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# Hi Ni Ni Ni Essay

686 words - 3 pages

2011-2012 Semester 1 Exam Question 1 is one of the most challenging ones, the suggested answer is provided below. Other questions of the exam are very much the “same” as that of previous years.
Q1. The customer specification of the thickness of a product is 1000±3 mm, i.e., a product is considered defective if its thickness is below 997 mm or above 1003 mm. The production process, when under control, produces the products with thicknesses following a normal distribution with a mean = 1000 mm and a standard deviation of 1 mm. However, due to wearing and tearing, the process mean thickness may change and deviate from the target 1000 mm while the standard deviation stays the same. If the process mean thickness drifts to a position at which the process produces more than 5% defective products, it is considered unacceptable and whenever found so the process should be stopped. A random sample of size 9 is taken from the process from ...view middle of the document...

If = 1000, = 1. Note that a product is defective if its x > 1003 or x < 997.
x > 1003 z = (1003 – 1000)/1 = 3 Prob(x > 1003) = Prob( z > 3) = 0.0013
x < 997 z = (997 – 1000)/1 = - 3 Prob( x < 997) = 0.0013
Thus, defective rate = 0.0013 + 0.0013 = 0.0026 or 0.26%

If = 1002, = 1
x > 1003 z = (1003 – 1002)/1 = 1 Prob(x > 1003) = Prob( z > 1) = 0.159
x < 997 z = (997 – 1002)/1 = - 5 Prob( x < 997) = Prob(z < -5) = 0.000
Thus, defective rate = 0.159 + 0.000 = 0.159 or 15.9%

B. Type II error: the actual process mean thickness shifts to a position at which the process produces more than 5% defective products and yet the sample taken fails to detect the fact evidence and the process is thus kept going.

C. First find the at which the process produces 5% defective rate. From the normal table, the corresponding z: z = 1.64 or z = -1.64

0.05
x
997 998.64 1000 1001.36 1003

Thus, when = 1003 – 1.64 * 1 = 1001.36 or = 997 + 1.64 * 1 = 998.64
The corresponding process would produce 5% defectives. Note that when the mean is at either level, one of the two tails defective rate is virtually 0.
Due to symmetry, take the upper side ( = 1001.36) to work with, the lower side will be symmetrical.
When shifts to 1001.36, and n = 9 (sample size), X follows a normal distribution with = 1001.36 and X = 1/9 = 1/3.
Let the upper cutting point be U
β = Prob(X falls in the non-rejection region | = 1001.36) = Prob(X ≤ U | = 1001.36)
= 0.01 (preset by the management)
Thus, (U – 1001.36)/(1/3) = -2.33 U = 1001.36 – 0.777 = 1000.583.

0.01
997 998.64 1000 1000.583 1001.36
Thus, the critical cutting points are U = 1000.583, L = 999.417.
i.e., non-rejection region: (999.417, 1000.587)

D. When = 1000,
Prob(X > 1000.583 | = 1000) = Prob[z > 0.583/(1/3)] = Prob( z > 1.749) = 0.04
Thus, = Prob(999.417 < X < 1000.583 | = 1000) = 2*0.04 = 0.08.

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