Business Statistics 2078 Board Question Paper

Step 1: Identify the given values

• Group A: $N_1 = 200, \bar{X}_1 = 25$
• Group B: $N_2 = 250, \bar{X}_2 = 10$
• Group C: $N_3 = 300, \bar{X}_3 = 15$

Step 2: Formula for combined arithmetic mean of three groups

$$\bar{X}_{123} = \frac{N_1\bar{X}_1 + N_2\bar{X}_2 + N_3\bar{X}_3}{N_1 + N_2 + N_3}$$

Step 3: Calculation

$$\bar{X}_{123} = \frac{200(25) + 250(10) + 300(15)}{200 + 250 + 300}$$

$$\bar{X}_{123} = \frac{5000 + 2500 + 4500}{750} = \frac{12,000}{750} = 16$$

Conclusion: The combined mean of the three groups is 16.

Step 1: Identify given values

• Coefficient of Variation ($C.V.$) = $20\%$
• Variance ($\sigma^2$) = $36 \implies \text{Standard Deviation } (\sigma) = \sqrt{36} = 6$

Step 2: Apply the Coefficient of Variation formula

$$C.V. = \frac{\sigma}{\bar{X}} \times 100$$

$$20 = \frac{6}{\bar{X}} \times 100$$

Step 3: Solve for Mean ($\bar{X}$)

$$20\bar{X} = 600 \implies \bar{X} = \frac{600}{20} = 30$$

Conclusion: The mean value of the distribution is 30.

Step 1: Identify given values

• $\text{Mean } (\bar{X}) = 45$
• $\text{Mode } (M_o) = 48$
• $\text{Standard Deviation } (\sigma) = 15$

Step 2: Karl Pearson’s formula based on Mode

$$S_k = \frac{\bar{X} - M_o}{\sigma}$$

Step 3: Calculation

$$S_k = \frac{45 - 48}{15} = \frac{-3}{15} = -0.2$$

Interpretation: The coefficient of skewness is -0.2, indicating that the distribution is negatively skewed (skewed to the left).

Step 1: Given probabilities

• $P(A) = 0.6$
• $P(B) = 0.56$
• $P(A \cup B) = 0.75$

Step 2: Addition Theorem of Probability for non-mutually exclusive (non-disjoint) events

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

Step 3: Rearrange and solve for $P(A \cap B)$

$$P(A \cap B) = P(A) + P(B) - P(A \cup B)$$

$$P(A \cap B) = 0.60 + 0.56 - 0.75 = 1.16 - 0.75 = 0.41$$

Conclusion: $P(A \cap B) = \mathbf{0.41}$.

Note on Question Text: In TU statistics nomenclature, the difference between the 90th and 10th percentiles is $(P_{90} - P_{10})$. Here $P_9$ represents the 9th decile ($D_9 = P_{90}$) and $P_{10}$ represents the 10th percentile.

Step 1: Given values

• Quartile Deviation ($Q.D.$) = $2$
• Percentile Range ($P_{90} - P_{10}$) = $8$

Step 2: Formula for Percentile Coefficient of Kurtosis ($\kappa$)

$$\kappa = \frac{Q.D.}{P_{90} - P_{10}}$$

$$\kappa = \frac{2}{8} = 0.25$$

Interpretation:

• For a normal (mesokurtic) distribution, $\kappa = 0.263$.
• Since $\kappa = 0.25 < 0.263$, the distribution is Platykurtic (flatter than a normal curve).

Step 1: Given regression coefficients

• $b_{yx} = -0.667 \approx -\frac{2}{3}$
• $b_{xy} = -0.75 = -\frac{3}{4}$

Step 2: Property of correlation coefficient and regression coefficients The correlation coefficient $r$ is the geometric mean of the two regression coefficients and must carry the same sign as both coefficients:

$$r = \pm \sqrt{b_{yx} \cdot b_{xy}}$$

Since both $b_{yx} < 0$ and $b_{xy} < 0$, $r$ must be negative.

Step 3: Calculation

$$r = -\sqrt{(-0.667) \times (-0.75)} = -\sqrt{0.50025} \approx -0.7073$$

(Using exact fractions: $r = -\sqrt{\frac{2}{3} \times \frac{3}{4}} = -\sqrt{\frac{1}{2}} = -\frac{1}{\sqrt{2}} \approx -0.7071$)

Conclusion: The coefficient of correlation $r$ is -0.707 (high degree of negative linear correlation).

