Business Statistics and Mathematics Solved Paper 2014, Punjab University, BCOM, ADC I

Business Statistics and Mathematics Solved Paper 2014, Punjab University, BCOM, ADC I
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In this Post, we are going to discuss the Paper of Business Statistics and Mathematics Solved Paper 2014, Punjab University, BCOM, ADCI in which Measures of Central Tendency, Measures of Dispersion, Correlation & Regression, Index Numbers, Matrix, Arithmetic Progression, Geometric Progression, Simultaneous Linear Equations, Annuity, Quadratic Equation is discussed and solved.

Other solved papers of Business Statistics & Mathematics

Solved by Iftikhar Ali, M.Sc Economics, MCOM Finance Lecturer Statistics, Finance and Accounting

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Business Statistics and Mathematics Solved Paper 2014, Punjab University, BCOM, ADC I

Section I Business Statistics

Q.1 Calculate Arithmetic Mean, Median and Coefficient of Variation.

Weekly Wages Rs.No of WorkersWeekly Wages Rs.No of Workers
0-406160-20045
40-8015200-24027
80-12022240-28013
120-16030280-3206

Solution:

ClassesFrequency (f)Xfxfx²C.f
0—4062012024006
40—8015609005400021
80—12022100220022000043
120—16030140420058800073
160—2004518081001458000118
200—2402722059401306800145
240—280132603380878800158
280—32063001800540000164
Sum164 266405048000 
 ∑f= ∑fx=∑fx²= 

    \[ \left( \mathbf{a} \right)\mathbf{\ A.M\ }\overline{\mathbf{X}}\mathbf{=}\frac{\mathbf{\sum fx}}{\mathbf{\sum f}}\mathbf{= \ }\frac{\mathbf{26640}}{\mathbf{164}}\mathbf{= 162.44\ }\ \]

    \[ \mathbf{Selection\ of\ class\ for\ median\ }\frac{\mathbf{n}}{\mathbf{2}}\mathbf{=}\frac{\mathbf{164}}{\mathbf{2}}\mathbf{= 82}\ \]

82 falls in C.F of 118 So L = 160, f= 45, n/2 = 82, h = 40, C = 73

    \[ \left( \mathbf{b} \right)\mathbf{Median = L + \ }\frac{\mathbf{h}}{\mathbf{f}}\left( \frac{\mathbf{n}}{\mathbf{2}}\mathbf{- \ C} \right)\ \]

    \[ \mathbf{Median = 160 +}\frac{\mathbf{40}}{\mathbf{45}}\left( \mathbf{82 - \ 73} \right)\ \]

    \[ \mathbf{Median = 168}\ \]

    \[ \left( \mathbf{c} \right)\mathbf{\ C.V = \ }\left( \frac{\mathbf{S.D}}{\mathbf{Mean}} \right)\mathbf{100}\ \]

    \[ \mathbf{S.D =}\sqrt{\left\lbrack \frac{\mathbf{\sum f}\mathbf{x}^{\mathbf{2}}}{\mathbf{\sum f}}\mathbf{- \ }\left( \frac{\mathbf{\sum fx}}{\mathbf{\sum f}} \right)^{\mathbf{2}} \right\rbrack}\ \]

    \[ \mathbf{S.D =}\sqrt{\left\lbrack \frac{\mathbf{5048000}}{\mathbf{164}}\mathbf{- \ }\left( \frac{\mathbf{26640}}{\mathbf{164}} \right)^{\mathbf{2}} \right\rbrack}\ \]

    \[ \mathbf{S.D =}\sqrt{\left\lbrack \mathbf{30780.48 - \ 26386.43} \right\rbrack}\ \]

    \[ \mathbf{S.D = 66.29}\ \]

    \[ \mathbf{Coefficient\ of\ Variation\ C.V = \ }\left( \frac{\mathbf{66.29}}{\mathbf{162.44}} \right)\mathbf{100 = 40.80}\ \]

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Q.2

X:        05, 06, 07, 08, 09, 10, 11, 12, 13, 14, 15

Y:        09, 07, 10, 03, 13, 11, 14, 10, 14, 12, 18

Required: Calculate Co-efficient of correlation and also the line of regression y on x.

