Question

The result of inspection of samples taken over the past 11 days is given below. The sample size is 100 per day. Day 1 2 3 4 5 6 7 8 9 10Defectives 7 9 9 11 7 8 0 11 13 2What are the upper and lower control limits?

Answer 1

UCL(p) = 0.157

LCL(p) = 0

Step-by-step explanation:

Number of samples (n) = 10

Size of sample (n) = 100

We compute the defective rate P as below

P = 7 + 9 + 9 + 11 + 7 + 8 + 0 + 11 + 13 + 2 / 10 * 100

P = 77 / 1000

p = 0.077

We derive σP / value of standard deviation of the sampling distribution as shown below

σP = √P * (1-P) / n

σP = √0.077 * (1 - 0.077) / 100

σP = √0.077 * 0.923/100

σP = √0.071071/100

σP = √0.00071071

σP = 0.02665

Now we calculate the Upper Control chart limit:

UCL(p) = P +Z*σP

UCL(p) = 0.077 + 3*0.02665

UCL(p) = 0.077 + 0.07995

UCL(p) = 0.15695

UCL(p) = 0.157

Now we calculate the Lower Control chart limit:

LCL(p) = P - Z*σP

LCL(p) = 0.077 - 3*0.02665

LCL(p) = 0.077 - 0.07995

LCL(p) = -0.00295 (Negative defect cannot go beyond Zero)

LCL(p) = 0