Probability Question For those whizkids out ther
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July 17, 2002 at 4:10 pm #29892
PROBMANParticipant@PROBMAN Include @PROBMAN in your post and this person will
be notified via email.The length of time for a part to be stamped is a random variable having an exponential distribution with a mean of four minutes.
What is the probability that a part is stamped in less than 3 minutes on at least 4 of the next 6 days?
Can anyone show me how to complete this problem?0July 17, 2002 at 4:42 pm #77316Take a look at the history of the outcomes. One way to look at it is that the long term relative frequency in effect becomes its probability.
0July 17, 2002 at 4:47 pm #77318
Ovidiu ContrasParticipant@OvidiuContras Include @OvidiuContras in your post and this person will
be notified via email.You can transform your data ,to get a normal distribution .Take advantage from the normal distribution and calculate your probabilities according to Z and the Standard Normal Table , for your transformed limits….
Hope this helps0July 17, 2002 at 5:27 pm #77323PROBMAN,
I will give it my best shot. Assuming, you do get at least 4 days of 6, there are 15 possible combinations of getting those 4 days of 6. On any given day, the probability of the time being less than 3 minutes is 0.433 (looked at poisson table). Therefore, the probability of being > 3 minutes for any day is .567 (1.433). Assuming that it takes you only 4 days out of 4, at best your probability is .433^4 or 3.5%. Worst case it takes you all 6 days to get 4 days that are conforming, your probability will be .567^2 * .433^4 for a probability of 1.1%.
To make a long story short, ASSUMING you can meet the specified time requirement, your probability of doing so should be between 1.1% and 3.5%.
That is my take on this. Anyone else?
Scott0July 17, 2002 at 5:38 pm #77324
Robert ButlerParticipant@rbutler Include @rbutler in your post and this person will
be notified via email.The exponential probability density function is:
p(x) = Theta*exp(Theta*x)
The average for an exponential distribution is 1/Theta
The variance for an exponential distribution is 1/Theta^2
Thus for a mean of 4 minutes we have
4 = 1/Theta so Theta = .25
The probability that the time for stamping is less than three minutes is
P{X<3} = 1 – 1/(exp(Theta*3) = 1 – 1/(exp(.25*3) = .53
So, regardless of the time interval chosen you can expect to see parts turned out in less than 3 minutes about 53% of the time.
Check BrownleeStatistical Theory and Methodology pp. 42 60 for more details.0July 17, 2002 at 7:44 pm #77333I used a less elegant method of simulating an answer.
I assumed that only one stamping opportunity occours per day……
————
Generate 6 (days) columns by 1000+ rows of random exponentially dist. data with mean of 4.0 minutes. I used MINITAB.
Copy data over to EXCEL and use COUNTIF formula to count number of cells (days) in a row which had a value < 3.0 minutes.
From here you can do pivot table of data and COUNTIF column and show frequency of each condition:Count of # Days of 6 <3 min to stamp
# Days of 6 <3 min to stamp
Total
Cumulative %
0
10
1%
1
76
9%
2
223
31%
3
307
62%
4
249
87%
5
115
98%
6
20
100%
Grand Total
1000
Looks like parts stamped <3 minutes 46 days of next 6 days about 38% of time (62% 03 days).
If you dont like pivot tables, copy data back to MINITAB and use probablity plot to eyeball the answer.
Thanks,
Carl
0July 29, 2002 at 9:24 am #77662Carl gets a gold star and gets to keep his job. For the rest of you…..
0July 30, 2002 at 4:19 pm #77715Thanks for the kudos!
Some of these questions are pretty thought provoking and force me to try new tools.
Carl
0July 30, 2002 at 9:26 pm #77733
Ashok KumarParticipant@AshokKumar Include @AshokKumar in your post and this person will
be notified via email.Only Robert Butler has the correct answer at least for the part of the question. The probability that a stamping will be produced in 3 minutes or less given that average stamping time is 4 minutes is = 1 EXP (3/4) = .527633447 = .53, say. For the purpose of responding to 4 out of 6 days part, we need to analyze this as a binomial distribution; viewing the population consisting of two types of stampings: one that are produced in three minutes or less (success) and the other that are produced in more than three minutes. Here the number of days is unimportant; number of stampings produced is. So, for r out of n stampings to be produced in 3 or less minutes (assuming average production time of 4 minutes and exponential distribution), the probability is: (n choose r) .(0.53)^ r . (0.47) ^ (nr), where read ^ as raised to the power of. Thus, the probability that any 5 of the 6 stampings produced would be have a production time of 3 or less minutes, the answer is: 6 * (0.53) ^5 * (0.47)^1 = 0.1159. The probability that 4 out of 6 will be produced in 3 or less minutes is 0.2594, and the probability for 3 out of 6 areproduced in 3 minutes or less will be 0.3096. Hope you can verify these numbers yourself if not dont hesitate to email me at [email protected].
0July 31, 2002 at 1:13 pm #77741Ashok,
You are correct and more precise than I was. I probably made Statisticians cringe by using brute force simulation versus the proper exponential/binomial formulas!
PROBMAN did not ask us about proportion of stampings completed in 3 minutes or less although if he had, your first answer was right on.
He did ask us for proportion of 6 stamping sets where 46 stampings were over 3 minutes. I think your numbers were exact per condition 06 and my simulated values were all within 1% of yours. Either way, the probability seems to be about 3839%.
Thanks
Carl
ps – Lots of ways to skin a cat unless of course you needed to skin a dog.
0July 15, 2003 at 9:02 pm #87981
phaedrusParticipant@phaedrus Include @phaedrus in your post and this person will
be notified via email.This is not a wellposed question. Do you mean what are the chances of a part being stamped in the first 3 minutes of at least 4 of the next 6 days? This is a very different question from asking for the prob that a machine that operated a certain length of time each day will stamp at least one item in less than 3 minutes on at least 4 of the next 6 days. None of the responses answered either of these 2 questions.
The first question is pretty simple. The answer is 1(1p)^66p*(1p)^5, where p=1exp(0.75). The second question is more difficult. I will work it out if you need the answer.0July 15, 2003 at 9:04 pm #87982
phaedrusParticipant@phaedrus Include @phaedrus in your post and this person will
be notified via email.that’s not what was asked in the original question. see my other message for further explanation of why i think not.
0September 26, 2003 at 7:58 pm #90374
CluelessParticipant@Clueless Include @Clueless in your post and this person will
be notified via email.Another question:
I work in a automated distribution center, our WMS and Conveyor systems are set to send 300 cartons per order to one shipping lane.(if the order has a total of 800 cases the order is split across 3 lanes)
Our defect is that the same catalog number is getting split across shipping lanes.The software only counts cases and then splits.The programmer is making adjustments, however he can only fix the problem if the catalog qty is less than the lane max qty.
What I have been asked to do is figure out what the probability is of a catalog # being split across multi lanes.
avg cases per order – 165 stdev 233
avg cases per cat# 7 stdev 11
avg cat#’s on order – 24 stdev 25
Max cases set at shipping lane:100, 200, 300, 400
Can someone point me in the right direction? I’m clueless.
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