Solution: Correlation Coefficient

Formula

    \[ \mathbf{r =}\frac{\frac{\mathbf{\sum xy}}{\mathbf{n}}\mathbf{- \ }\left( \frac{\mathbf{\sum x}}{\mathbf{n}} \right)\left( \frac{\mathbf{\sum y}}{\mathbf{n}} \right)}{\mathbf{Sx \times Sy}}\ \]

S/NoXYXYX²Y²
159452581
267423649
37107049100
48324649
591311781169
61011110100121
71114154121196
81210120144100
91314182169196
101412168196144
111518270225324
     
SUM110121130212101489
 ∑X =∑Y =∑XY =∑X² =∑Y² =

    \[ \overline{\mathbf{X}}\mathbf{=}\frac{\mathbf{\sum X}}{\mathbf{n}}\mathbf{=}\frac{\mathbf{110}}{\mathbf{11}}\mathbf{= 10\ ,\ }\overline{\mathbf{Y}}\mathbf{=}\frac{\mathbf{\sum Y}}{\mathbf{n}}\mathbf{=}\frac{\mathbf{121}}{\mathbf{11}}\mathbf{= 11\ }\ \]

    \[ \mathbf{Sx =}\sqrt{\left\lbrack \frac{\mathbf{\sum}\mathbf{x}^{\mathbf{2}}}{\mathbf{n}}\mathbf{- \ }\left( \frac{\mathbf{\sum x}}{\mathbf{n}} \right)^{\mathbf{2}} \right\rbrack}\ \]

    \[ \mathbf{Sx =}\sqrt{\left\lbrack \frac{\mathbf{1210}}{\mathbf{11}}\mathbf{- \ }\left( \mathbf{10} \right)^{\mathbf{2}} \right\rbrack}\ \]

    \[ \mathbf{Sx =}\sqrt{\mathbf{110 - 100}}\ \]

    \[ \mathbf{Sx =}\sqrt{\mathbf{10}}\ \]

    \[ \mathbf{Sx = 3.162}\ \]

    \[ \mathbf{Sy =}\sqrt{\left\lbrack \frac{\mathbf{\sum}\mathbf{y}^{\mathbf{2}}}{\mathbf{n}}\mathbf{- \ }\left( \frac{\mathbf{\sum y}}{\mathbf{n}} \right)^{\mathbf{2}} \right\rbrack}\ \]

    \[ \mathbf{Sy =}\sqrt{\left\lbrack \frac{\mathbf{1489}}{\mathbf{11}}\mathbf{- \ }\left( \mathbf{11} \right)^{\mathbf{2}} \right\rbrack}\mathbf{\ }\ \]

    \[ (\mathbf{Sy =}\sqrt{\mathbf{135.36 - 121}}\ \]

    \[ \mathbf{Sy =}\sqrt{\mathbf{14.36}}\ \]

    \[ \mathbf{Sy = 3.78}\ \]

    \[ \mathbf{r =}\frac{\frac{\mathbf{\sum xy}}{\mathbf{n}}\mathbf{- \ }\left( \frac{\mathbf{\sum x}}{\mathbf{n}} \right)\left( \frac{\mathbf{\sum y}}{\mathbf{n}} \right)}{\mathbf{Sx \times Sy}}\ \]

    \[ \mathbf{r =}\frac{\frac{\mathbf{1302}}{\mathbf{11}}\mathbf{- \ }\left( \mathbf{10} \right)\left( \mathbf{11} \right)}{\left( \mathbf{3.162} \right)\mathbf{\times (3.78)}}\ \]

    \[ \mathbf{r =}\frac{\mathbf{118.36\ -\ 110}}{\mathbf{11.95}}\ \]

    \[ \mathbf{r =}\frac{\mathbf{8.36}}{\mathbf{55.95}}\ \]

    \[ \mathbf{r = 0.7}) \textbf{(Positive Correlation)} \]

Line of Regression Y on X

    \[ \widehat{\mathbf{Y}}\mathbf{= a + bx}\ \]

Where:

    \[ \mathbf{a =}\overline{\mathbf{Y}}\mathbf{- b}\overline{\mathbf{X}}\mathbf{\ \ \&\ }\mathbf{b}{\mathbf{yx}}\mathbf{=}\mathbf{r}{\mathbf{xy}}\mathbf{=}\frac{\mathbf{Sy}}{\mathbf{Sx}}\mathbf{\ \ \ }\ \]

    \[ \mathbf{b}_{\mathbf{yx}}\mathbf{= 0.7 =}\frac{\mathbf{3.78}}{\mathbf{3.162}}\mathbf{= 0.8368}\ \]

    \[ \mathbf{a = 11 - (0.8368)(10)}\ \]

    \[ \mathbf{a = 2.632}\ \]

    \[ \widehat{\mathbf{Y}}\mathbf{= a + bx}\ \]

    \[ \widehat{\mathbf{Y}}\mathbf{= 2.632 + 0.8368}\mathbf{x}\ \]

