A detergent company periodically tests its products for variations in the fill weight. To do this, the company uses x-bar and R charts. Eight samples, each containing 10 detergent boxes, were selected from different batches. The data in fill weight (in ounces) of the eight samples is presented below. Sample Sample Mean Sample Range
1 5.8 0.42
2 6.1 0.38
3 16.02 0.08
4 15.95 0.15
5 16.12 0.42
6 6.18 0.23
7 5.87 0.36
8 16.2 0.4

If A2 for the sample size is 0.3080, the lower controllimit for the x-bar chart is

10.9361

Step-by-step explanation:

The lower control limit for xbar chart is

xdoublebar-A2(Rbar)

We are given that A2=0.308.

xdoublebar=sumxbar/k

Rbar=sumR/k

xbar      R

5.8        0.42

6.1         0.38

16.02    0.08

15.95    0.15

16.12     0.42

6.18      0.23

5.87     0.36

16.2      0.4

Xdoublebar=(5.8+6.1+16.02+15.95+16.12+6.18+5.87+16.2)/8

Xdoublebar=88.24/8

Xdoublebar=11.03

Rbar=(0.42+0.38+0.08+0.15+0.42+0.23+0.36+0.4)/8

Rbar=2.44/8

Rbar=0.305

The lower control limit for the x-bar chart is

LCL=xdoublebar-A2(Rbar)

LCL=11.03-0.308*0.305

LCL=11.03-0.0939

LCL=10.9361

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b

only 10/60 isn't equivalent

hope this : )