The surface finish of a piece of steel is measured by a number called the R2. Th
ID: 3204132 • Letter: T
Question
The surface finish of a piece of steel is measured by a number called the R2. The lower the number, yurface of the steel. 1 of 4 line, the hardness of 9 individual pieces was tested. The parts were randomly ae finishis assumed to benormally distributed. The measured values obtained Find the mean of the above values Find the median of the above values Find the standard deviation of the above values What is the Range of the Values? What is the "statistical Range (Mean l.3std Deviations What is the range of surface finish values that wil contain 95% of the production parts hinta+/- 47.5N?Explanation / Answer
Mean = Sum of observations/ Count of observations
Mean = (0.9 + 0.85 + 0.68 + 0.75 + 0.51 + 0.8 + 0.85 + 0.81 + 0.92 / 9) = 0.7856
Variance
Step 1: Add them up
0.9 + 0.85 + 0.68 + 0.75 + 0.51 + 0.8 + 0.85 + 0.81 + 0.92 = 7.07
Step 2: Square your answer
7.07*7.07 =49.9849
…and divide by the number of items. We have 9 items , 49.9849/9 = 5.5539
Set this number aside for a moment.
Step 3: Take your set of original numbers from Step 1, and square them individually this time
0.9^2 + 0.85^2 + 0.68^2 + 0.75^2 + 0.51^2 + 0.8^2 + 0.85^2 + 0.81^2 + 0.92^2 = 5.6825
Step 4: Subtract the amount in Step 2 from the amount in Step 3
5.6825 - 5.5539 = 0.1286
Step 5: Subtract 1 from the number of items in your data set, 9 - 1 = 8
Step 6: Divide the number in Step 4 by the number in Step 5. This gives you the variance
0.1286 / 8 = 0.0161
Step 7: Take the square root of your answer from Step 6. This gives you the standard deviation
0.1268
Mean = 0.7856
variance = 0.0161
standard deviation=0.1268
Median: 0.81
Quartile Q1: 0.715
Quartile Q2: 0.81
Quartile Q3: 0.875
Minimum: 0.51
Quartile Q1: 0.715
Median: 0.81
Quartile Q3: 0.875
Maximum: 0.92
Range: 0.41
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