A manufacturing process is designed to produce bolts with a 0.5-inch diameter. Once each day, a random sample of 36 bolts is selected and the bolt diameters recorded. If the resulting sample mean is less than 0.490 inches or greater than 0.510 inches, the process is shut down for adjustment. The standard deviation for diameter is 0.02 inches. What is the probability that the manufacturing line will be shut down unnecessarily

Answer is 0.0027.

Refer below for the explanation.

Step-by-step explanation:

As per the question,

0.5 in diameter,

36 bolts,

Standard deviation for diameter is 0.02 inches,

Resulting sample mean is less than 0.490 inches or greater than 0.510 inches.

Probability (b < 0.49) = Probability

(a< (0.49 - 0.50) / (0.02 / squareroot 36)

= Probability (a< - 3) = 0.00135

Probability = 2 x 0.00135

= 0.0027

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.