Question

Each day, a factory counts the number of machine failures. In a given day there may be multiple machine failures (nonconformances). The total number of machine failures over a 25 day period is 45. Compute the control limits for this process.

solve with excel and please show formulas.

Answer 1

Answer:

Here, we will find the Upper Control Limit (UCL) and Lower Control Limit (LCL) of C- Chart as mentioned below:

As specifically mentioned in the question, we will solve this problem by using MS-Excel as mentioned below:

Step 1: First, prepare the following table with the exact no. of row and column:

Where C-Bar = 45 Failures / 25 Days = 1.8 Failures / Day

Step 2: Now, type the formula for UCL and LCL as mentioned in the below screenshot:

UCL = C- + 3 X SQRT ( C-)

LCL = C- - 3 X SQRT ( C-)

Hence, we get the UCL and LCL as mentioned below:

(Note: Negative LCL is considered as '0')

Answer 2

Answer:

Here, we will find the Upper Control Limit (UCL) and Lower Control Limit (LCL) of C- Chart as mentioned below:

As specifically mentioned in the question, we will solve this problem by using MS-Excel as mentioned below:

Step 1: First, prepare the following table with the exact no. of row and column:

A 1 4. 5 C-Bar (CL) | 1.80 UCL LCL

Where C-Bar = 45 Failures / 25 Days = 1.8 Failures / Day

Step 2: Now, type the formula for UCL and LCL as mentioned in the below screenshot:

UCL LCL 1 =$B$1+3*SQRT($B$1) =$B$1-3*SQRT($B$1) 5

UCL = C- + 3 X SQRT ( C-)

LCL = C- - 3 X SQRT ( C-)

Hence, we get the UCL and LCL as mentioned below:

1 4 5 А C-Bar (CL) UCL LCL | в 1.8 5.82 -2.22

(Note: Negative LCL is considered as '0')

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