Question
An industrial sewing machine uses ball bearings that are targeted to have a diameter of .75 inches. The lower and upper specifications limits under which the ball bearing can operate are .74 inches and .76 inches, respectively. Past experience has indicated that the actual diameter of the ball bearings is approximately normally distributed with a mean of .753 inches and a standard deviation of .004 inches. What is the probability that a ball bearing will not meet the operating specifications?
Explanation / Answer
An industrial sewing machine uses ball bearings that are targeted to have a diameter of .75 inches. The lower and upper specifications limits under which the ball bearing can operate are .74 inches and .76 inches, respectively. Past experience has indicated that the actual diameter of the ball bearings is approximately normally distributed with a mean of .753 inches and a standard deviation of .004 inches. What is the probability that a ball bearing will not meet the operating specifications? In this case, any ball bearing that is 0.76 inches are rejected. Thus, we need to calculate the probablity of these areas. Let X be the diameter of ball bearing X~Normal (0.753. 0.004^2) First, we compute the z value for both 0.74 and 0.76 z value of 0.74 = (0.74 - 0.753)/0.004 = -3.25 z value of 0.76 = (0.76 - 0.753)/0.004 = 1.75 Using statistical table, we can find that: P(z < -3.25) = 0.00058 P(z < 1.75) = 0.95994 Thus, we are interested in the area where z< -3.25 and z>1.75 P(meeting requirement) = P(X< 0.76) - P(X < 0.74) = 0.95994 - 0.00058 = 0.95936 Thus, P(Not meeting requirement) = 1 - 0.95936 = 0.04064 Thus, 4.06% of the ball bearing will not meet the operating specifications. The probability is 0.04064. Hope this helps!