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

Here, we are going to solve the Paper of Business Statistics and Mathematics, Solved Paper 2010, 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. Solved Paper 2007 Punjab University , Solved Paper 2008 and Solved Paper 2009 have already posted.

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

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

Section I Business Statistics

Q.1 Calculate A.M, Harmonic Mean, Standard Deviation and Coefficient of Variation

Solution

\[ \left( \mathbf{a} \right)\mathbf{\ A.M\ }\overline{\mathbf{X}}\mathbf{=}\frac{\mathbf{\sum fx}}{\mathbf{\sum f}}\mathbf{= \ }\frac{\mathbf{79528}}{\mathbf{500}}\mathbf{= 159.056\ }\ \]

\[ \left( \mathbf{b} \right)\mathbf{\ Harmonic\ Mean\ H.M =}\frac{\mathbf{\sum f}}{\mathbf{\sum(}\frac{\mathbf{f}}{\mathbf{X}}\mathbf{)}}\mathbf{= \ }\frac{\mathbf{500}}{\mathbf{3.181811164}}\mathbf{= 157.143\ }\ \]

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

\[ \mathbf{S.D =}\sqrt{\left\lbrack \frac{\mathbf{\sum f}\mathbf{x}^{\mathbf{2}}}{\mathbf{\sum f}}\mathbf{- \ }\left( \frac{\mathbf{\sum fx}}{\mathbf{\sum f}} \right)^{\mathbf{2}} \right\rbrack}\mathbf{=}\sqrt{\left\lbrack \frac{\mathbf{12795669}}{\mathbf{500}}\mathbf{- \ }\left( \frac{\mathbf{79528}}{\mathbf{500}} \right)^{\mathbf{2}} \right\rbrack}\ \]

\[ \mathbf{S.D =}\sqrt{\left\lbrack \mathbf{25591.338 – \ 25298.82} \right\rbrack}\ \]

\[ \textbf{S.D = 17.10} \]

\[ \mathbf{Coefficient\ of\ Variation\ C.V = \ }\left( \frac{\mathbf{17.10}}{\mathbf{159.056}} \right)\mathbf{100 = 10.75\%}\ \]

Q.2

X:        16, 72, 73, 63. 83. 80. 66 66. 74, 62

Y:        40, 52, 43, 49, 61, 58, 44, 58, 50, 45

Required: Calculate coefficient of correlation and comment on the answers.

Solution:

Formula:

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

\[ \overline{\mathbf{X}}\mathbf{=}\frac{\mathbf{\sum X}}{\mathbf{n}}\mathbf{=}\frac{\mathbf{655}}{\mathbf{10}}\mathbf{= 65.5\ ,\ }\overline{\mathbf{Y}}\mathbf{=}\frac{\mathbf{\sum Y}}{\mathbf{n}}\mathbf{=}\frac{\mathbf{500}}{\mathbf{10}}\mathbf{= 50\ }\ \]

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

\[ \mathbf{Sx =}\sqrt{\left\lbrack \frac{\mathbf{46059}}{\mathbf{10}}\mathbf{- \ }\left( \mathbf{65.5} \right)^{\mathbf{2}} \right\rbrack}\ \]

\[ \mathbf{Sx =}\sqrt{\mathbf{4605.9 – 4290.25}}\ \]

\[ \mathbf{Sx =}\sqrt{\mathbf{315.65}}\ \]

\[ \mathbf{Sx = 17.76}\ \]

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

\[ \mathbf{Sy =}\sqrt{\left\lbrack \frac{\mathbf{25464}}{\mathbf{10}}\mathbf{- \ }\left( \mathbf{50} \right)^{\mathbf{2}} \right\rbrack}\mathbf{\ }\ \]

\[ \mathbf{Sy =}\sqrt{\mathbf{2546.4 – 2500}}\ \]

\[ \mathbf{Sy =}\sqrt{\mathbf{46.4}}\ \]

\[ \mathbf{Sy = 6.811}\ \]

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

\[ \mathbf{r =}\frac{\frac{\mathbf{33535}}{\mathbf{10}}\mathbf{- \ }\left( \mathbf{65.5} \right)\left( \mathbf{50} \right)}{\left( \mathbf{17.76} \right)\mathbf{\times (6.811)}}\ \]

\[ \mathbf{r =}\frac{\mathbf{3353.5\ –\ 3275}}{\mathbf{120.96}}\ \]

\[ \mathbf{r =}\frac{\mathbf{78.5}}{\mathbf{120.96}}\ \]

\[ \mathbf{r = 0.6489}\ \]

Comment: Correlation Coefficient “r” 0.6489 indicates that there is a positive relation between two variables “X” and “Y”.

