Using this partial Z Table, how many units from a months production run are expected to not satisfy customer requirements for the following process?
Upper specification limit: 7.2 Lower specification limit: 4.3 Mean of the process: 5.9 Standard Deviation: 0.65 Monthly production: 450 units
To determine how many units are expected to not satisfy customer requirements, we need to calculate the Z-scores for both the Upper Specification Limit (USL) and the Lower Specification Limit (LSL) using the given mean and standard deviation. The Z-score for the USL (7.2) is (7.2 - 5.9) / 0.65 = 2. The Z-score for the LSL (4.3) is (4.3 - 5.9) / 0.65 = -2. From the Z-table, the area to the left of Z = 2 is 0.97725 and the area to the left of Z = -2 is 0.02275. The area within the specifications (between the Z-scores) is 0.97725 - 0.02275 = 0.9545. The area outside the specifications (not satisfying customer requirements) is 1 - 0.9545 = 0.0455. To find the number of units expected to not satisfy customer requirements: 0.0455 * 450 = 20.475, which is approximately 10 units as we are considering whole units. Therefore, the correct answer is 10.
I think this question doesn't have the right answer option of 13 as available on another forum. The correct answer should be 13. * Find Z score @ UCL and @ LCL. viz Area left of UCL = 0.97725 and Area left of LCL = 0.00695 * Subtracting the two will give the area under the curve. 0.9703 * 1 minus the above area will give the failure zone. i.e. 0.0297 * multiply with the N i.e. 450 X 0.0297 = 13.36 > This is no. of units NOT to satisfy customer requirement.
Apply Z score formula. Z=(UCL-mean)/SD => Z=2. From there found P(Z)=0.97725. Need to find outside of normal curve= 1-P = 0.02275. Multiply 0.02275 with 450 you get 10.2375. Hence the answer is 10.
Me too.. pls help out anyone
I need full solution plz