Question

Word problem with math/statistics

Asked by: johntmcii

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?

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Asked On
2008-10-29 at 08:04:06ID23858192
Tags

Math / statistics

Topic

Math & Science

Participating Experts
3
Points
500
Comments
10

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Answers

 

by: Infinity08Posted on 2008-10-29 at 08:23:03ID: 22832135

>> (a) Estimate the probability that the machine will overflow an 8 oz cup
>>  = .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 ...

 

by: johntmciiPosted on 2008-10-29 at 08:36:47ID: 22832335

so would you mind checking my work please?

(a) = .1586

(b) = .8414

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

 

by: aburrPosted on 2008-10-29 at 08:37:23ID: 22832341

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.
---
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.

 

by: johntmciiPosted on 2008-10-29 at 08:40:35ID: 22832367

This is all regarding normal probability distributions (normal curves)

 

by: Infinity08Posted on 2008-10-29 at 08:47:24ID: 22832436

>> (a) = .1586

>> (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 ?

 

by: johntmciiPosted on 2008-10-29 at 08:49:08ID: 22832462

850*.1586 = 134.81 or 135?

 

by: Infinity08Posted on 2008-10-29 at 08:52:43ID: 22832507

>> 850*.1586 = 134.81

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

 

by: deightonPosted on 2008-10-29 at 09:08:15ID: 22832700

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.

 

by: deightonPosted on 2008-10-29 at 09:09:45ID: 22832714

,although if people wanted not to round, it'd be fine with me.

 

by: johntmciiPosted on 2008-10-29 at 09:14:16ID: 22832768

now i am confused again

20120131-EE-VQP-002

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