BUSINESS

STATISTIC

Submitted To,

Dr. Maya M. Ahmed

Prof. Of JKSHIM

Nitte.

Submitted By,

SHETTY MANJUNATH

1st MBA

‘C’ Section

Date-18-12-2013 Roll No.-14

Problem Is Taken From My SSC (10th) and HSC(12th) Scores They Are As Follows:

Scores

Subject

SSC(x)

HSC(y)

English/English 53 61

Hindi / Hindi 63 65

Maths / Account 135 75

Social Science / Economic

83 71

Marathi / Organization In

Commerce

63 69

Solutions

1. Suitable bar diagram to represent the data:

1 2 3 4 50

20

40

60

80

100

120

140

160

SSCHSC

2. Calculation of mean.

Mean of SSC

x=∑xn

3975

= 79.4

Mean of HSC

y=∑ yn

=3415 = 68.2

SSC Marks(x) HSC Marks(y)

53 61

63 65

135 75

83 71

63 69

∑x=397 ∑y=341

2.Calculation of standard deviation

SSC Marks(x)

HSC Marks(y)

x-x (x-x)2 y-y (y-y)2

53 61 -26.4 696.96 -7.2 51.84

63 65 -16.4 268.96 -3.2 10.24

135 75 55.6 3091.36 6.8 46.24

83 71 3.6 12.96 2.8 7.84

63 69 -16.4 268.96 0.8 0.64

∑x=397 ∑y=341 4339.2 116.8

x=79.4y=68.2

Its calculation showed in above while calculating mean.

For the data of SSC, standard deviation is as follows

S .D of=√∑ (x−x )2

n

=√ 4339.25

=29.46

From the data of HSC, standard deviation is calculated as follows.

S .D of y=√∑ ( y− y )2

n

=√ 116.85

=4.83

3. Computation of KarlPearson’s coefficient of correlation by using data of HSC and SSC scores are as follows,

X Y U=x-x V=y-y u2 v2 uv

53 61 -26.4 -7.2 696.96 51.84 190.08

63 65 -16.4 -3.2 268.96 10.24 52.48

135 75 55.6 6.8 3091.36 46.24 378.08

83 71 3.6 2.8 12.96 7.84 10.08

63 69 -16.4 0.8 268.96 0.64 -13.12

0 0 4339.2 116.8 617.6

x=79.4y=68.2

Its calculation shown in above while calculating mean.

r(uv)= n∑uv−∑u∑v

√n¿¿¿

¿

5 (617.6 )−0 (0 )

√5¿¿¿ ¿¿

=3088

√21696∗584

= 30883559.56

=0.868

5. Computation Regression equation by using data of SSC and HSC scores are as follows,

X Y dx=x-x dy=y-y dx2 dy2 dxdy

53 61 -26.4 -7.2 696.96 51.84 190.08

63 65 -16.4 -3.2 268.96 10.24 52.48

135 75 55.6 6.8 3091.36 46.24 378.08

83 71 3.6 2.8 12.96 7.84 10.08

63 69 -16.4 0.8 268.96 0.64 -13.12

0 0 4339.2 116.8 617.6

x=79.4y=68.2

its calculation shown in above while calculating mean.

Formula forcalculating regression

byx=∑dxdy

∑d x2

= 617.64339.2

=0.142

The line of regression of y on x is

y-y =byx (x-x)

y-68.2=(0.142) (x-79.4)

y= 0142x+56.925

bxy=∑dxdy

∑d y2

=617.6116.8

= 5.288

the line of regression of x on y is,

x-x= bxy(y-y ¿¿

x-79.4 =5.288 (y-68.2)

x= 5.288y-281.242

Source: SSC(10th) And HSC(12th) marks

Conclusion:

THE ABOVE DATA IS ABOUT THE MARKS OF SSC AND HSC WHERE WE TAKEN X AS SSC AND Y IS HSC. THROUGH THIS DATA WE HAVE COME TO KNOW ABOUT MEAN, STANDARD DEVIATION,REGRESSION AND BAR DIAGRAM FROM THIS WE CAME TO KNOW ABOUT EACH INDIVIDUAL PERFORMANCE AND WHERE THEY ARE STANDING.