Question

From the following data, calculate the control limits for the mean and range chart.

Sample No.	1	2	3	4	5	6	7	8	9	10
Sample Observations	50	21	50	48	46	55	45	50	47	56
	55	50	53	53	50	51	48	56	53	53
	52	53	48	50	44	56	53	54	549	55
	49	50	52	51	48	47	48	53	52	54
	54	46	47	53	47	51	51	47	54	52

Answer 1

Sample No.	Sample Observations					`sum"X"`	`bar"X" = (sumx)/5`	`"R" = "x"_"max" - "x"_"min"`
	I	II	III	IV	V
1	50	55	52	49	54	260	52	55 – 49 = 6
2	51	50	53	50	46	250	50	53 – 46 = 7
3	50	53	48	52	47	250	50	53 – 47 = 6
4	48	53	50	51	53	255	51	53 – 48 = 5
5	46	50	44	48	47	235	47	50 – 44 = 6
6	55	51	56	47	51	260	52	56 – 47 = 9
7	45	48	53	48	51	245	49	53 – 50 = 8
8	50	56	54	53	47	270	54	57 – 50 = 7
9	47	53	49	52	54	255	51	54 – 47 = 7
10	56	53	55	54	52	270	54	56 – 52 = 4
Total							`sum"X"` = 510	`sum"R"` = 65

The control limits for `bar"X"` chart is

`\overset{==}{"X"} = (sumbar"X")/"Number od samples" = 510/10` = 51

`bar"R" = (sum"R")/"n" = 65/10` = 6.5

UCL = `\overset{==}{"X"} + "A"_2 bar"R"`

= 51 + 0.577(6.5)

= 51 + 3.7505

= 54.7505

= 54.75

CL = `\overset{==}{"X"}` = 51

UCL = `\overset{==}{"X"} - "A"_2 bar"R"`

= 51 – 0.577(6.5)

= 51 – 3.7505

= 47.2495

= 47.25

The control limits for Range chart is

UCL = `"D"_4bar"R"`

= 2.114(6.5)

= 13.741

CL = `bar"R"` = 6.5

LCL = `"D"_3bar"R"` = 0(6.5) = 0