Standard Deviation and Variance
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 This topic has 9 replies, 2 voices, and was last updated 1 month, 3 weeks ago by Robert Butler.

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February 14, 2019 at 1:31 pm #236386
guptaneeraj8888Participant@guptaneeraj8888 Include @guptaneeraj8888 in your post and this person will
be notified via email.According to a manager it takes an average weekday commute of 39 minutes with a Standard Deviation of 7 minutes for the employees to get to work when they use their personal vehicles for their office commute while management set a policy of not more than 40 minutes for their daily oneway commute. A survey conducted one day on 70 employees showed an average of 34 minutes commuting time using the metro public transportation system with a Standard Deviation of 21 minutes. For the employees choosing to increase their chances to come on time using personal transportation their variation should be reduced to ___________?
Note: Please help and let me know a general idea to approach such type of problems.
0February 14, 2019 at 1:36 pm #236387
Katie BarryParticipant@KatieBarry Include @KatieBarry in your post and this person will
be notified via email.@guptaneeraj8888 It appears that this is a homework question.
What do YOU think you should do? Why or why not? The more details you provide, the more likely you are to get a response. The iSixSigma audience is helpful, but they like to see that someone is putting forth a goodfaith effort. They are not here to do your homework for you.
0February 14, 2019 at 2:04 pm #236390
guptaneeraj8888Participant@guptaneeraj8888 Include @guptaneeraj8888 in your post and this person will
be notified via email.@KatieBarry this is not a home work question. I am not asking for a complete solution but a general approach. May be my understanding of concept is weak but that not make my effort less. I have tried several approach which includes following :
I have tried by calculating the z score for the people using personal transport by assuming n= 70.
Also tried by plotting all the data on a normal distribution curve.
Is my approach is correct or I need to go through a different route?
0February 14, 2019 at 2:06 pm #236391
Robert ButlerParticipant@rbutler Include @rbutler in your post and this person will
be notified via email.That has to be one of the worst homework problem statements I think I’ve ever seen. As written there are all kinds of answers and solutions to the problem.
1. If the reported standard deviation is the standard deviation of the mean and not just the sample standard deviation then there isn’t much of anything you can do with the statement concerning personal vehicles.
2. If the standard deviations are the sample standard deviations then we are assuming that somehow employees taking public transportation have some kind of control over how that mode of transportation functions.If we assume the standard deviations are sample standard deviations and we assume the employees have total control of traffic conditions and how public transportation functions then I guess you have the makings of a homework problem one could address.
Questions for you:
1. What is the basic expression for a confidence limit around a mean?
2. If we pretend the standard deviations are sample standard deviations and we pretend the average represents the central tendency of a reasonably symmetric time distribution then how would you determine the range of actual arrival times (95%) around the mean?
3. As the question is written, what is the sample size we need to use to discuss the results of an individual employee?
4. Given that you have identified the 95% limits for the mean and that you have identified the sample size associated with an individual you should be able to calculate the ranges of variation associated with various employees. I will give you one hint – some of the employees will have to employ time travel in order to meet the requirement.
2February 14, 2019 at 3:20 pm #236394
guptaneeraj8888Participant@guptaneeraj8888 Include @guptaneeraj8888 in your post and this person will
be notified via email.I have calculated the Z score as it represent the probability by considering X=70
Z= 7039/7= 4.42
Now, in order to calculate variance for personal transportation I have done 1Zscore= 14.42=3.42.
So we should deduct 3.42 in order to increase their chances to come on time using personal transportation.
Am i correct?0February 14, 2019 at 4:45 pm #236396
Robert ButlerParticipant@rbutler Include @rbutler in your post and this person will
be notified via email.I guess I don’t understand why you are bothering with Z scores. To rephrase your question – I have an estimated average travel time of 39 minutes to get to work with a sample standard deviation of 7 minutes.
The question that is being asked is: what kind of reduction in variation of individual travel time must an employee make in order to meet (the absolutely absurd) management target of a maximum of 40 minutes travel time.
The basic equation to consider is management time = meantime +(2*std/sqrt(n))
Therefore, for one person n = 1. If all employees have a time of 39 minutes then to have 40 minutes travel time the standard deviation would have to drop from 7 minutes to 30 seconds.
Now, 39 is just the mean of the population. The 95% CI for that mean – assuming symmetry with respect to the distribution of arrival time – is going to be 39 minutes +2*7. Thus, at the extremes you have someone arriving at work after 25 minutes of travel and someone arriving after 53 minutes of travel.
The poor soul who needs 53 minutes to get to work can reduce the variation in his travel time to 0 and he will still need a time machine at the entrance to the workplace to take him back in time 13 minutes in order to meet the target set by management.
As for the gal who needs only 25 minutes to get to work, she can stop off for a coffee, do a little window shopping, and still show up at the gate with an elapsed travel time of less than 40 minutes.
In short, the management target is based on wishful thinking and has no connection to what is happening in the real world.
1February 14, 2019 at 10:15 pm #236405
StrayerParticipant@Straydog Include @Straydog in your post and this person will
be notified via email.Here’s a simple thought. It’s an interesting statistics question but useless. Employers don’t care as long you arrive on time and would only have a policy for a 40 minute commute time if that was mandated by government regulation or union contract, which is highly unlikely. Employees do care, but how many of them would do the math or care about such fine distinctions? We don’t waste our time on useless questions and whoever asked you to answer this one is clueless.
1February 15, 2019 at 1:10 pm #236407
Mike CarnellParticipant@MikeCarnell Include @MikeCarnell in your post and this person will
be notified via email.@guptaneeraj8888 If you are operating in the US this type of question may be borderline illegal in the US. In the US you cannot even ask them how they get to work bus, car, etc.
0August 8, 2022 at 1:02 am #258050
Theresa_5566Participant@Theresa_5566 Include @Theresa_5566 in your post and this person will
be notified via email.Still a bit confused with this question. Tried many ways still not able to get the right answer (3.5 minutes).
Where did you get the following formula (is 2 an approx z score you have used):
management time = meantime +(2*std/sqrt(n))
0August 8, 2022 at 9:07 am #258054
Robert ButlerParticipant@rbutler Include @rbutler in your post and this person will
be notified via email.The expression is the equation for expressing the 95% confidence limit around the mean (See for example Statistical Methods 7th Edition – Snedecor and Cochran – pp5153) . If you have an infinite sample size the 2 will become 1.96 but, unless you happen to know the exact sample size for the mean (which you don’t) the usual practice, for an estimate of the 95% limits, is to just use 2.
As for 3.5 being the correct answer – it can’t be.
Given – the average commute time for personal vehicles is 39 minutes
– management has a target of 40 minutes
– the difference between 39 minutes and 40 minutes is 1 minute.
– the standard deviation associated with 39 minutes is 7
– therefore the existing standard deviation is well in excess of 1 and a reduction of the standard deviation to 3.5 is still well in excess of 1. As I noted previously – to even have a hope of hitting an average target of 40 (given the current average is 39 minutes) you would need a standard deviation of 30 seconds and that only applies to the grand average – not to the reductions needed for individual employees.0 
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