Q.3. Calculate Price Index Numbers using Laspeyre’s, Paasche’s, Fisher’s and Marshall’s formulae for 2001 taking 2000 as base year from the following data:

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Commodity20002001
PriceQuantityPriceQuantity
Wheat3011032112
Rice4010038110
Jawar25502280
Maize10401550

Solution:

Commodity20002001PoqoP1q1P1qoPoq1
Price PoQuantity qoPrice P1Quantity q13300358435203360
Wheat30110321124000418038004400
Rice40100381101250176011002000
Jawar25502280400750600500
Maize10401550895010274902010260
     3300358435203360
     ∑ Poqo=∑ P1q1=∑P1q0=∑ Poq1=

    \[ \left( \mathbf{i} \right)\mathbf{Laspeyr}\mathbf{e}^{\mathbf{'}}\mathbf{s\ Index}\mathbf{\ 2001}\mathbf{=}\frac{\mathbf{\sum}\mathbf{p}{\mathbf{1}}\mathbf{q}{\mathbf{0}}}{\mathbf{\sum}\mathbf{p}{\mathbf{0}}\mathbf{q}{\mathbf{0}}}\mathbf{\times \ 100}\ \]

    \[ \mathbf{Laspeyr}\mathbf{e}^{\mathbf{'}}\mathbf{s\ Index}\mathbf{\ 2001}\mathbf{=}\frac{\mathbf{3520}}{\mathbf{3300}}\mathbf{\ \times \ 100 = \ 106.67}\ \]

    \[ \left( \mathbf{ii} \right)\mathbf{Paasch}\mathbf{e}^{\mathbf{'}}\mathbf{s\ Index\ 2001}\mathbf{=}\frac{\mathbf{\sum}\mathbf{p}{\mathbf{1}}\mathbf{q}{\mathbf{1}}}{\mathbf{\sum}\mathbf{p}{\mathbf{0}}\mathbf{q}{\mathbf{1}}}\mathbf{\times \ 100}\ \]

    \[ \mathbf{Paasch}\mathbf{e}^{\mathbf{'}}\mathbf{s\ Index\ 2001}\mathbf{=}\frac{\mathbf{3584}}{\mathbf{3360}}\mathbf{\times \ 100 = 106.67}\ \]

    \[ \left( \mathbf{iii} \right)\mathbf{Fishe}\mathbf{r}^{\mathbf{'}}\mathbf{s\ Ideal\ Index\ 2001 =}\sqrt{\mathbf{L \times P}}\ \]

    \[ \mathbf{Fishe}\mathbf{r}^{\mathbf{'}}\mathbf{s\ Ideal\ Index\ 2001 =}\sqrt{\mathbf{106.67 \times 106.67}}\ \]

    \[ \mathbf{Fishe}\mathbf{r}^{\mathbf{'}}\mathbf{s\ Ideal\ Index\ 2001 = 106.67}\ \]

    \[ \left( \mathbf{iv} \right)\mathbf{\ Marsha}\mathbf{l}^{\mathbf{'}}\mathbf{s\ Index\ 2001 =}\left( \frac{\mathbf{\sum}\mathbf{p}{\mathbf{1}}\mathbf{q}{\mathbf{0}}\mathbf{+ \sum}\mathbf{p}{\mathbf{1}}\mathbf{q}{\mathbf{1}}}{\mathbf{\sum}\mathbf{p}{\mathbf{0}}\mathbf{q}{\mathbf{0}}\mathbf{+ \sum}\mathbf{p}{\mathbf{0}}\mathbf{q}{\mathbf{1}}} \right)\mathbf{\times 100}\ \]

    \[ \mathbf{Marsha}\mathbf{l}^{\mathbf{'}}\mathbf{s\ Index\ 2001 =}\left( \frac{\mathbf{3520 + 3584}}{\mathbf{3300 + 3360}} \right)\mathbf{\times 100}\ \]

    \[ \mathbf{Marsha}\mathbf{l}^{\mathbf{'}}\mathbf{s\ Index\ 2001 =}\left( \frac{\mathbf{7104}}{\mathbf{6660}} \right)\mathbf{\times 100}\ \]

    \[ \mathbf{Marsha}\mathbf{l}^{\mathbf{'}}\mathbf{s\ Index\ 2001 = 106.67}\ \]

Q.4. Test the hypothesis that hair color and eye color are independent. The table value of Chi-square at 4 degree of freedom at 5% level of significance is 9.49.

