An industrial sewing machine uses ball bearings that are targeted to have a diameter of 0.75 inch. The lower and upper specification limits under which the ball bearings can operate are 0.74 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 0.753 inch and a standard deviation of 0.004 inch. What is the probability that a ball bearing is:. a. between the target and the actual mean?
b. between the lower specification limit and the target?
c. above the upper specification limit?d. below the lower specification limit?

(a) Probability that a ball bearing is between the target and the actual mean is 0.2734.

(b) Probability that a ball bearing is between the lower specification limit and the target is 0.226.

(c) Probability that a ball bearing is above the upper specification limit is 0.0401.

(d) Probability that a ball bearing is below the lower specification limit is 0.0006.

Step-by-step explanation:

We are given that an industrial sewing machine uses ball bearings that are targeted to have a diameter of 0.75 inch. The lower and upper specification limits under which the ball bearings can operate are 0.74 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 0.753 inch and a standard deviation of 0.004 inch.

Let X = diameter of the ball bearings

SO, X ~ Normal()

The z-score probability distribution for normal distribution is given by;

                                Z  =    ~ N(0,1)

where, = population mean = 0.753 inch

            = standard deviation = 0.004 inch

(a) Probability that a ball bearing is between the target and the actual mean is given by = P(0.75 < X < 0.753) = P(X < 0.753 inch) - P(X 0.75 inch)

      P(X < 0.753) = P( < ) = P(Z < 0) = 0.50

      P(X 0.75) = P( ) = P(Z -0.75) = 1 - P(Z < 0.75)

                                                             = 1 - 0.7734 = 0.2266

The above probability is calculated by looking at the value of x = 0 and x = 0.75 in the z table which has an area of 0.50 and 0.7734 respectively.

Therefore, P(0.75 inch < X < 0.753 inch) = 0.50 - 0.2266 = 0.2734.

(b) Probability that a ball bearing is between the  lower specification limit and the target is given by = P(0.74 < X < 0.75) = P(X < 0.75 inch) - P(X 0.74 inch)

      P(X < 0.75) = P( < ) = P(Z < -0.75) = 1 - P(Z 0.75)

                                                            = 1 - 0.7734 = 0.2266

      P(X 0.74) = P( ) = P(Z -3.25) = 1 - P(Z < 3.25)

                                                             = 1 - 0.9994 = 0.0006

The above probability is calculated by looking at the value of x = 0.75 and x = 3.25 in the z table which has an area of 0.7734 and 0.9994 respectively.

Therefore, P(0.74 inch < X < 0.75 inch) = 0.2266 - 0.0006 = 0.226.

(c) Probability that a ball bearing is above the upper specification limit is given by = P(X > 0.76 inch)

      P(X > 0.76) = P( > ) = P(Z > -1.75) = 1 - P(Z 1.75)

                                                            = 1 - 0.95994 = 0.0401

The above probability is calculated by looking at the value of x = 1.75 in the z table which has an area of 0.95994.

(d) Probability that a ball bearing is below the lower specification limit is given by = P(X < 0.74 inch)

      P(X < 0.74) = P( < ) = P(Z < -3.25) = 1 - P(Z 3.25)

                                                            = 1 - 0.9994 = 0.0006

The above probability is calculated by looking at the value of x = 3.25 in the z table which has an area of 0.9994.

answer: ok

step-by-step explanation: