Statistical Analysis of Process A and Process B

1)
Process A Diameter in (mm) Process B Diameter in (mm)
45.03 45.01
44.98 44.99
45.01 45.01
45.04 45.01
44.96 44.99
45.02 44.98
44.98 45.02
45.02 45.02
45.03 44.99
44.99 44.98
a),b) & c)
Mean is defined as,
μ= 1
n ∑
i=1
n
xi
For process A,
μA = 1
10 ∑
i=1
n
xi= 450.06
10 =45.006
Var A = 1
10 ∑
i=1
n
(xi−μ A )
Var A =0.02403
σ A= √Var A =0.155
For process B,
μB= 1
10 ∑
i=1
n
xi= 450
10 =45
VarB = 1
10 ∑
i=1
n
( xi −μB )
VarB =0.01404
σ B= √Var B=0.118
d)
The manufacturer should prefer process B as the deviation is less from the mean in the process B as
compared to process A.
2)

a),b) The lowest value = 20.41 and the highest value = 21.16
Class Interval Data Frequency
20.4 – 20.50 20.41 1
20.5 – 20.60 20.57, 20.53, 20.5, 20.56 4
20.6 – 20.70 20.63, 20.65, 20.65, 20.61, 20.6,
20.68
6
20.7 – 20.80 20.74, 20.75, 20.71, 20.73, 20.75,
20.79, 20.78, 20.77, 20.72, 20.75
10
20.8 – 20.90 20.86, 20.82, 20.89, 20.89, 20.88,
20.8, 20.82, 20.86, 20.88
9
20.9 – 21.00 20.9, 20.94, 20.97, 20.92, 20.9,
20.94
6
21.00 – 21.10 21.03, 21.08, 21.07 3
21.10 – 21.20 21.16 1
c)
20.4 –
20.50 20.5 –
20.60 20.6 –
20.70 20.7 –
20.80 20.8 –
20.90 20.9 –
21.00 21.00 –
21.10 21.10 –
21.20
0
2
4
6
8
10
12
Histogram
d)
The median is 20.785
e)

20.4 –
20.50 20.5 –
20.60 20.6 –
20.70 20.7 –
20.80 20.8 –
20.90 20.9 –
21.00 21.00 –
21.10 21.10 –
21.20
0
2
4
6
8
10
12
Frequency
f)
The standard deviation is 0.16707
g)
Pie Chart
20.4 – 20.50 20.5 – 20.60 20.6 – 20.70 20.7 – 20.80
20.8 – 20.90 20.9 – 21.00 21.00 – 21.10 21.10 – 21.20
3)
i)
a) The probability it is polished in H is 2/3.
b) Component is trimmed on either on F or G = (1/3 + 1/3) = 2/3
c) The probability is = (1/4 + 1/4)*(1/3)*(2/3) = 1/9