johntmcii
asked on
Word problem with math/statistics
Any help here?
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?
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ASKER
This is all regarding normal probability distributions (normal curves)
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ASKER
850*.1586 = 134.81 or 135?
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well 134.81 don't make sense for this case where there are 850 actual cups, so I'd give the answer as 135.
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.
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.
,although if people wanted not to round, it'd be fine with me.
ASKER
now i am confused again
ASKER
(a) = .1586
(b) = .8414
(c) = .000186588 (lost on this one?)