Q.3. Test the association

The value of Chi-Square at 5% level of significance 4 d.f is 9.488

Solution:

Solution:

(i) Testing the Hypothesis

Ho: There is no Association between Attributes A’s and B’s.

H1: There is Association between Attributes A’s and B’s.

(ii) Level of Significance = α=0.05

(iii) Test Statistics is: \[ \mathbf{\chi}\mathbf{²}\mathbf{= \ }\mathbf{\sum}\frac{{\mathbf{(}\mathbf{fo}\mathbf{-}\mathbf{fe}\mathbf{)}}^{\mathbf{2}}}{\mathbf{fe}}\ \]

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

(v) Calculation of χ²:

(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)=9.488

The Critical Region χ²cal>9.488

(vii) Conclusion: The calculated value of χ² is 0.86107 is less than the tabulated value of χ² 9.488 or 0.86107 falls in the acceptance region. We accept the Null Hypothesis Ho that there is no association between Attributes A’s and B’s.

Q.4. Compute index number for 2002 from the following data taking 2000 as base using:

(i) Laspeyre’s (ii) Paasche’s (iii) Fisher’s Method

Solution:

\[ \left( \mathbf{i} \right)\mathbf{Laspeyr}\mathbf{e}^{\mathbf{‘}}\mathbf{s\ Index}\mathbf{\ 2002}\mathbf{=}\frac{\mathbf{\sum}\mathbf{p}_{\mathbf{1}}\mathbf{q}_{\mathbf{0}}}{\mathbf{\sum}\mathbf{p}_{\mathbf{0}}\mathbf{q}_{\mathbf{0}}}\mathbf{\times \ 100}\ \]

\[ \mathbf{Laspeyr}\mathbf{e}^{\mathbf{‘}}\mathbf{s\ Index}\mathbf{\ 2002}\mathbf{=}\frac{\mathbf{3570}}{\mathbf{2900}}\mathbf{\ \times \ 100 = \ 123.103}\ \]

\[ \left( \mathbf{ii} \right)\mathbf{Paasch}\mathbf{e}^{\mathbf{‘}}\mathbf{s\ Index\ 2002}\mathbf{=}\frac{\mathbf{\sum}\mathbf{p}_{\mathbf{1}}\mathbf{q}_{\mathbf{1}}}{\mathbf{\sum}\mathbf{p}_{\mathbf{0}}\mathbf{q}_{\mathbf{1}}}\mathbf{\times \ 100}\ \]

\[ \mathbf{Paasch}\mathbf{e}^{\mathbf{‘}}\mathbf{s\ Index\ 2002}\mathbf{=}\frac{\mathbf{3420}}{\mathbf{2840}}\mathbf{\times \ 100 = 120.42}\ \]

\[ \left( \mathbf{iii} \right)\mathbf{Fishe}\mathbf{r}^{\mathbf{‘}}\mathbf{s\ Ideal\ Index\ 2002 =}\sqrt{\mathbf{L \times P}}\ \]

\[ \mathbf{Fishe}\mathbf{r}^{\mathbf{‘}}\mathbf{s\ Ideal\ Index\ 2002 =}\sqrt{\mathbf{123.103 \times 120.42}}\ \]

\[ \mathbf{Fishe}\mathbf{r}^{\mathbf{‘}}\mathbf{s\ Ideal\ Index\ 2002 = 121.75}\ \]

Section II Business Mathematics

Q.5: If:

\[ \mathbf{A = \ }\begin{bmatrix}\mathbf{1} & \mathbf{4} & \mathbf{3} \\\mathbf{2} & \mathbf{1} & \mathbf{8} \\\mathbf{1} & \mathbf{1} & \mathbf{2} \\\end{bmatrix}\mathbf{\ and\ B = \ }\begin{bmatrix}\mathbf{2} & \mathbf{1} & \mathbf{2} \\\mathbf{0} & \mathbf{- 4} & \mathbf{8} \\\mathbf{6} & \mathbf{1} & \mathbf{4} \\\end{bmatrix}\mathbf{\ }\ \]

