Question

Computer upgrades have a nominal time of 80 minutes. samples of five observations each have been taken, and the results are as listed. sample 1 2 3 4 5 6 79.2 80.5 79.6 78.9 80.5 79.7 78.8 78.7 79.6 79.4 79.6 80.6 80.0 81.0 80.4 79.7 80.4 80.5 78.4 80.4 80.3 79.4 80.8 80.0 81.0 80.1 80.8 80.6 78.8 81.1 factors for three-sigma control limits for formula179.mml and r charts factors for r charts number of observations in subgroup, n factor for formula180.mml chart, a2 lower control limit, d3 upper control limit, d4 2 1.88 0 3.27 3 1.02 0 2.57 4 0.73 0 2.28 5 0.58 0 2.11 6 0.48 0 2.00 7 0.42 0.08 1.92 8 0.37 0.14 1.86 9 0.34 0.18 1.82 10 0.31 0.22 1.78 11 0.29 0.26 1.74 12 0.27 0.28 1.72 13 0.25 0.31 1.69 14 0.24 0.33 1.67 15 0.22 0.35 1.65 16 0.21 0.36 1.64 17 0.20 0.38 1.62 18 0.19 0.39 1.61 19 0.19 0.40 1.60 20 0.18 0.41 1.59
a. using factors from above table, determine upper and lower control limits for mean and range charts. (round your intermediate calculations and final answers to 2 decimal places. leave no cells blank - be certain to enter "0" wherever required.) mean chart range chart ucl 81.06 4.01 lcl 78.86 0.00 g

Answer 1

the answer:
the upper and lower control limits (uclim and lclim) for mean formula is 

for the mean chart
uclim= x+A2xR
where x = sum(of the value)  / number of each value
and for 
lclim=x+A2xR

R is the range such that R= Xmax - Xmin

in the case of the sample 1: S1
the data are: 
79.2    78.8   80.0   78.4   81.0

the mean is   x1 = (79.2  +  78.8  +  80.0  + 78.4  + 81.0)  / 5=  79.48
its range is    R 1= 81.0 -78.4 = 2.6

we can do the same method for finding the mean chart and range for all samples
S2: x2= 80.14  ,  R2=2.3
S3: x3= 80.14  ,  R3=1.2
S4: x4= 79.60  ,  R4=1.7
S5: x5= 80.02  ,  R5=2.0
S6: x6=80.38   ,  R6=1.4

therefore the average value is   X= sum( x1+x2+...+x6) / 6 = 79.96
and R=sum(R1+R2+...+R6)/6=1.87
finally
range chart uclim =D4xR=3.95  and lclim is always equal to 0, because D3=0

we can say that the process is not in control.