Given Information:
Population mean = μ = 0.5
Population standard deviation = σ = 0.02
Sample size n = 36
Required Information:
Probability that the manufacturing line will be shut down unnecessarily = ?
Answer:
P = 0.9975
Explanation:
The sample mean will be same as population mean μs = 0.5
The sample standard deviation is given by
σs = σ/√n
σs = 0.02/√36
σs = 0.0033
Now we can calculate the probability that the process will shut down for adjustment.
P(x < 0.490) = (x - μs)/σs
P(x < 0.490) = (0.490 - 0.50)/0.0033
P(x < 0.490) = -3.03
The corresponding z-score from the z table is 0.00122
P(x > 0.510) = (x - μs)/σs
P(x > 0.510) = (0.510 - 0.50)/0.0033
P(x > 0.510) = 3.03
As you can see the distribution is symmetric, therefore,
P( 0.510 < x < 0.490) = 2*0.00122
P( 0.510 < x < 0.490) = 0.00244
Notice that it was asked in the question to find the probability that the manufacturing line will be shut down unnecessarily, which means
P = 1 - P( 0.510 < x < 0.490)
P = 1 - 0.00244
P = 0.9975
Therefore, the probability that the manufacturing line will be shut down unnecessarily is 0.9975.
