1. Data Analysis in Excel ACADs (08-006) Covered Keywords average, median, min, max, standard deviation, variable, varp, standardize, normal distribution, norminv, normsinv Description Supporting Material 1.1.1.21.1.1.4

2. Data Analysis in Excel Analysis of Uncertainty

3. Learning Objectives Learn to use statistical Excel functions: average, median, min, max, stdev, var, varp, standardize, normdist, norminv, normsinv

4. General Excel Behavior - Analyzes the range of cells you specify - Skips blank cells

5. Mean Excel =AVERAGE(cellrange) =AVERAGE(B72:B81) Example: SamplePopulation

6. Mode Value that occurs most often in discretized data ExcelExample: =MODE(cellrange) =MODE(B2:B81) If tie, reports first value in list

7. Median The middle value in sorted data Excel =MEDIAN(cellrange) =MEDIAN(D2:D81) Example: Note: When using this command, there is no need to sort the data first.

8. Maximum, Minimum, and Range Excel Example: =MIN(cellrange) =MIN(D2:D81) =MAX(cellrange) =MAX(D2:D81) There is no explicit command to find the range. However, it can be easily calculated. = MAX(D2:D81) - MIN(D2:D81)

9. Standard Deviation and Variance Population Sample Excel =STDEVP(cellrange) =STDEV(cellrange) =VARP(cellrange) =VAR(cellrange) Variance =  2 Variance = s 2

10. Review: The Normal Distribution The normal distribution is sometimes called the “Gauss” curve. mean x RF Relative Frequency

11. Standard Normal Cumulative Distribution Excel Example: =NORMSDIST(z) =NORMSDIST(1.0) =0.8413 area from minus infinity to z NOT 0 to z, like Z-table

12. Exam Grade Histogram

13. Excel Example Normal distribution with  =5,  =0.2 Find area from 4.8 to 5.4  Solution 1: =STANDARDIZE(4.8,5,0.2)Gives -1 =STANDARDIZE(5.4,5,0.2)Gives 2 =NORMSDIST(2)-NORMSDIST(-1) = 0.8186  Solution 2: =NORMDIST(5.4,5,0.2,TRUE)- NORMDIST(4.8,5,0.2,TRUE) = 0.8186

14. Inverse Problem Given ,  and probability, find x =NORMINV(prob,mean,stddev) Given probability, find z =NORMSINV(prob) Note: The probability is the area under the curve from minus infinity to x (or z)

15. Inverse Problem: Example 1 A batch of bolts have length  =5.00 mm,  =0.20 mm. 99% of the bolts are shorter than what length? Solution 1: =NORMINV(0.99,5,0.2) gives 5.47 mm Solution 2: =NORMSINV(0.99) = 2.33 5.00+0.20*2.33 = 5.47 mm

16. Inverse Problem: Example 2 A batch of bolts have length  =5.00 mm,  =0.20 mm. The bolt length is specified as 5.00 mm  tolerance. What is the value of the tolerance such that 99% of the bolts are encompassed? Solution: =NORMINV(0.995,5,0.2) = 5.52 mm =NORMINV(0.005,5,0.2) = 4.48 mm Tolerance = 5.52 - 5.00 = 0.52 mm Note: It is symmetrical; therefore 0.5% on either side

17. Bolt Specification