Probability Question -For those whiz-kids out ther

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• #29892

PROBMAN
Participant

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?

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#77316

Taylor
Member

Take 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.

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#77318

Ovidiu Contras
Participant

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 helps

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#77323

mcleod
Member

PROBMAN,
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?
Scott

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#77324

Robert Butler
Participant

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 Brownlee-Statistical Theory and Methodology pp. 42- 60 for more details.

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#77333

Carl H
Participant

I 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 4-6 days of next 6 days about 38% of time (62% 0-3 days).
If you dont like pivot tables, copy data back to MINITAB and use probablity plot to eyeball the answer.
Thanks,
Carl

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#77662

MJH
Participant

Carl gets a gold star and gets to keep his job.  For the rest of you…..

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#77715

Carl H
Participant

Thanks for the kudos!
Some of these questions are pretty thought provoking and force me to try new tools.
Carl

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#77733

Ashok Kumar
Participant

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) ^ (n-r), 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 dont hesitate to e-mail me at [email protected].

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#77741

Carl H
Participant

Ashok,
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!
He did ask us for proportion of 6 stamping sets where 4-6 stampings were over 3 minutes.  I think your numbers were exact per condition 0-6 and my simulated values were all within 1% of yours.  Either way, the probability seems to be about 38-39%.
Thanks
Carl
ps – Lots of ways to skin a cat unless of course you needed to skin a dog.

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#87981

phaedrus
Participant

This is not a well-posed 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-(1-p)^6-6p*(1-p)^5, where p=1-exp(-0.75).  The second question is more difficult.  I will work it out if you need the answer.

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#87982

phaedrus
Participant

that’s not what was asked in the original question. see my other message for further explanation of why i think not.

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#90374

Clueless
Participant

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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