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Eye ColourHair Colour 
Light BlackDark BlackBrownTotal
Blue26211360
Black25422188
Brown19181552
Total708149200

Solution

(i) Testing the Hypothesis

Ho: There is no Association between Attributes Eyes Colour and Hair Colour.

H1: There is Association between Attributes Eyes Colour and Hair Colour.

(ii) Level of Significance = α=0.05

(iii) Test Statistics is:

    \[ \mathbf{\chi ² = \ \sum}\frac{\mathbf{(fo - fe)}^{\mathbf{2}}}{\mathbf{fe}}\ \]

(iv) To find χ² first calculate expected frequencies fe:

Calculation of expected frequencies (fe)

AttributesAttributes 
Light BlackDark BlackBrownTotal 
Blue

    \[ \frac{60 \times 70}{200} = 21\ \]

    \[ \frac{60 \times 81}{200} = 24.3\ \]

    \[ \frac{60 \times 49}{200} = 14.7\ \]

60 
Black

    \[ \frac{88 \times 70}{200} = 30.8\ \]

    \[ \frac{88 \times 81}{200} = 35.64\ \]

    \[ \frac{88 \times 49}{200} = 21.56\ \]

88 
Brown

    \[ \frac{52 \times 70}{200} = 18.2\ \]

    \[ \frac{52 \times 81}{200} = 21.06\ \]

    \[ \frac{52 \times 49}{200} = 12.74\ \]

52 
Total708149200 

(v) Calculation of χ²:

Table B. Computation of χ²

fofefo-fe(fo-fe)²

    \[ \frac{\mathbf{(fo - fe)}^{\mathbf{2}}}{\mathbf{fe}}\ \]

26215251.190
2124.3-3.310.890.448
1314.7-1.72.890.197
2530.8-5.833.641.092
4235.646.3640.44961.135
2121.56-0.560.31360.015
1918.20.80.640.035
1821.06-3.069.36360.445
1512.742.265.10760.401
  4.958
    

    \[ \mathbf{\sum}\frac{\left( \mathbf{fo - fe} \right)^{\mathbf{2}}}{\mathbf{fe}}\mathbf{=}\ \]

(vi) Critical Region: Degree of Freedom d.f= (R-1)(C-1)

So d.f= (3-1)(3-1)=4

The Value of Tabulated χ²(0.05,4)=9.49

The Critical Region χ²cal>9.49

(vii) Conclusion: The calculated value of χ² is 4.958 is less than the tabulated value of χ² 9.49 or 4.958 falls in the acceptance region. We accept the Null Hypothesis Ho that there is no association between Eyes Colour and Hair Colour.

Section II Business Mathematics

Q.5 Calculate (i) (A+B), (ii) 2A – 3B (iii) AB from the following matrix A and B

    \[ \textbf{Q.5} If\ A = \begin{bmatrix}2 & - 3 & 4 \\1 & 5 & - 2 \\4 & 2 & 6 \\\end{bmatrix},\ B = \begin{bmatrix}1 & - 2 & 3 \\4 & - 5 & - 6 \\7 & 8 & 9 \\\end{bmatrix}\ Calculate\ (i)A + B\ (ii)2A - 3B\ (iii)AB\ \]

Solution

    \[ \begin{enumerate}\def\labelenumi{(\roman{enumi})}\item A + B = \ \begin{bmatrix}2 + 1 & - 3 + ( - 2) & 4 + 3 \\1 + 4 & 5 + ( - 5) & - 2 + ( - 6) \\4 + 7 & 2 + 8 &6 + 9 \\\end{bmatrix}\ \]

    \[ A + B = \ \begin{bmatrix}3 & - 5 & 7 \\5 & 0 & - 8 \\11 & 10 & 15 \\\end{bmatrix}\ \]

