Complete Question
Assume a production process produces items with a mean weight of 10 ounces.
Through process design improvements, the process standard deviation can be reduced to 0.04. Assume the process control remains the same, with weights less than 9.88 or greater than 10.12 ounces being classified as defects. Calculate the probability of a defect.
Answer:
0.0027
Step-by-step explanation:
We solve using the formula for z score
z = (x-μ)/σ, where
x is the raw score
μ is the population mean = 10 ounces
σ is the population standard deviation = 0.04 ounces
Hence:
For x < 9.88 ounces
z = 9.88 - 10/0.04
z = -3
Probability value from Z-Table:
P(x < 9.88) = 0.0013499
For x > 10.12 ounces
z = 10.12 - 10/0.04
z = 3
Probability value from Z-Table:
P(x<10.12) = 0.99865
P(x>10.12) = 1 - P(x<10.12)
= 0.0013499
The probability of a defect is calculated as:
P(x = 10.12) or P(x = 9.88)
P(z = 3) or P(z = -3)
= 2P
= 2(0.0013499)
= 0.0026998
Approximately = 0.0027
Therefore, the Probability of the defect is 0.0027