Step 1: Compute sums of prices Base year (2020) price $= P_0$; Current year (2021) price $= P_1$.

Step 2: Formula for Simple Aggregative Price Index

$$P_{01} = \frac{\sum P_1}{\sum P_0} \times 100$$

Step 3: Calculation

$$P_{01} = \frac{97}{75} \times 100 = 129.33$$

Conclusion: The simple aggregative price index number for 2021 is 129.33, indicating a 29.33% increase in general price level compared to 2020.

A time series consists of four fundamental components that account for variations over time:

1. Secular Trend ($T$): The general, smooth, long-term tendency of the data to grow or decline over a prolonged period of time (e.g., population growth, technological adoption).
2. Seasonal Variations ($S$): Short-term regular fluctuations that occur periodically within a fixed timeframe of one year or less, driven by weather, seasons, festivals, or social customs (e.g., umbrella sales in monsoon, Dashain shopping).
3. Cyclical Variations ($C$): Medium-to-long-term wave-like oscillatory movements recurring over a period greater than one year, typically corresponding to economic business cycles (Prosperity, Recession, Depression, Recovery).
4. Irregular / Random Variations ($I$): Unpredictable, erratic, and non-recurring fluctuations caused by unforeseen exogenous shocks such as earthquakes, floods, wars, political strikes, or pandemics.

Mathematical Representation:

• Additive Model: $Y = T + S + C + I$
• Multiplicative Model: $Y = T \times S \times C \times I$

Let the determinant be:

$$|A| = \begin{vmatrix} 1 & 5 & 6 \\ 2 & 3 & 7 \\ 3 & 5 & 6 \end{vmatrix}$$

Expansion along Row 1 ($R_1$):

$$|A| = 1 \begin{vmatrix} 3 & 7 \\ 5 & 6 \end{vmatrix} - 5 \begin{vmatrix} 2 & 7 \\ 3 & 6 \end{vmatrix} + 6 \begin{vmatrix} 2 & 3 \\ 3 & 5 \end{vmatrix}$$

Evaluate $2 \times 2$ determinants:

1. $1 \times [(3 \times 6) - (7 \times 5)] = 1 \times [18 - 35] = 1 \times (-17) = -17$
2. $-5 \times [(2 \times 6) - (7 \times 3)] = -5 \times [12 - 21] = -5 \times (-9) = +45$
3. $+6 \times [(2 \times 5) - (3 \times 3)] = 6 \times [10 - 9] = 6 \times 1 = +6$

Summing the terms:

$$|A| = -17 + 45 + 6 = 34$$

Conclusion: The value of the determinant is 34.

Step 1: State the given matrices

$$A = \begin{bmatrix} 7 & 8 \\ 9 & 10 \\ 2 & 5 \end{bmatrix}, \quad B = \begin{bmatrix} 6 & 3 \\ 3 & 2 \\ 7 & 5 \end{bmatrix}$$

Both matrices are of order $3 \times 2$.

Step 2: Compute the matrix difference $(A - B)$

$$A - B = \begin{bmatrix} 7 - 6 & 8 - 3 \\ 9 - 3 & 10 - 2 \\ 2 - 7 & 5 - 5 \end{bmatrix} = \begin{bmatrix} 1 & 5 \\ 6 & 8 \\ -5 & 0 \end{bmatrix}$$

Step 3: Scalar multiplication by 7

$$7(A - B) = 7 \begin{bmatrix} 1 & 5 \\ 6 & 8 \\ -5 & 0 \end{bmatrix} = \begin{bmatrix} 7(1) & 7(5) \\ 7(6) & 7(8) \\ 7(-5) & 7(0) \end{bmatrix} = \begin{bmatrix} 7 & 35 \\ 42 & 56 \\ -35 & 0 \end{bmatrix}$$

Conclusion:

$$7(A - B) = \mathbf{\begin{bmatrix} 7 & 35 \\ 42 & 56 \\ -35 & 0 \end{bmatrix}}$$

Step 1: Set up the calculation table Let assumed mean $A = 45$, class width $h = 10$. Then $d = \frac{m - 45}{10}$, where $m$ is the mid-value.

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Step 2: Calculate Arithmetic Mean ($\bar{X}$)

$$\bar{X} = A + \left(\frac{\sum fd}{N}\right) \times h = 45 + \left(\frac{-17}{100}\right) \times 10 = 45 - 1.7 = \mathbf{43.30}$$

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Step 3: Calculate Mode ($M_o$) The highest frequency is $f_1 = 25$, which lies in the modal class 40 - 50.

