Vending Machine: Soft drinks
u (mean) = 7.6 ; o(sd)= .4
(a) Estimate the probability that the machine will overflow an 8 oz cup
= .1586 probability that the machine will overflow an 8 oz c.
(b) Estimate the probability that the machine will not overflow an 8 oz cup
= .8585 probability that the machine will not overflow an 8 oz cup
(c) The machine just been loaded with 850 cups. How many of these do you expect will overflow when served?
There are a few minor difficulties with the numbers. u is the mean of what?
The probability of not overflowing and of overflowing do not add up to 1.
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The probability of an overflow is 0.15. That is, if 100 cups were filled the most likely number of overflows would be 15. So one would expect, in your case, 850 x 0.1586 or 135 to overflow. What a mess. The janitor will insist that the machine be recallibrated.
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>> = .1586 probability that the machine will overflow an 8 oz c.
Correct. Using the cumulative distribution function for normal distributions :
Phi(x) = [ 1 + erf((x - mu) / (sigma * sqrt(2))) ] / 2
with x = 8, mu = 7.6 and sigma = 0.4
>> (b) Estimate the probability that the machine will not overflow an 8 oz cup
>> = .8585 probability that the machine will not overflow an 8 oz cup
This is the inverse of the result from (a), so 1 - 0.1586 = ...
>> (c) The machine just been loaded with 850 cups. How many of these do you expect will overflow when served?
Use the probability calculated in (a) for this ...