# Control Chart with Range Calculation

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- This topic has 13 replies, 6 voices, and was last updated 2 years, 7 months ago by Chris Seider.

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- May 16, 2017 at 10:10 am #55730

John PetersParticipant@johnpeters123**Include @johnpeters123 in your post and this person will**

be notified via email.I’m new to Lean Six Sigma and have been given a question (as seen below) and would really appreciate any help available.

Question: When using a control chart what would an average of 95.75 show in terms of the range when there was a sample of 20.

a) 1.0

b) 0.8

c) 0.5

d) 0.6Any help would be appreciated and an explanation on how it was worked out would be amazing!

Thanks in advance.John

0May 16, 2017 at 10:38 am #201396

Katie BarryKeymaster@KatieBarry**Include @KatieBarry in your post and this person will**

be notified via email.@johnpeters123 You’re more likely to get helpful responses if you first explain what you think/why.

The more details you provide, the better. The iSixSigma audience is helpful, but they like to see that someone is putting forth a good-faith effort first.

0May 16, 2017 at 10:47 am #201397

John PetersParticipant@johnpeters123**Include @johnpeters123 in your post and this person will**

be notified via email.Hi Katie,

Yes, sorry i do understand that. But I have no idea how it could be answered with the information provided

.

When it states range I would assume that it would refer to the span of the data for the upper and lower control limits. However this just states that the ‘average’ is 95.75 with a sample of 20… so i’m completely unsure how you would go about workout out the range just from these figures.Thanks anyway!

John0May 16, 2017 at 11:18 am #201398

Katie BarryKeymaster@KatieBarry**Include @KatieBarry in your post and this person will**

be notified via email.@johnpeters123 What you just wrote is exactly what I meant! :)

0May 16, 2017 at 12:15 pm #201399

John PetersParticipant@johnpeters123**Include @johnpeters123 in your post and this person will**

be notified via email.Hi Katie,

Thank you, so what you’re saying is it can’t be answered based on the information provided?

Thanks again!

0May 16, 2017 at 1:29 pm #201400

Chris SeiderParticipant@cseider**Include @cseider in your post and this person will**

be notified via email.it looks like homework here but….

think of the following. just because you know the center of your data, why would you presume you can describe the variation?

look at the formulas for your consideration.

0May 16, 2017 at 1:33 pm #201401

John PetersParticipant@johnpeters123**Include @johnpeters123 in your post and this person will**

be notified via email.Wow.

0May 16, 2017 at 1:33 pm #201402

Katie BarryKeymaster@KatieBarry**Include @KatieBarry in your post and this person will**

be notified via email.@johnpeters123 I didn’t answer your question. I suggested a better way of presenting your question that made it more likely you’ll get a reply.

0May 16, 2017 at 1:35 pm #201403

John PetersParticipant@johnpeters123**Include @johnpeters123 in your post and this person will**

be notified via email.Sorry, accidentally hit send. Thanks for your help.

I’ll look into it.

No it’s not homework, it’s just revision for my green belt exam.Thanks

0May 16, 2017 at 1:49 pm #201404

Robert ButlerParticipant@rbutler**Include @rbutler in your post and this person will**

be notified via email.If there is a way to answer that question I will have to admit I can’t think of a way to do it. You are correct – the usual problem of this type is to give a min and a max, compute their average and then divide the result by d2 for 20 samples (d2 comes from tables of percentage points of the distribution of the relative range) For 20 samples that would be 3.735.

I would say that if taken at face value – 95.75 is the mean and 20 is the sample size then there is no way to estimate the standard deviation and even if you did so the statement/question “what would an average of 95.75 show in terms of the range…” doesn’t make any sense.

A real stretch – If we assume the question implies that the average range is 95.75 then the estimate of the standard deviation based on 20 samples would be 25.6 but that number isn’t close to any of the choices and it still begs the question – what does the phrase “show in terms of the range” mean.

My guess would be that at least one part of the question went missing in action when it was transcribed.

0May 16, 2017 at 7:34 pm #201407

MBBinWIParticipant@MBBinWI**Include @MBBinWI in your post and this person will**

be notified via email.@rbutler – or the preparer of the question isn’t all that knowledgeable.

0May 17, 2017 at 5:12 am #201413

Robert ButlerParticipant@rbutler**Include @rbutler in your post and this person will**

be notified via email.@MBBinWI Actually I started to post a comment with respect to the overall quality of the test and what it might suggest relative to the quality of the training received. I find this question particularly worrisome since it seems to be the second poorly presented question in the material the OP is using.

0May 22, 2017 at 5:09 pm #201435

Mike CarnellParticipant@Mike-Carnell**Include @Mike-Carnell in your post and this person will**

be notified via email.@johnpeters123 That is what happens when you listen to Katie. She has been at this a very long time and knows most of the people who contribute and what will get them to help.

Your sincerity goes a very long way as well.

0May 23, 2017 at 8:37 am #201443

Chris SeiderParticipant@cseider**Include @cseider in your post and this person will**

be notified via email.@mbbinwi I’m shocked you’d imply someone put together a test question without understanding nuances and breadth of experience. :)

Having created customer “loved” BB, MBB, and GB exams I can tell you it’s not easy or trivial.

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