• $L = 40$
• $f_1 = 25$
• $f_0 = 17$ (preceding frequency)
• $f_2 = 18$ (succeeding frequency)
• $h = 10$$M_o = L + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$$
  $$M_o = 40 + \left(\frac{25 - 17}{2(25) - 17 - 18}\right) \times 10 = 40 + \left(\frac{8}{50 - 35}\right) \times 10 = 40 + \frac{80}{15} = 40 + 5.33 = \mathbf{45.33}$$

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Step 4: Calculate Standard Deviation ($\sigma$)

$$\sigma = h \times \sqrt{\frac{\sum fd^2}{N} - \left(\frac{\sum fd}{N}\right)^2}$$

$$\sigma = 10 \times \sqrt{\frac{303}{100} - \left(\frac{-17}{100}\right)^2} = 10 \times \sqrt{3.03 - (-0.17)^2} = 10 \times \sqrt{3.03 - 0.0289} = 10 \times \sqrt{3.0011} = 10 \times 1.7324 = \mathbf{17.324}$$

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Step 5: Calculate Karl Pearson’s Coefficient of Skewness ($S_k$)

$$S_k = \frac{\bar{X} - M_o}{\sigma} = \frac{43.30 - 45.33}{17.324} = \frac{-2.03}{17.324} = \mathbf{-0.117}$$

(Alternative check using Median: $\frac{N}{2} = 50$, Median class 40-50, $M_d = 40 + \frac{50-40}{25} \times 10 = 44$. Then $S_k = \frac{3(\bar{X} - M_d)}{\sigma} = \frac{3(43.3 - 44)}{17.324} = \frac{-2.1}{17.324} = -0.121$).

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Interpretation: The Karl Pearson’s coefficient of skewness is -0.117. Since $S_k < 0$, the distribution is slightly negatively skewed, meaning the tail of the distribution extends further to the left, and the majority of observations are concentrated toward higher values.

We will solve the system using Cramer’s Rule (Determinant Method).

Step 1: Write the coefficient determinant ($D$)

$$D = \begin{vmatrix} 2 & 2 & 1 \\ 4 & 2 & 3 \\ 1 & 3 & 4 \end{vmatrix}$$

Expanding along Row 1 ($R_1$):

$$D = 2 \begin{vmatrix} 2 & 3 \\ 3 & 4 \end{vmatrix} - 2 \begin{vmatrix} 4 & 3 \\ 1 & 4 \end{vmatrix} + 1 \begin{vmatrix} 4 & 2 \\ 1 & 3 \end{vmatrix}$$

$$D = 2(8 - 9) - 2(16 - 3) + 1(12 - 2)$$

$$D = 2(-1) - 2(13) + 1(10) = -2 - 26 + 10 = -18$$

Since $D = -18 \ne 0$, a unique solution exists.

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Step 2: Calculate $D_x$ (replace 1st column with constant terms $[5, 10, 7]^T$)

$$D_x = \begin{vmatrix} 5 & 2 & 1 \\ 10 & 2 & 3 \\ 7 & 3 & 4 \end{vmatrix}$$

Expanding along Row 1:

$$D_x = 5(8 - 9) - 2(40 - 21) + 1(30 - 14)$$

$$D_x = 5(-1) - 2(19) + 1(16) = -5 - 38 + 16 = -27$$

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Step 3: Calculate $D_y$ (replace 2nd column with constant terms $[5, 10, 7]^T$)

$$D_y = \begin{vmatrix} 2 & 5 & 1 \\ 4 & 10 & 3 \\ 1 & 7 & 4 \end{vmatrix}$$

Expanding along Row 1:

$$D_y = 2(40 - 21) - 5(16 - 3) + 1(28 - 10)$$

$$D_y = 2(19) - 5(13) + 1(18) = 38 - 65 + 18 = -9$$

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Step 4: Calculate $D_z$ (replace 3rd column with constant terms $[5, 10, 7]^T$)

$$D_z = \begin{vmatrix} 2 & 2 & 5 \\ 4 & 2 & 10 \\ 1 & 3 & 7 \end{vmatrix}$$

Expanding along Row 1:

