Creating a probability distribution in Excel
Question
I was given three data points and told to make a distribution. Here are the instructions I was given to complete the task:
The bottom 20th percentile will yield 13 units this month. The median of the data shows we should produce 23 units this month. Best case, the top 20th percentile shows us producing 30 units this month.
This is in reference to trying to find out the % likelihood of producing a certain amount of a part in the month.
I tried making a distribution like so:
However, I am looking to produce a probability distribution like below:
The data I am working with (that is summarized from my boss' instructions above) is below:
| Serial Number | Median Projected Finish Date | Median In July | Best Case Projected Finish Date | Best In July | Worst Case Projected Finish Date | Worst In July |
|:-------------:|:----------------------------:|:--------------:|:-------------------------------:|:------------:|:--------------------------------:|:-------------:|
| 8473 | 7/18/2019 | 1 | 6/28/2019 | 1 | 8/2/2019 | 0 |
| 11963 | 6/30/2019 | 1 | 6/28/2019 | 1 | 7/28/2019 | 1 |
| 15165 | 6/27/2019 | 1 | 6/27/2019 | 1 | 6/28/2019 | 1 |
| 28023 | 7/1/2019 | 1 | 6/29/2019 | 1 | 7/3/2019 | 1 |
| 14355 | 9/1/2019 | 0 | 7/11/2019 | 1 | 9/13/2019 | 0 |
| 14388 | 7/3/2019 | 1 | 7/1/2019 | 1 | 7/7/2019 | 1 |
| 796 | 7/18/2019 | 1 | 6/28/2019 | 1 | 8/2/2019 | 0 |
| 20574 | 7/11/2019 | 1 | 7/9/2019 | 1 | 7/19/2019 | 1 |
| 6518 | 7/11/2019 | 1 | 7/9/2019 | 1 | 7/19/2019 | 1 |
| 19969 | 7/11/2019 | 1 | 7/9/2019 | 1 | 7/18/2019 | 1 |
| 10244 | 7/11/2019 | 1 | 7/9/2019 | 1 | 7/19/2019 | 1 |
| 9980 | 7/11/2019 | 1 | 7/9/2019 | 1 | 7/19/2019 | 1 |
| 26056 | 7/11/2019 | 1 | 7/9/2019 | 1 | 7/19/2019 | 1 |
| 8849 | 7/18/2019 | 1 | 7/2/2019 | 1 | 8/2/2019 | 0 |
| 7409 | 7/11/2019 | 1 | 7/9/2019 | 1 | 7/19/2019 | 1 |
| 1386 | 7/11/2019 | 1 | 7/9/2019 | 1 | 7/19/2019 | 1 |
| 13971 | 7/17/2019 | 1 | 6/27/2019 | 1 | 8/1/2019 | 0 |
| 21974 | 7/25/2019 | 1 | 7/19/2019 | 1 | 8/12/2019 | 0 |
| 20546 | 7/25/2019 | 1 | 7/19/2019 | 1 | 8/12/2019 | 0 |
| 10055 | 6/30/2019 | 1 | 6/27/2019 | 1 | 7/28/2019 | 1 |
| 22766 | 7/17/2019 | 1 | 6/27/2019 | 1 | 8/1/2019 | 0 |
| 12679 | 7/18/2019 | 1 | 7/2/2019 | 1 | 8/2/2019 | 0 |
| 28837 | 7/26/2019 | 1 | 6/30/2019 | 1 | 8/14/2019 | 0 |
| 12509 | 7/31/2019 | 1 | 7/4/2019 | 1 | 8/18/2019 | 0 |
| 1624 | 8/5/2019 | 0 | 7/29/2019 | 1 | 8/21/2019 | 0 |
| 5689 | 8/1/2019 | 0 | 7/4/2019 | 1 | 8/19/2019 | 0 |
| 29315 | 8/2/2019 | 0 | 7/5/2019 | 1 | 8/29/2019 | 0 |
| 10618 | 8/2/2019 | 0 | 7/5/2019 | 1 | 8/29/2019 | 0 |
| 16235 | 8/2/2019 | 0 | 7/5/2019 | 1 | 8/29/2019 | 0 |
| 12079 | 8/2/2019 | 0 | 7/5/2019 | 1 | 8/29/2019 | 0 |
| | | 23 | | 30 | | 13 |
The data source above is an outline of the unique identifiers of a part, the projected completion date of that serial number, the projected completion date if the steps to complete it are in the 20% best / worst times, and an "in July" column to see if that finishes in the month of july (used to sum totals in the last row).
I have tried following this resource but was unable to do so. How can I create a probability distribution with my data?
The goal would be to find out what is the % likelihood of finishing x number of units in the given month (say in July 2019). I can only work with the data above. Being able to have a graph that showed (or would come close to showing) there is an 11% probability of finishing 32 units in 2019 (arbitrary). Ideally, the probability distribution would show us what is the probability of finishing x number of units. I imagine the floor would be 0 and the ceiling would be 30 (as there are only 30 items listed).
1 answers
solution1
1 ACCPTED 2019-06-29 13:19:25
You can view the date at which each unit is completed as a random variable. For each unit, you are given three probabilities. For the first unit, P(U 1 < 6/28) = 0.2, P(U 1 > 8/2) = 0.2, and P(U 1 < 7/18) = 0.5, where 7/18 is the median.
If we assume that U 1 is normally distributed, then its median and average are equal to 7/18, and the probabilities P(U 1 < k 1 ) = 0.2 and P(U 1 > k 2 ) = 0.2 must be for k 1 and k 2 equally distant to the mean 7/18. This is not the case for U 1 , indicating that U 1 is most likely not normally distributed. You may want to consider other probability distributions that are skewed and where the median is at 0.5 probability. There are the Exponentially modified Gaussian distribution , the Skew normal distribution , and many others. Whatever knowledge you may have about the production of units may help selecting a probability distribution.
Assume we want to use a normal distribution. Instead of working with dates, we will work with day numbers with day 1 being 7/1. We need to estimate for each unit, the parameters of its normal distribution, that is, the mean and standard deviation given the three probability points we have. Because the normal distribution is symetrical, the mean is the middle day between the worst/best 20% days. For U 1 the mean would be m 1 = (33 - (-2)) / 2 + -2 = 15.5. We know that P(U 1 > 33) = 0.2. This occurs for a N(0;1) > 0.84. So, the standard deviation s 1 = (33 - 15.5) / 0.84 = 20.8. Knowing m 1 and s 1 , we can calculate the probability that U 1 will be completed in July, that is, P(U 1 ~ N(m 1 ;s 1 ) < 32). The same estimation is done for all N units. This gives us N probabilities, that is, the probabilities that the units be completed in July.
To calculate the probability that R of the N units will be completed in July, refer to the following answers.
- How to determine probability of 0 to N events occurring given probability of each of those N events?
- Probability of exactly k out of n events occuring
- Probability-generating function
Finally, it is assumed that the production of units are independent. If this is not the case (for example, two units that depend on a common supplier of parts), then the calculated probabilities might not be good. But I think the best improvement would be to find a distribution that reprensent your data better than the normal distribution.
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