The process standard deviation is 0.15, and the process control is set at plus o
ID: 3128847 • Letter: T
Question
The process standard deviation is 0.15, and the process control is set at plus or minus one standard deviation. Units with weights less than 9.85 or greater than 10.15 ounces will be classified as defects. Calculate the probability of a defect.
0.5248
0.7421
0.3174
0.500
The process standard deviation is 0.15, and the process control is set at plus or minus one standard deviation. Units with weights less than 9.85 or greater than 10.15 ounces will be classified as defects. Calculate the expected number of defects for a 1000-unit production.
524.8
742.1
317.4
500
Through process design improvements, the process standard deviation can be reduced to 0.05. Assume the process control remains the same, with weights less than 9.85 or greater than 10.15 ounces being classified as defects. Calculate the probability of a defect.
0.2224
0.9974
0.0013
0.0026
Through process design improvements, the process standard deviation can be reduced to 0.05. Assume the process control remains the same, with weights less than 9.85 or greater than 10.15 ounces being classified as defects. Calculate the expected number of defects for a 1000-unit production.
54.25
99.74
2.6
1.3
A.0.5248
B.0.7421
C.0.3174
D.0.500
Explanation / Answer
1.
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 9.85
x2 = upper bound = 10.15
u = mean = 10
s = standard deviation = 0.15
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -1
z2 = upper z score = (x2 - u) / s = 1
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.158655254
P(z < z2) = 0.841344746
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.682689492
Thus, those outside this interval is the complement = 0.3174 [ANSWER, c]
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2.
E(x) = 0.3174*1000 = 317.4 [ANSWER, C]
**********************
3.
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 9.85
x2 = upper bound = 10.15
u = mean = 10
s = standard deviation = 0.05
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -3
z2 = upper z score = (x2 - u) / s = 3
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.001349898
P(z < z2) = 0.998650102
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.997300204
Thus, those outside this interval is the complement = 0.0026 [ANSWER, D]
***********************
4.
E(x) = 0.0026*1000 = 2.6 [ANSWER, C]
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