1. Homework solution#1 Q1: Suppose you have a sample from Palestine University and the distribution of the sample as: MedicineDentistEngineeringArtsCommerce Male3532401327 Female3733492729

2. Homework solution#1 Calculate the mean for male and female at the university. Solution: The above data is categorical data so there is no meaning of mean but we can calculate the proportion distribution of the Male and Female at the population “Palestine University” as:

3. Homework solution#1 For Male: For Female:

4. Homework solution#1 1.Plot bar chart for male and female separately.

5. Homework solution#1 1.Plot bar chart for male and female separately.

6. Homework solution#1 1.Plot companying bar chart over all the university

7. Homework solution#1 Plot pie chart for male

8. Homework solution#1 Plot pie chart for male

9. Homework solution#1 Plot pie chart for Female

10. Homework solution#1 Plot pie chart for Female

11. Homework solution#1 Q2: You have a data set {7, 3, 8, 11, 6, 7, 5, 2, 0} Calculate: mean –median- mean deviation-interquartile range- variance –standard deviation

12. Homework solution#1 Mean Median: 1. data will be sorted as: {0, 2, 3, 5, 6, 7, 7, 8, and 11}, here the number of data is 9 so it is even so the median divided the data and here is 6.

13. Homework solution#1 1.

14. Homework solution#1 Interquartile range =Q3-Q1 Q3 position = 0.75*10=7.5 Q1 position = 0.25*(9+1) = 2.5 IQR= Q3-Q1=7.5-2.5=5

15. Homework solution#1 Variance:

16. Homework solution#1 Q3: You have a data set {13, 23, 27, 5, 16, 11, 9, 4} Plot box plot. 1. Sort the data as {4, 5, 9, 11, 13, 16, 23, 27} Minimum = 4, maximum = 27 Q1 position = 0.25(8+1)= 2.24

17. Homework solution#1 2. Q2 position = 0.5(9)= 4.5 Q3 position = 0.75(9) = 6.75

18. Homework solution#1 Box plot

19. Homework solution#1 Calculate: mean –median- mean deviation-interquartile range- variance standard deviation – coefficient of variation Median = Q2 = 4.5

20. Homework solution#1 Interquartile range = Q3-Q1 = 19.5-7=12.5 Coefficient of Variation

21. Homework solution#1 Suppose you have the tabulated data set: AgeStudent 0 – 223 3 – 533 6 – 8 9 – 1168 12 – 1419 15 – 1710 18 – 201 Total217

22. Homework solution#1 Plot Histograms – Polygon – Ogive. From the plotting explain the relation between histogram and polygon

23. Homework solution#1 Histogram AgeStudentclass midpointR.Frq. 0 00 0 – 22310.105991 3 – 53340.152074 6 – 86370.290323 9 – 1168100.313364 12 – 1419130.087558 15 – 1710160.046083 18 – 201190.004608 Total217 1

24. Homework solution#1 Histogram

25. Homework solution#1 Polygon The relation is polygon is the mid of column bars.

26. Homework solution#1 Ogive

27. Homework solution#1 Calculate mean – median - interquartile range- variance – standard deviation AgeStudentclass midpointf.x 0000 0 – 2231 3 – 5334132 6 – 8637441 9 – 116810680 12 – 141913247 15 – 171016160 18 – 20119 Total217 1702

28. Homework solution#1 1.Mean Median AgeStudent Freqclass midpointCum Freq. 0000 0 – 2231 3 – 533456 6 – 8637119 9 – 116810187 12 – 141913206 15 – 171016216 18 – 20119217 Total217

29. Homework solution#1 Step 1: Form the cumulative frequency table. Step (2): Step (3): median class is (6-8) Step (4): Lm =6, Fm-1 = 56, fm =63, Cm = 2 Step (5)

30. Homework solution#1 IQR = Q3 – Q1 Q3 Step (2): Step (3): Third quartile class (9-11) Step (4): Lm =9, Fm-1 = 119, fm =10, Cm = 2

31. Homework solution#1 Step (2): Step (3): Third quartile class (3-5) Step (4): Lm =3, Fm-1 = 23, fm =4, Cm = 2 IQR = 31.6 -19.5 = 12.1

32. Homework solution#1 Variance: AgeStudentxX^2fxfx^2 000000 0 – 22311 3 – 533416132528 6 – 8637494413087 9 – 1168101006806800 12 – 1419131692473211 15 – 1710162561602560 18 – 2011936119361 Total217 1702 1657 0

33. Homework solution#1 Variance: Standard Deviation =

34. Any Questions?