    \[ \begin{enumerate}\def\labelenumi{(\roman{enumi})}\setcounter{enumi}{1}\item 2A = \ \begin{bmatrix}4 & - 6 & 8 \\2 & 10 & - 4 \\8 & 4 & 12 \\\end{bmatrix} 3B = \ \begin{bmatrix}3 & - 6 & 9 \\12 & - 15 & - 18 \\21 & 24 & 27 \\\end{bmatrix}\ \]

    \[ 2A - 3B = \ \begin{bmatrix}4 - 3 & - 6 - ( - 6) & 8 - 9 \\2 - 12 & 10 - ( - 15) & - 4 - ( - 18) \\8 - 21 & 4 - 24 & 12 - 27 \\\end{bmatrix}\ \]

    \[ 2A - 3B = \ \begin{bmatrix}1 & 0 & - 1 \\10 & 25 & 14 \\13 & - 20 & - 15 \\\end{bmatrix}\ \]

    \[ (iii) AB = \ \begin{bmatrix}2 \times 1 + ( - 3)4 + 4 \times 7 & 2( - 2) + ( - 3)( - 5) + 4 \times 8 & 2 \times 3 + ( - 3) - 6 + 4 \times 9 \\1 \times 1 + 5 \times 4 + ( - 2) \times 7 & 1( - 2) + 5( - 5) + ( - 2) \times 8 & 1 \times 3 + 5( - 6) + ( - 2)9 \\4 \times 1 + 2 \times 4 + 6 \times 7 & 4( - 2) + 2( - 5) + 6 \times 8 & 4 \times 3 + 2( - 6) +6 \times 9 \\\end{bmatrix}\ \]

    \[ AB = \ \begin{bmatrix}2 - 12 + 28 & - 4 + 15 + 32 & 6 + 18 + 36 \\1 + 20 - 14 & - 2 - 25 - 16 & 3 - 30 - 18 \\4 + 8 + 42 & - 8 - 10 + 48 & 12 - 12 + 54 \\\end{bmatrix}\ \]

    \[ AB = \ \begin{bmatrix}18 & 43 & 60 \\7 & - 43 & - 45 \\54 & 30 & 54 \\\end{bmatrix}\ \]

Q.6 (a) Solve the following quadratic equation: 2x² + 15x + 18 = 0

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Solution (a)

    \[ {2x}^{2} + 15x + 18 = 0\ \]

    \[ \Here\ a = 2,\ b = 15\ \&\ c = 18\ \]

    \[ \x = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}\ \]

    \[ \x = \frac{- (15) \pm \sqrt{{(15)}^{2} - 4(2)(18)}}{2(2)}\ \]

    \[ \x = \frac{- 15 \pm \sqrt{225 - 144}}{4}\ \]

    \[ \x = \frac{- 15 \pm 9}{4}\ \]

    \[ \x = \frac{- 15 + 9}{4} = \frac{- 6}{4} = - 1.5\ \]

    \[ \x = \frac{- 15 - 9}{4} = \frac{- 24}{4} = - 6\ \]

    \[ So x={-1.5, -6} \]

(b) The difference of two numbers is 33. The larger number is one more than three times the smaller number. Find the numbers.

Solution (b)

Let the smaller number be x

Larger number is one more than 3 times the smaller number = 3x+1

Equation will be:

(3x + 1) – x = 33

3x + 1 – x = 33

2x + 1 = 33

2x = 33 – 1

2x = 32

X=32/2

X= 16

Hence

Smaller Number = x = 16

Larger Number = 3x + 1 = 3(16) + 1 = 49

Prove

3x + 1 – x = 33

49 – 16 = 33

33 = 33

Q.7 (a) The sum of 10 terms of an A.P; whose last term is 28, is 145. Find the first term and the common difference.

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Solution:

    \[ \mathbf{Here\ n\ = \ 10,\ }\mathbf{a}{\mathbf{10}}\mathbf{= \ 28,\ }\mathbf{S}{\mathbf{10}}\mathbf{\ = \ 145\ } \]

    \[ \mathbf{S}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{n}}{\mathbf{2}}\mathbf{\lbrack}\mathbf{a}_{\mathbf{1}}\mathbf{+}\mathbf{a}_{\mathbf{n}}\mathbf{\rbrack} \]

    \[ \mathbf{S}{\mathbf{10}}\mathbf{=}\frac{\mathbf{10}}{\mathbf{2}}\mathbf{\lbrack}\mathbf{a}{\mathbf{1}}\mathbf{+ 28\rbrack} \]

    \[ \mathbf{145 = 5\lbrack}\mathbf{a}_{\mathbf{1}}\mathbf{+ 28\rbrack} \]

