Download Devor Chapter 3

Using the Standard Normal Distribution
to Solve SPC Problems
Table A.1: “Areas Under the Normal Curve”
Standard Normal Distribution
-5
-4
-3
-2
-1
0
1
2
3
4
5
Z
ISE 327 Ch. 3 Homework
1
Applications of the Normal Distribution
 A certain machine makes electrical resistors having a
mean resistance of 40 ohms and a standard deviation of
2 ohms. What percentage of the resistors will have a
resistance less than 44 ohms?
 Solution: X is normally distributed with μ = 40 and σ = 2 and x = 44
X 
Z

Z
44  40
2
-5
0
5
P(X<44) = P(Z< +2.0) = 0.9772
Therefore, we conclude that 97.72% will have a resistance less than 44
ohms.
What percentage will have a resistance greater than 44 ohms?
ISE 327 Ch. 3 Homework
2
The Normal Distribution “In Reverse”
Given a normal distribution
with μ = 40 and σ = 6, find the
value of X for which 45% of the
area under the normal curve is
to the left of X.
\\\
-5
45%
0
5
k = -0.125
1) If P(Z < k) = 0.45, then
k = -0.125 (table)
2) Z = / (x-µ) / σ
3) -0.125 = (X – 40)/6
4) X- 40 = 6 * (-0.125)
Therefore, X = 39.25
-5
0
5
X = 39.25
ISE 327 Ch. 3 Homework
3
Problem 3.9a Part 1
Specifications 0.500 ± 0.025
LSL = 0.475
USL = 0.525
Normal distribution
μx = 0.505 and σx = 0.0065
Find the percent that are within
specifications.
-5
0
LSL = 0.475
5
USL = 0.525
Step 1 Find the corresponding z values.
Z = (x-µ) / σ
ZL = (-0.475-0.505)/0.0065
ZL = -4.62
Z = (x-µ) / σ
ZU = (0.525-0.505)/0.0065
ZU = 3.08
-5
0
ZL = -4.62
ISE 327 Ch. 3 Homework
5
ZU = 3.08
4
Problem 3.9a Part 2
Specifications 0.500 ± 0.025
LSL = 0.475
USL = 0.525
Normal distribution
μx = 0.505 and σx = 0.0065
Find the percent that are within specifications.
Step 2 Find the area under the curve between the
LSL and USL.
-5
0
LSL = 0.475
5
USL = 0.525
P(Z  3.08) - P(Z  -4.62)
= 0.99897 – 0.0
= 0.99897
Therefore 99.897% are within specifications.
What percent are outside of specifications?
99.897%
-5
0
ZL = -4.62
ISE 327 Ch. 3 Homework
5
ZU = 3.08
5
Problem 3.9b
Specifications 0.500 ± 0.025
LSL = 0.475
USL = 0.525
Normal distribution
Process variability reduced to 0.003
μx = 0.505 and σx = 0.003
Find the percent that are within specifications.
Z = (x-µ) / σ
ZL = (-0.475-0.505)/0.003
ZL = -10.0
Similarly, ZU = 6.667
-5
0
5
USL = 0.525
LSL = 0.475
P(Z  6.667) - P(Z  -10.0)
Approximately 100%
Therefore 100% are within specifications.
~ 100%
-5
ZL = -10.0
ISE 327 Ch. 3 Homework
0
5
ZU = 6.667
6