Five data entry operators work at the data processing department of the Birmingham Bank. Each day for 30 days, the number of defective records in a sample of 300 records typed by these operators has been noted, as follows:

Sample No.

No. Defectives

Sample No.

No. Defectives

Sample No.

No. Defectives

1

8

11

7

21

17

2

4

12

5

22

13

3

19

13

16

23

6

4

10

14

3

24

6

5

12

15

12

25

13

6

9

16

8

26

11

7

11

17

13

27

15

8

10

18

4

28

7

9

6

19

18

29

13

10

14

20

17

30

4

a) Establish 3σ upper and lower control limits.

UCLp = (enter your response as a number between 0 and 1, rounded to three decimal places).

LCPp =(enter your response as a number between 0 and 1, rounded to three decimal places).

b) Why can the lower limit not be a negative number? (I actually know this one but I wanna get all the points down)

c) The industry standards for the upper and lower control limits are 0.10 and 0.01, respectively. What does this imply about Birmingham Bank's own standards?

Step-by-step explanation:

a)

b) Hence, Lower control limit cannot be a negative number as percent defective cannot be a negative number. As such, No. Percent of defective records cannot be a negative number.

see below

step-by-step explanation:

7.   there are 10 data points in the set.   the middle point is between 5 and 6.   the middle point in the top half is 8   and the middle point in the bottom half is 3. to find the iqr , take point 8 and subtract point 3

mid term   point 8 = 100   point 3 = 88

iqr = 100-88 = 12

final   point 8 = 93   point 3 = 78

iqr = 93 - 78 = 15

the final has the greatest iqr

6.   b - there is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical .   there is an outlier for the males that skews the data

5.   the   iqr is a better measure of spread for the movies and basketball games.   since the basketball game has an outlier, we do not want to use standard deviation for measurement.  

3.   the exam median is much higher than the class median.   the median line is farther to the right on the exams

2.   there is no way to tell what the means are from the box and whisker plots.  

so firstly, we need to isolate the y variables to be able to solve these inequalities. firstly, add 0.5x on both sides of the first inequality and subtract 1.5x on both sides of the second inequality:

now since the slope is positive for the first inequality, this means that the line going upwards belongs to the first inequality, and the line going downwards belongs to the second inequality.

next, since y ≥ in the first inequality, this means that the shaded region will be above the first inequality's line, thus shading regions a and b.

next, since y ≤ in the second inequality, this means that the shaded region is going to be below this line, thus shading regions a and c.

since both lines shade region a, this means that region a is the solution.

step-by-step explanation:

mark brainliest!