Question

A bottling plant produces bottles of 500 ml at the production rate of 10,000 bottles per hour. An inspector randomly picks up 100 bottles every 15 minutes and checks for quality. With visual inspection he finds a bottle either good or defective. In last 8 samples, the number of defective bottles that he found are: 0, 0, 2, 1, 0,1, 2 and 0. What are the upper and lower control limits of the corresponding p-chart?

a 0.82,0.18
b. 0.099.0
C.12.1, 0
d. 1.31, 0.06
e.none of these

Answer 1

e.none of these

Explanation:

Computations For CC for Fraction defective

Sample No d p=d/100

1 0 0

2 0 0

3 2 0.02

4 1 0.01

5 0 0

6 1 0.01

7 2 0.02

8 0 0

Total 0.06

`\overline{p} =(1)/(k)\sum p=(0.06)/(8) =0.0075`

`\overline{q} =1- 0.0075= 0.9925`

3 sigma control limits for p chart are given by:

`\overline{p} \pm 3\sqrt{\overline{p}\overline{q}/n}\Rightarrow 0.0075\pm 3\sqrt{(0.0075*0.9925)/(100)}`

`= 0.0075\pm0.0259\Rightarrow ( 0.0334,0 )`

hence option e is correct