Abottling 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

e.none of these

Step-by-step 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

3 sigma control limits for p chart are given by:

hence option e is correct

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17