This is the assembly tolerances questions from the quality module practice problems. We are given that a supplier delivers a part for which the acceptable measurement range is between 110.45 and 110.55 centimeters. A random sample of 100 parts finds that the average part measures 110.5 centimeters and the standard deviation is 0.05 centimeters. The first question asks us what the supplier's capability score is. Recall that the capability score is defined as the upper specification limit minus the lower specification limit, divided by six times the standard deviation. Plugging in the numbers from above this is 110.55 minus 110.45 divided by 6 times 0.05, or 0.1 divided by 0.3, which is one-third, or 0.33. The second question asks us what percentage of deliveries are outside the specification limits. For this question we are assuming that the distribution of the part measurements follows the normal distribution. Then what we are interested in is the percentage of the deliveries that are in the two tails, below 110.45 and above 110.55. To solve this you can use the norm des to function in Excel. This tells you for a given value X the area under the curve to the left of that value X. Put another way, this tells us the probability that a parts measurement will be less than or equal to X. So to find the probability that the part is too small. In Excel use the formula NORM.DIST 110.45 for X, 110.5 for the mean, 0.05 for the standard deviation and 1 to indicate that we want the cumulative distribution, this gives us 0.15865525. To find the probability that the part is too large which is the area to the right of 110.55. We use the formula 1 minus NORM.DIST 110.55 now for X. And again, 110.5 for the mean, 0.05 for the standard deviation, and 1 to indicate that we want the cumulative distribution. Again, we get 0.15865525. Then the defect probability is simply the sum of these two values, which is 0.3173. Finally, if we want a capability score of 1.67. We use the capability score equation where the standard deviation is now unknown. We have 1.67 equals 110.55 minus 110.45 divided by 6 times the standard deviation. Solving for the standard deviation we have that this equals 0.1 divided by 6, times 1.67 which equals 0.01.