Question

An industrial sewing machine uses ball bearings that are targeted to have a diameter of 0.74 inch. The lower and upper specification limits under which the ball bearings can operate are 0.72 inch and 0.76 ​inch, respectively. Past experience has indicated that the actual diameter of the ball bearings is approximately normally​ distributed, with a mean of 0.743 inch and a standard deviation of 0.005 inch.

A. What is the probability that a ball bearing is between the target and the actual​ mean?

B. What is the probability that a ball bearing is between the lower specification limit and the target?

C. What is the probability that a ball bearing is above the upper specification limit?

D. What is the probability that a ball bearing is below the lower specification limit?

Answer 1

a) The probability that a ball bearing is between the target and the actual​ mean is computed here as:

P( 0.74 < X < 0.743)

Converting it to a standard normal variable, we have here:

Getting it from the standard normal tables, we have here:

Therefore 0.2257 is the required probability here.

b) The probability that a ball bearing is between the lower specification limit and the target is computed here as:

P( 0.72 < X < 0.74)

Converting it to a standard normal variable, we have here:

Getting it from the standard normal tables, we have here:

Therefore 0.2743 is the required probability here.

c) The probability here is computed as:

P(X > 76)

Converting it to a standard normal variable, we have here:

Getting it from the standard normal tables, we have here:

Therefore 0.0003 is the required probability here.

d) The probability that a ball bearing is below the lower specification limit is computed here as:

P(X < 72)

Converting it to a standard normal variable, we have here:

P(Z < -4.6) as got from part b).

= 0 is the required probability here.