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A set of data from a process has 8 readings per sample and 50 samples. The mean of the 50 sample means is 12.62. The mean of the 50 ranges is 0.18.Find the control limits for a median chart.
Suggested Answer: A
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37 comments · newest first
AU
Augustine
10 months ago
The mean of the ranges is 0.18.
WI
Wilbert
11 months ago
Totally agree with option B!
SH
Shay
11 months ago
Wait, how do we even get those limits? Sounds off.
VO
Vonda
11 months ago
I think the control limits should be around 12.54 and 12.70.
CO
Cordell
11 months ago
The mean of the sample means is 12.62.
SA
Salena
12 months ago
I’m leaning towards D because none of the options seem to match what I expected for the control limits.
LI
Linn
12 months ago
This question feels similar to one we practiced, where we had to find limits using sample means. I think the answer might be B.
HO
Hollis
12 months ago
I think the control limits are based on the mean and the range, but I can't recall the exact formula we used.
AN
Anastacia
1 year ago
I remember we calculated control limits for median charts in class, but I'm not sure how to apply the mean and range here.
TW
Twanna
1 year ago
This seems straightforward enough. I just need to plug the values into the right formulas and do the math. I'm feeling good about being able to solve this one.
AL
Alease
1 year ago
Let's see, the question gives us the mean of the 50 sample means and the mean of the 50 ranges. I think I can use those values to find the control limits for the median chart.
LA
Launa
1 year ago
Hmm, I'm a bit unsure about this one. I know we covered median charts in class, but I'm not totally confident in how to apply the calculations here.
JA
Jaime
1 year ago
Okay, this looks like a question on control charts. I'll need to remember the formulas for calculating the control limits for a median chart.
MA
Marta
2 years ago
Ah, a classic median chart question. I can already smell the trick answer. Time to put on my thinking cap.
CH
Cherrie
2 years ago
I believe it's C) 0.02 and 0.33.
MY
Myra
2 years ago
I'm leaning towards B) 12.54 and 12.70.
KA
Katy
2 years ago
I think the answer is A) 12.52 and 12.72.
LO
Lorriane
2 years ago
Alright, time to break out the statistical formulas! This is my kind of question. I think I've got it figured out.
SH
Sherron
2 years ago
I agree with Glendora, A) seems to be the correct answer based on the given data.
GL
Glendora
2 years ago
But the mean of the 50 sample means is 12.62, so I think A) makes more sense.
MA
Malika
2 years ago
Did anyone else notice the sample size is a bit odd? 50 samples with 8 readings each? Feels like there might be a trick question here.
LO
Lorean
2 years ago
I'm not sure, but it does seem like there could be a trick in the question.
BL
Blythe
2 years ago
I think the answer might be A) 12.52 and 12.72.
AF
Afton
2 years ago
I agree, that sample size does seem unusual.
DU
Dusti
2 years ago
This seems too easy. Are you sure there's not a catch? I'm going to go with D just to be safe.
VI
Virgina
2 years ago
Wait, why are we looking at the mean of the ranges? Shouldn't we be focusing on the sample means? I'm leaning towards A.
HU
Huey
2 years ago
Yes, focusing on the mean of the ranges helps determine the control limits for the median chart.
WE
Weldon
2 years ago
I believe the correct answer is A) 12.52 and 12.72.
ME
Melina
2 years ago
I agree, the mean of the ranges is important for setting control limits.
MO
Mollie
2 years ago
I think we need to calculate the control limits based on the mean of the ranges.
LE
Lewis
2 years ago
I disagree, I believe the answer is B) 12.54 and 12.70.
GL
Glendora
2 years ago
I think the answer is A) 12.52 and 12.72.
RO
Roslyn
2 years ago
Hmm, the data seems pretty straightforward. I'm thinking option B might be the right answer here.
TA
Tawna
2 years ago
I'm leaning towards option A actually, it seems more accurate to me.
SH
Shonda
2 years ago
I'm not so sure, maybe we should reconsider our options.
BR
Brett
2 years ago
I think so too, let's go with option B.
DO
Dolores
2 years ago
I agree, option B does seem like the most reasonable choice.