$$D_z = 2(14 - 30) - 2(28 - 10) + 5(12 - 2)$$

$$D_z = 2(-16) - 2(18) + 5(10) = -32 - 36 + 50 = -18$$

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Step 5: Apply Cramer’s Rule

$$x = \frac{D_x}{D} = \frac{-27}{-18} = \frac{3}{2} = \mathbf{1.5}$$

$$y = \frac{D_y}{D} = \frac{-9}{-18} = \frac{1}{2} = \mathbf{0.5}$$

$$z = \frac{D_z}{D} = \frac{-18}{-18} = \mathbf{1}$$

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Verification: Substitute into equation 1: $2(1.5) + 2(0.5) + 1 = 3 + 1 + 1 = 5$ (Correct).

To decide which program is easier to teach, we compare their Coefficient of Variation ($C.V.$). The program with the lower C.V. has greater homogeneity of age and will be easier to teach.

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1. Calculations for Evening MBS Program ($X_1$)

Sample size $n_1 = 10$. Observations: $24, 30, 28, 23, 25, 22, 26, 27, 28, 25$.

• Mean ($\bar{X}_1$): $\bar{X}_1 = \frac{\sum X_1}{n_1} = \frac{258}{10} = 25.8$ years.
• Standard Deviation ($\sigma_1$): $\sigma_1 = \sqrt{\frac{\sum x_1^2}{n_1}} = \sqrt{\frac{55.6}{10}} = \sqrt{5.56} \approx 2.358$ years.
• Coefficient of Variation ($C.V._1$):
  $$C.V._1 = \frac{\sigma_1}{\bar{X}_1} \times 100 = \frac{2.358}{25.8} \times 100 = \mathbf{9.14\%}$$

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2. Calculations for Morning MBS Program ($X_2$)

Sample size $n_2 = 10$. Observations: $28, 27, 34, 33, 29, 27, 28, 29, 33, 26$.

• Mean ($\bar{X}_2$): $\bar{X}_2 = \frac{\sum X_2}{n_2} = \frac{294}{10} = 29.4$ years.
• Standard Deviation ($\sigma_2$): $\sigma_2 = \sqrt{\frac{\sum x_2^2}{n_2}} = \sqrt{\frac{74.4}{10}} = \sqrt{7.44} \approx 2.728$ years.
• Coefficient of Variation ($C.V._2$):
  $$C.V._2 = \frac{\sigma_2}{\bar{X}_2} \times 100 = \frac{2.728}{29.4} \times 100 = \mathbf{9.28\%}$$

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Conclusion & Recommendation:

• $C.V.\text{ (Evening MBS)} = 9.14\%$
• $C.V.\text{ (Morning MBS)} = 9.28\%$

Since the Evening MBS program has a lower coefficient of variation ($9.14\% < 9.28\%$), its students’ ages are more homogeneous (less dispersed). Therefore, the Evening MBS program will be easier to teach based on the homogeneity criterion.

Step 1: Set up the index number computation table Let Base Year (2020) values be $P_0, Q_0$ and Current Year (2021) values be $P_1, Q_1$.

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Step 2: Laspeyre’s Price Index Number ($P_{01}^L$)

$$P_{01}^L = \frac{\sum P_1 Q_0}{\sum P_0 Q_0} \times 100 = \frac{38,500}{35,000} \times 100 = \mathbf{110.00}$$

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Step 3: Paasche’s Price Index Number ($P_{01}^P$)

$$P_{01}^P = \frac{\sum P_1 Q_1}{\sum P_0 Q_1} \times 100 = \frac{38,400}{34,500} \times 100 = \mathbf{111.30}$$

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Step 4: Fisher’s Ideal Price Index Number ($P_{01}^F$) Fisher’s Index is the geometric mean of Laspeyre’s and Paasche’s indices:

$$P_{01}^F = \sqrt{P_{01}^L \times P_{01}^P} = \sqrt{110.00 \times 111.30} = \sqrt{12,243} = \mathbf{110.65}$$

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Interpretation: According to Fisher’s Ideal Index, the cost of the consumption basket for the labor class increased by 10.65% between 2020 and 2021.

