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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