probability that a random chosen bottle will contain less that 19.7oz
of pop?
(c) If the operator wants only 2.5% of all bottles to contain less that 20oz of pop, what should the operator
select to be the mean?

Answer 1

a) As the normal distribution is symmetric about its mean, therefore the percentage of bottles that contain less than 20oz that is less than the mean value would be given as 50%. Therfore 50% is the required percentage here.

b) The probability that the bottle contain less than 19.7 oz is computed here as:
P(X < 19.7)

Converting it to a standard normal variable, we have here:

Getting it from the standard normal tables, we have here:

Therefore 0.1587 is the required probability here.

c) Let the mean selected be K here.

From standard normal tables, we have here:
P( Z < -1.96) = 0.025

Therefore the z score for 20 here should be 1.96

So the mean is computed here as:

therefore 20.588 oz is the required new mean here.