Question

The time to replace vehicle wiper blades at a service center was monitored using a mean and a range chart. six samples of n = 20 observations have been obtained and the sample means and ranges computed:

Answer 1

control limits for X (mean time) = 3.1 +/- 0.081 or (3.181 ; 3.019)

control limits for R (range) = (0.1845 ; 0.711)

Step-by-step explanation:

Sample Mean Range

1 3.06 0.42

2 3.15 0.50

3 3.11 0.41

4 3.13 0.46

5 3.06 0.46

6 3.09 0.45

average 3.10 0.45

to determine the upper and lower limits we need to find z in table for R charts, and look at the 20th value for A₂, D₃ and D₄:

• z for A₂ (n = 20) = 0.18
• z for D₃ (n = 20) = 0.41
• z for D₄ (n = 20) = 1.59

control limits for X = average mean +/- z(A₂) x average range

control limits for X = 3.1 +/- (0.18 x 0.45)

control limits for X = 3.1 +/- 0.081 or (3.181 ; 3.019)

now the control limits for R (range):

lower control limit = D₃ x average range = 0.41 x 0.45 = 0.1845

upper control limit = D₄ x average range = 1.59 x 0.45 = 0.711

control limits for R (range) = (0.1845 ; 0.711)

Answer 2

Step-by-step explanation:

The complete question statement is given;

The time to replace vehicle wiper blades at a service center was monitored using a mean and a range chart. six samples of n = 20 observations have been obtained and the sample means and ranges computed:

Sample Mean Range Sample Mean Range

1 3.06 .42

2 3.15 .50

3 3.11 .41

4 3.13 .46

5 3.06 .46

6 3.09 .45

Determine the upper limits for the mean

Determine the lower control limit for the mean

Solution

To determine, the upper limit and the lower limit, we need to first find the average of the mean for six sample

Mean average = 3.1

Range average = 0.45

Now for the upper limit; the formula is

= Mean Average + 0.18 (Range average)

= 3.1 + 0.18 (0.45)

=3.181

Now for the lower limit; the formula is

= Mean Average - 0.18 (Range average)

= 3.1 - 0.18 (0.45)

=3.01