Calculate (i) A+B (ii) 2A – 3B (iii) AB

Solution

\[ \mathbf{A + B = \ }\begin{bmatrix} \mathbf{1 + 2} & \mathbf{4 + 1} & \mathbf{3 + 2} \\\mathbf{2 + 0} & \mathbf{1 + ( – 4)} & \mathbf{8 + 8} \\\mathbf{1 + 6} & \mathbf{1 + 1} & \mathbf{2 + 4} \\\end{bmatrix}\ \]

(i) \[ \mathbf{A + B = \ }\begin{bmatrix}\mathbf{3} & \mathbf{5} & \mathbf{5} \\\mathbf{2} & \mathbf{- 3} & \mathbf{16} \\\mathbf{7} & \mathbf{2} & \mathbf{6} \\\end{bmatrix}\ \]

(ii) \[ \mathbf{2}\mathbf{A = \ }\begin{bmatrix} \mathbf{2} & \mathbf{8} & \mathbf{6} \\\mathbf{4} & \mathbf{2} & \mathbf{16} \\\mathbf{2} & \mathbf{2} & \mathbf{4} \\\end{bmatrix}\ \]

\[ \mathbf{3}\mathbf{B = \ }\begin{bmatrix} \mathbf{6} & \mathbf{3} & \mathbf{6} \\\mathbf{0} & \mathbf{- 12} & \mathbf{24} \\\mathbf{18} & \mathbf{3} & \mathbf{12} \\\end{bmatrix}\ \]

\[ \mathbf{2}\mathbf{A – 3}\mathbf{B = \ }\begin{bmatrix}\mathbf{2 – 6} & \mathbf{8 – 3} & \mathbf{6 – 6} \\\mathbf{4 – 0} & \mathbf{2 + 12} & \mathbf{16 – 24} \\\mathbf{2 – 18} & \mathbf{2 – 3} & \mathbf{4 – 12} \\\end{bmatrix}\ \]

\[ \mathbf{2}\mathbf{A – 3}\mathbf{B = \ }\begin{bmatrix}\mathbf{- 4} & \mathbf{5} & \mathbf{0} \\\mathbf{4} & \mathbf{14} & \mathbf{- 8} \\\mathbf{- 16} & \mathbf{- 1} & \mathbf{- 8} \\\end{bmatrix}\ \]

(iii) \[ \mathbf{AB = \ }\begin{bmatrix}
\mathbf{1 \times 2 + 4 \times 0 + 3 \times 6} & \mathbf{1 \times 1 + 4( – 4) + 3 \times 1} & \mathbf{1 \times 2 + 4 \times 8 + 3 \times 4} \\
\mathbf{2 \times 2 + 1 \times 0 + 8 \times 6} & \mathbf{2 \times 1 + 1( – 4) + 8 \times 1} & \mathbf{2 \times 2 + 1 \times 8 + 8 \times 4} \\
\mathbf{1 \times 2 + 1 \times 0 + 2 \times 6} & \mathbf{1 \times 1 + 1( – 4) + 2 \times 1} & \mathbf{1 \times 2 + 1 \times 8 + 2 \times 4} \\
\end{bmatrix}\ \]

\[ \mathbf{AB = \ }\begin{bmatrix}\mathbf{2 + 0 + 18} & \mathbf{1 – 16 + 3} & \mathbf{2 + 32 + 12} \\\mathbf{4 + 0 + 48} & \mathbf{2 – 4 + 8} & \mathbf{4 + 8 + 32} \\\mathbf{2 + 0 + 12} & \mathbf{1 – 4 + 2} & \mathbf{2 + 8 + 8} \\\end{bmatrix}\ \]

\[ \mathbf{AB = \ }\begin{bmatrix}\mathbf{20} & \mathbf{- 12} & \mathbf{46} \\\mathbf{52} & \mathbf{6} & \mathbf{44} \\\mathbf{14} & \mathbf{- 1} & \mathbf{18} \\\end{bmatrix}\ \]

Q.6 (a) Solve the following simultaneous equations.