    \[ \mathbf{a}_{\mathbf{1}}\mathbf{+ 28 =}\frac{\mathbf{145}}{\mathbf{5}} \]

    \[ \mathbf{a}_{\mathbf{1}}\mathbf{+ 28 = 29} \]

    \[ \mathbf{a}_{\mathbf{1}}\mathbf{= 29 - 28} \]

    \[ \mathbf{a}_{\mathbf{1}}\mathbf{= 1} \]

    \[ \textbf{Common Difference} \]

    \[ \mathbf{a}{\mathbf{n}}\mathbf{=}\mathbf{a}{\mathbf{1}}\mathbf{+}\left( \mathbf{n - 1} \right)\mathbf{d} \]

    \[ \mathbf{a}{\mathbf{10}}\mathbf{=}\mathbf{a}{\mathbf{1}}\mathbf{+}\left( \mathbf{10 - 1} \right)\mathbf{d} \]

    \[ \mathbf{28 = 1 +}\left( \mathbf{10 - 1} \right)\mathbf{d} \]

    \[ \mathbf{28 = 1 + 9}\mathbf{d} \]

    \[ \mathbf{9}\mathbf{d = 28 - 1} \]

    \[ \mathbf{9}\mathbf{d = 27} \]

    \[ \mathbf{d =}\frac{\mathbf{27}}{\mathbf{9}} \]

    \[ \mathbf{d = 3} \]

(b) Find the sum of the series: 1, 1/2, 1/4, 1/8, 1/16,……….to infinity cannot exceed 2.

Solution:

    \[ \mathbf{Here\ a = 1,\ r\ =}\frac{\mathbf{1}}{\mathbf{2}} \]

    \[ \mathbf{S}_{\mathbf{\infty}}\mathbf{=}\frac{\mathbf{a}}{\mathbf{1 - r}} \]

    \[ \mathbf{S}_{\mathbf{\infty}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{1 -}\frac{\mathbf{1}}{\mathbf{2}}} \]

    \[ \mathbf{S}_{\mathbf{\infty}}\mathbf{=}\frac{\mathbf{1}}{\frac{\mathbf{1}}{\mathbf{2}}} \]

    \[ \mathbf{S}_{\mathbf{\infty}}\mathbf{= 1 \times}\frac{\mathbf{2}}{\mathbf{1}} \]

    \[ \mathbf{S}_{\mathbf{\infty}}\mathbf{= 2} \]

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Q.8 Find out the effective rate of interest equivalent to the nominal rate of 8% p.a. Compounded quarterly.

Solution:

Quarterly Case

    \[ \mathbf{Here\ i = 8\%\ =}\frac{\mathbf{8}}{\mathbf{4}}\mathbf{= 2\% = \ 0.02,\ n = 1 \times 4 = 4} \]

    \[ \mathbf{Effective\ Rate\ of\ Interest\ = \ }\left( \mathbf{1 + i} \right)^{\mathbf{n}}\mathbf{- 1} \]

    \[ \mathbf{Effective\ Rate\ of\ Interest\ = \ }\left( \mathbf{1 + 0.02} \right)^{\mathbf{4}}\mathbf{- 1} \]

    \[ \mathbf{Effective\ Rate\ of\ Interest\ =}\mathbf{0.0824} \]

    \[ \mathbf{Effective\ Rate\ of\ Interest\ =}\mathbf{8.24\%}\ \]

You may also interested in the following:

Business Statistics and Mathematics Solved Paper 2013, Punjab University, BCOM, ADC I

Business Statistics and Mathematics Solved Paper 2012, Punjab University, BCOM,ADC I

Business Statistics & Mathematics, Solved Paper 2011, Punjab University, BCOM,ADCI

Business Statistics & Mathematics, Solved Paper 2010, Punjab University, BCOM, ADCI

Business Statistics & Mathematics, Solved Paper 2009, Punjab University, BCOM, ADCI

Business Mathematics and Statistics, Solved Paper 2008, Punjab University, BCOM,ADCI

Business Mathematics and Statistics, Solved Paper 2007, Punjab University, BCOM,ADCI

Introduction to Statistics Basic Important Concepts

Measures of Central Tendency, Arithmetic Mean, Median, Mode, Harmonic, Geometric Mean

Correlation Coefficient, Properties, Types, Important Formulas for Correlation Coefficient

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