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The following data was collected from the manufacturing of an auto component. It represents the diameter (in mm) of that component. What is the UCL for a control chart using this data (z=3)? show working please
Sample Obs 1 Obs 2 Obs 3 Obs 4
1 10 12 12 14
2 12 11 13 16
3 11 13 14 14
4 11 10 7 8
5 13 12 14 13

Sagot :

Answer:

UCL = 14.624

Explanation:

Data Given:

z= 3

And beside the value of z, we are given:

Number of samples = 5

And observations.

First of all, we have to calculate the mean of all the samples.

Sample 1 Mean = [tex]\frac{10+12+12+14}{4}[/tex] = 12

Sample 2 Mean = [tex]\frac{12+11+13+16}{4}[/tex]  = 13

Sample 3 Mean = [tex]\frac{11+13+14+14}{4}[/tex] = 13

Sample 4 Mean = [tex]\frac{11+10+7+8}{4}[/tex] = 9

Sample 5 Mean = [tex]\frac{13+12+14+13}{4}[/tex] = 13

Now, add all the means and find out the mean of the sample size.

Mean of the sample size = [tex]\frac{12+13+9+13+13}{5}[/tex] = [tex]\frac{60}{5}[/tex] = 12

Secondly, we have to find out the maximum and minimum of the observations in each sample.

Sample 1 (max, min) = (14,10)

Sample 2 (max, min) = (16,11)

Sample 3 (max, min) = (14,11)

Sample 4 (max, min) = (11,7)

Sample 5 (max, min) = (14,12)

Thirdly, we need to calculate the Range of each sample.

Range = Max - Min

Range 1 = 14-10 = 4

Range 2 = 16-11 = 5

Range 3 = 14-11 = 3

Range 4 = 11-7 = 4

Range 5 = 14-12 = 2

ΣRange = 18

Now, sum all the ranges and divide it by the number of samples n.

(Range-bar) = ΣRange/n

Range-bar = 18/5

Range-bar = 3.6

Now, we need value of A2 for the sample size of 4. So, it can find out from the table of control chart constants which is attached for your reference.

A2 = 0.729

Now, we have all the data required to calculate the UCL.

UCL = (Mean of the sample size) + (A2 * Range-bar)

UCL = (5) + (0.729x 3.6)

UCL = 14.624

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