2x + y = -7

3x + 2y = -12

Solution

Multiply equation (i) by 2, we get

\[ \mathbf{4}\mathbf{x\ + \ 2}\mathbf{y\ = \ – 14}\ \]

\[ \mathbf{\pm 3}\mathbf{x\ \pm \ 2}\mathbf{y\ = \ \mp 12}\ \]

\[ \mathbf{x\ = \ – 2}\ \]

Put X = -2 in any equation to get y:

\[ \mathbf{2}\mathbf{x\ + \ y\ = \ – 7}\ \]

\[ \mathbf{2( – 2)\ + \ y\ = \ – 7}\ \]

\[ \mathbf{- 4\ + \ y\ = \ – 7}\ \]

\[ \mathbf{y\ = \ – 7\ + \ 4}\ \]

\[ \mathbf{y\ = \ – 3}\ \]

\[ \textbf{So x= -2, y = -3} \]

(b) Solve the quadratic equation 6x² – 5x – 6 = 0

Solution

\[ \mathbf{x =}\frac{\mathbf{- b \pm}\sqrt{\mathbf{b}^{\mathbf{2}}\mathbf{- 4}\mathbf{ac}}}{\mathbf{2}\mathbf{a}}\mathbf{\ where\ a = 6,\ b = \ – 5\ \&\ c = \ – 6}\ \]

\[ \mathbf{x =}\frac{\mathbf{- ( – 5) \pm}\sqrt{\left( \mathbf{- 5} \right)^{\mathbf{2}}\mathbf{- 4(6)( – 6)}}}{\mathbf{2(6)}}\ \]

\[ \mathbf{x =}\frac{\mathbf{5 \pm}\sqrt{\mathbf{25}\mathbf{+ 144}}}{\mathbf{12}}\ \]

\[ \mathbf{x =}\frac{\mathbf{5 \pm}\sqrt{\mathbf{169}}}{\mathbf{12}}\ \]

\[ \mathbf{x =}\frac{\mathbf{5 \pm 13}}{\mathbf{12}}\ \]

\[ \mathbf{x =}\frac{\mathbf{5 + 13}}{\mathbf{12}}\mathbf{=}\frac{\mathbf{18}}{\mathbf{12}}\mathbf{=}\frac{\mathbf{3}}{\mathbf{2}}\ \]

\[ \mathbf{x =}\frac{\mathbf{5 – 13}}{\mathbf{12}}\mathbf{=}\frac{\mathbf{8}}{\mathbf{12}}\mathbf{=}\frac{\mathbf{2}}{\mathbf{3}}\ \]

Q.7 Find out the compound amount and compound interest at the end of 3 years on a sum of Rs. 20,000 borrowed at 6% compounded annually.

Solution

\[ \mathbf{P\ = \ 20,000,\ n = 3,\ i = 0.06}\ \]

\[ \mathbf{F.V\ or\ Amount = \ P}\mathbf{(1 + i)}^{\mathbf{n}}\mathbf{\ }\ \]

\[ \mathbf{F.V\ or\ Amount = \ }\mathbf{20,000(1 + 0.06)}^{\mathbf{3}}\mathbf{\ }\ \]

\[ \mathbf{F.V\ or\ Amount = \ 20,000(1.191016)\ }\ \]

\[ \mathbf{F.V\ or\ Amount = 23820.32}\ \]

\[ \mathbf{Compound\ Interest = 23820.32 – 20000 = 3820.32}\ \]

Q.8 A 90 days Rs. 4,000, 7% interest bearing note dated April 4, was discounted on May 4, at a discount rate of 8%. What was the discounted value of the note? (Take 360 days in the year)

Solution

Total Days = 90

Days of April = 27

Days of May = 30

Days of June = 30

Days of July = 3

Maturity Date 3^rd July

P = 4000, N = 3 months, I = (7/12)=0.5834 = (0.5834/100) = 0.005834

S.I = PIN

S.I = 4000 x 3 x 0.005834 = 70

Amount or future value of Note = 4000 + 70 = 4070

Discounted Date 4^th May

Days of May = 27

Days of June = 30

Days of July = 3

Total Days = 60

Discount rate = 8% = 0.08

Principal Amount = 4070

Interest = 4070 x 0.08 = 325.6

\[ \mathbf{Interest = \ }\frac{\mathbf{325.6}}{\mathbf{12}}\mathbf{\ }\mathbf{\times}\mathbf{\ 2\ or\ 325.6}\left( \frac{\mathbf{60}}{\mathbf{360}} \right)\mathbf{= 54.26}\ \]

Discounted Value of interest bearing Note = 4070 – 54.26 = 4015.74

You may also like to visit following links:

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