II SEMESTER BBA APRIL/MAY 2015QUANTITATIVE METHODS FOR BUSINESS IITIME 3 HRSMARKS - 70Dr. SriramappaPhone No - 9448412626 SECTION A1.a. In plural sense it means a collection of numerical facts.b. Tabulation is the systematic arrangement of statistical data in columns and rows, the purpose of the table is to simplify the presentation of data & facilitate comparison.c. Statistical data is graphically presented as the presentation of data is attractive & effective, simple & understandable, useful in comparison & interpretation. d. If both the variables vary in the same direction it is said to be positive & if the variables are varying in opposite direction it is negative correlation.e. Limitations of median are it is more affected by fluctuations in sample, it has to be computed through interpolation in case of continuous serious, it may not be the true representative when the variations are more in values.f. According to A L Bowley Dispersion is the measure of variation of itemsg. Consumer price index represents the increase in price paid by the ultimate consumer for a specified basket of goods and services.SECTION B2. The following main aspects are included i. Subject matter of statistics Statistical methods & Applied Statisticsii. Nature of Statistics Statistics as science & as an art

3. Guidelines for tabulation are i. Table number ii. Title of the table iii. Caption iv. Stubs v. Body vi. Head note

4. Presentation of data relating to coffee habitsSex MaleFemaleTotal

City CDNCDTotalCDNCDTotalCDNCDTotal

A402060535404555100

B2530551530454060100

Total655011520658585115200

5. Calculation of MedianCIFCF

0-1044

10-201216

20-302440

30-403676

40-502096

50-6016112

60-708120

70-805125

125

Median class = N/2th item= 125/2th item= 62.5th item lies in CI 30-40Median = L + h/f [N/2 C]= 30 + 10/36 [125/2 40]= 30 + 0.278 [22.4]=36.236. r = 0.6x = 1.5y = 2.00x = 10y = 20 x on yx-x = r [x/ y] [y-y] x-10 = 0.6 [1.5/2] [y-20]x-10 = 0.45[y-20]x-10 = 0.45y 9x=0.45y+1y on xy-y = r [y/ x] [x-x]y-20 = 0.6 [2/1.5] [x-10]y-20 = 0.8[x-10]y-20 = 0.8x 8y=0.8x+12

SECTION C7. Calculation of Mean, Median & ModeCIFXFXCF

0-101457014

10-20231534537

20-30352587572

30-40203570092

40-50845360100

50-60555275105

1052625

Mean = fx / n= 2625/105=25Median Class = n/2 = 105/2 = 52.5 lies in 20-30= L + [h/f] [n/2 c]= 20 + 10/35 [52.5 37]= 20 + 0.29[15.5]= 24.43Mode = L + [h (f1-f0) / 2f1-f0-f2]20 + 10(35-23) / 2 x 35 23 -20= 24.44

8. Computation of coefficient of variation for A & Bxd=x-60d2xd=x-60d2

44-1625648-12144

80204007515225

761625654-636

48-121446000

52-8646339

721214469981

72121447212144

51-98151-981

600057-39

54-63666636

6099152561515765

For Student AX = x /N = 609/10 = 60.9 = d2/n (d/n)2= 1525/10 (9/10)2= 152.5 0.81 =151.69= 12.32CV = ( / x) 100= (12.32/60.9) 100= 20.23%For Student BX = x /N = 615/10 = 61.5 = d2/n (d/n)2= 765/10 (15/10)2= 76.5 2.25 =74.25 = 8.617CV = ( / x) 100= (8.617/61.5) 100= 14.01%Here coefficient of variation is less for student B, so B is more consistent student & if prize is based on consistency B should get the prize.

9. Correlation coefficientPass percentage = 480/800*100 = 60%Percentage of failure = 40%xdxdx2ydydy2dxdy

600040101000

50-101005020400-200

50-101005020400-200

802040020-10100-200

903090010-20400-600

751522525-525-75

405451725195151425-1275

dx = x-60 dy = y-30 R= dxdy [(dx)(dy)] / N dx2 (dx)2/n x dy2 (dy)2/nR = -1275 (45x15)/61725 (45)2/6 x 1425 (15)2/6= -1275-112.5 / 13875.5 x 1387.5 = -1387.5/1387.5 = -1 The correlation is perfect negative correlation

10. Calculation of Fishers Ideal Index numberP0Q0P1Q1P1Q0P0Q0P1Q1P0Q1

A85020261000400520208

B21561090306020

C12022540205025

D2105850204016

E140330120409030

1300510760299

Fishers Ideal Index P01= (P1Q0 / P0Q0) X (P1Q1 / P0Q1) X 100= (1300 / 510) X (760 / 299) X 100= [2.55 x 2.54] x 100= 6.48 x 100= 2.5456 x 100= 254.56TRT = (P1Q0 / P0Q0) X (P1Q1 / P0Q1) X [(P0Q1 / P1Q1) X (P0Q0/ P1Q0)]= (1300 / 510) X (760 / 299) X (299 / 760) X (510/ 1300) = 1 = 1FRT = P01 X Q01 = P1Q1 / P0Q0= (P1Q0 / P0Q0) X (P1Q1 / P0Q1) X [(P0Q1 / P0Q0) X (P1Q1/ P1Q0)]= (1300 / 510) X (760 / 299) X (299 / 510) X (760/ 1300)` = (760 / 510) X (760 / 510) = 760/510

11. Calculation of Co-efficient of Pearsons Skewness Wagesfmdd2fdFd2

270-28012275-24-2448

280-29018285-11-1818

290-300352950000

300-31042305114242

310-3205031524100200

320-3304532539135405

330-3402033541680320

340-350834552540200

3551233

X = A + (fd /n) x i= 295 + 355/230 x 10 = Rs.310.43SD = (fd2 /n - (fd /n)2 X i = 1233/230 (355/230)2 x100= 5.3609 2.3823 x 100= Rs.17.26M0 = L + [ h(f1-f0) / 2f1-f0-f2]= 310 + 10 [(50-42) / 2(50)42 45] = 310 + 80/13= 316.1538Sk = X M0 / = 310.43 316.15 / 17.26= -0.33