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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?

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?

(a) = .1586

(b) = .8414

(c) = .000186588 (lost on this one?)

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.

>> (b) = .8414

Correct.

>> (c) = .000186588 (lost on this one?)

If you have 850 cups, and there is a probability of 0.1586 for each them to overflow, then how many of the 850 cups are expected to overflow ?

Correct. Rounding is not necessary (and if it is, then rounding up or down depends on what the value will be used for).

also, distribution is binomial, so

variance is (n.p.(p-1)) = 113.43

so s.d for number of cup overflows = 10.65, so for 3 standard deviations either side you've got between 102 and 167 failures being likely.

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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 ...