MI
Unit 2Errors in MeasurementQ1) What is mean?A1) The most probable value of measured variable is the arithmetic mean of the number of readings taken. The best approximation is made when the number of readings of the same quantity are very large. Theoretically an infinite number of readings would give the best result although in practice only a finite number of measurements can be made. The arithmetic mean is given by
Q2) What is standard deviation?A2) The standard deviation of an infinite number of data is defined as the square root of the sum of the individual deviations squared, divided by the number of readings.Thus, the standard deviation is
In practice however the number of observations is finite. When the number of observations is greater than 20 S.D is denoted by symbol 
while if it less than 20 the symbol used is s. The Standard deviation of finite number of data is given by:
Q3) A set of independent current measurements were taken by six observers and were recorded as 12A, 12.2 A 12.5 A 13.1A and 12.9 A and 12.4 A. Calculate:(a) Arithmetic mean (b) Deviation from mean (c) Average deviation (d) Standard deviation A3)
Q4) Explain six sigma estimation?
Figure 1. Six Sigma Components of VariationThe variation in process can result due to the Actual Process Variation and the Variation from Measurement System. The Actual Process Variation is resulted because of Controllable Factors and/or Uncontrollable Factors.
Q5) Explain measurement analysis?A5) Measurement system errors can be due toAccuracy – The difference between the average of observed values and the standardRepeatability – Variation in measurement when a person measures the same unit repeatedly with the same measuring gage (or tool)Reproducibility - Variation in measurement when two or more persons measure the same unit using the same measuring gage (or tool)Stability - Variation in measurement when the same person measures the same unit using the same measuring gage (or tool) over an extended period of time.Linearity – The consistency of the measurement across the entire range of the measuring gage. Q6) What is Cp and Cpk?A6)Cp= Process Capability. A simple and straightforward indicator of process capability.
Cpk= Process Capability Index. Adjustment of Cp for the effect of non-centered distribution.Cpk is an index which measures how close a process is running to its specification limits, relative to the natural variability of the process. The larger the index, the less likely it is that any item will be outside the specs.” Q7) Determine the mean for the data values 5, 3,7, 8, 4, 9.A7):
Q8) Explain the steps in standard deviation?Calculate the mean (simple average of the numbers). For each number: subtract the mean. Square the result. Add up all of the squared results. Divide this sum by one less than the number of data points (N - 1). ... Take the square root of this value to obtain the sample standard deviation. Q9) 20 crystals from a solution and measure the length of each crystal in millimetres. Here is your data:9, 2, 5, 4, 12, 7, 8, 11, 9, 3, 7, 4, 12, 5, 4, 10, 9, 6, 9, 4Calculate the sample standard deviation of the length of the crystals.
Q10) What are the applications of mean?A10) Mean, median & mode shows different perspective of same data. Mean gives average of the data. All data is given equal importance. It is used in case where all data is important. E.g. Average salary of employees in an organization.
Where
x1,x2……xn = readings or variates or samples n = number or readings. |
S.D = |
while if it less than 20 the symbol used is s. The Standard deviation of finite number of data is given by:s = √d1 2 + d2 2 + d3 2 + ………………… + dn2 / n-1 = √ d2 / n-1 |
(a) Arithmetic mean
(b) Deviations : d1 = x1 - d2 = x2 - d3 = x3 - d4 = x4 – d5 = x5 - d6 = x6 - ( c ) Average deviation = D = 0.283 A ( d ) Standard deviation S = √ S=0.399 A |
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Figure 2. Six Sigma Process Variation
Figure 3. Observed Variation vs True Variation |
Cpk= Process Capability Index. Adjustment of Cp for the effect of non-centered distribution.Cpk is an index which measures how close a process is running to its specification limits, relative to the natural variability of the process. The larger the index, the less likely it is that any item will be outside the specs.” Q7) Determine the mean for the data values 5, 3,7, 8, 4, 9.A7):
Given data values are 5, 3, 7, 8, 4, 9. We know that the procedure to calculate the mean deviation. First, find the mean for the given data: Mean, µ = ( 5+3+7+8+4+9)/6 µ = 36/6 µ = 6 Therefore, the mean value is 6. |
A9)
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