Fve data entry operators work at the data processing department of the Binmingha
ID: 412354 • Letter: F
Question
Fve data entry operators work at the data processing department of the Binmingham Bank. Each day for 30 days, the number of defective records in a sampla of 350 records typed by these operators has been noted, as follows: Sample No. Sample No. Sample Defectives No Defectives eNo Deectives 18 10 10 15 16 17 18 19 20 25 26 27 12 15 19 15 13 12 30 a) Establish 3o upper and wer contral limits UCL,-- ](enter your response “·number between 0 and 1 rounded to three decrmsi pieces). L CLp = enfer your respons a5 a number between 0 and 1, rounded to three dec-ral places b) Why can the lower control limit not be a negative number? O A. O B. ° C. Since the percent of defective records cannot be a negative number. Since the percent of defective records is always a positive number Since th upper control limit cannot be a negative number upper control limit is positive. D. Since the c) The industry standard for the upper control limit is D.10. What daes this imply about Birmingham Bank's own standards? The industry standard is the standard at Birmingham Bank.Explanation / Answer
sample no
no defectives (np)
sample size (n)
Fraction defective (p=np/n)
1
6
350
0.0171
2
5
350
0.0143
3
20
350
0.0571
4
10
350
0.0286
5
10
350
0.0286
6
9
350
0.0257
7
12
350
0.0343
8
8
350
0.0229
9
6
350
0.0171
10
12
350
0.0343
11
5
350
0.0143
12
5
350
0.0143
13
16
350
0.0457
14
5
350
0.0143
15
10
350
0.0286
16
9
350
0.0257
17
11
350
0.0314
18
4
350
0.0114
19
19
350
0.0543
20
15
350
0.0429
21
18
350
0.0514
22
13
350
0.0371
23
5
350
0.0143
24
7
350
0.0200
25
12
350
0.0343
26
11
350
0.0314
27
15
350
0.0429
28
6
350
0.0171
29
13
350
0.0371
30
2
350
0.0057
total
299
10500
0.854286
p bar = np/ n
p bar = 299/10500
p bar =
0.03
(no. of samples, k=30 samples)
n bar = n / k
n bar = 10500/30
n bar =
350
UCLp = p bar + 3 [(p bar(1-p bar))/n bar]
UCLp = 0.03+3*(0.03*(1-0.03))/350)
UCLp
0.055
LCLp = p bar - 3 [(p bar(1-p bar))/n bar]
LCLp = 0.03-3*(0.03*(1-0.03))/350)
LCLp
0.002
b. Lower control limit cannot be negative as the percent of defective records cannot be a negative number.
ans. : A.
c. Birmingham bank's UCL = 0.0551, or 0.1 if we round it off to 1 decimal place. Hence, the bank's UCL is equal to the industry standard.
The industry standard is EQUAL TO the standard at Birmingham Bank.
sample no
no defectives (np)
sample size (n)
Fraction defective (p=np/n)
1
6
350
0.0171
2
5
350
0.0143
3
20
350
0.0571
4
10
350
0.0286
5
10
350
0.0286
6
9
350
0.0257
7
12
350
0.0343
8
8
350
0.0229
9
6
350
0.0171
10
12
350
0.0343
11
5
350
0.0143
12
5
350
0.0143
13
16
350
0.0457
14
5
350
0.0143
15
10
350
0.0286
16
9
350
0.0257
17
11
350
0.0314
18
4
350
0.0114
19
19
350
0.0543
20
15
350
0.0429
21
18
350
0.0514
22
13
350
0.0371
23
5
350
0.0143
24
7
350
0.0200
25
12
350
0.0343
26
11
350
0.0314
27
15
350
0.0429
28
6
350
0.0171
29
13
350
0.0371
30
2
350
0.0057
total
299
10500
0.854286
p bar = np/ n
p bar = 299/10500
p bar =
0.03
(no. of samples, k=30 samples)
n bar = n / k
n bar = 10500/30
n bar =
350
UCLp = p bar + 3 [(p bar(1-p bar))/n bar]
UCLp = 0.03+3*(0.03*(1-0.03))/350)
UCLp
0.055
LCLp = p bar - 3 [(p bar(1-p bar))/n bar]
LCLp = 0.03-3*(0.03*(1-0.03))/350)
LCLp
0.002
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