Part (a): Probability Problem (5 Marks)

Let:

• $P(B) =$ Probability boy gets scholarship $= 0.9 \implies P(B') = 1 - 0.9 = 0.1$
• $P(G) =$ Probability girl gets scholarship $= 0.8 \implies P(G') = 1 - 0.8 = 0.2$

Assuming independence between the two candidates:

• Probability that neither gets scholarship:
  $$P(B' \cap G') = P(B') \times P(G') = 0.1 \times 0.2 = 0.02$$
• Probability that at least one gets scholarship:
  $$P(\text{At least one}) = 1 - P(B' \cap G') = 1 - 0.02 = \mathbf{0.98}$$

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Part (b): Decision Theory Criteria (5 Marks)

(Note: Criteria in TU exams refer to: (i) Maximin (Pessimistic), (ii) Maximax (Optimistic), and (iii) Minimax Regret).

Let strategies be $S_1, S_2, S_3, S_4, S_5$ (labeled in rows $S_{10}$ to $S_{14}$):

(i) Maximin Criterion (Pessimistic):

• Minimum payoffs: $\{400, 380, 360, 340, 320\}$.
• Maximum of these minimums = $\mathbf{400}$ for strategy $S_{10}$.
• Decision: Select $S_{10}$.

(ii) Maximax Criterion (Optimistic):

• Maximum payoffs: $\{400, 440, 480, 520, 560\}$.
• Maximum of these maximums = $\mathbf{560}$ for strategy $S_{14}$.
• Decision: Select $S_{14}$.

(iii) Minimax Regret (Opportunity Loss) Criterion:

Find maximum payoff in each state of nature column:

• Column Max: $N_0 = 400, N_1 = 440, N_2 = 480, N_3 = 520, N_4 = 560$.

Regret Table $(\text{Regret} = \text{Column Max} - \text{Payoff})$:

• Minimum of maximum regrets = $\min(160, 120, 80, 60, 80) = \mathbf{60}$ corresponding to $S_{13}$.
• Decision: Select $S_{13}$.

Step 1: Convert inequalities into boundary equations

1. Line 1: $x + y = 30$

   • When $x = 0 \implies y = 30 \implies (0, 30)$
   • When $y = 0 \implies x = 30 \implies (30, 0)$
   • Testing $(0, 0)$: $0 + 0 = 0 \le 30$ (True, region includes origin).

2. Line 2: $x + 2y = 40$

   • When $x = 0 \implies y = 20 \implies (0, 20)$
   • When $y = 0 \implies x = 40 \implies (40, 0)$
   • Testing $(0, 0)$: $0 + 2(0) = 0 \le 40$ (True, region includes origin).

3. Non-negativity: $x \ge 0, y \ge 0$ restricts the solution to the First Quadrant.

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Step 2: Find the point of intersection between Line 1 and Line 2 Subtract Line 1 from Line 2:

$$(x + 2y) - (x + y) = 40 - 30$$

$$y = 10$$

Substitute $y = 10$ into Line 1:

$$x + 10 = 30 \implies x = 20$$

Intersection point $B = (20, 10)$.

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Step 3: Identify the Feasible Region The feasible region is bounded by the vertices:

• $O = (0, 0)$
• $A = (30, 0)$
• $B = (20, 10)$
• $C = (0, 20)$

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Step 4: Evaluate Objective Function $Z = 30x + 50y$ at Corner Points

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Conclusion: The maximum value of the objective function is $Z_{max} = 1100$, which occurs at $x = 20$ and $y = 10$.

Criteria for Testing Normality

A continuous distribution is approximately Normal if:

1. Skewness: $\beta_1 = \frac{\mu_3^2}{\mu_2^3} = 0$ (or $\gamma_1 = \sqrt{\beta_1} = 0$).
2. Kurtosis: $\beta_2 = \frac{\mu_4}{\mu_2^2} = 3$ (or $\gamma_2 = \beta_2 - 3 = 0$).

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Step 1: Computations for Distribution X

Mid-points $m$: $25, 35, 45, 55, 65, 75, 85$. Let assumed mean $A = 55$, class width $h = 10$. Step deviation $d = \frac{m - 55}{10}$.

Raw Moments about Assumed Mean ($A=55$):

• $\mu_1' = \frac{\sum f_x d}{N} \times h = \frac{-70}{200} \times 10 = -3.5$
• $\mu_2' = \frac{\sum f_x d^2}{N} \times h^2 = \frac{448}{200} \times 100 = 224$
• $\mu_3' = \frac{\sum f_x d^3}{N} \times h^3 = \frac{-358}{200} \times 1000 = -1790$
• $\mu_4' = \frac{\sum f_x d^4}{N} \times h^4 = \frac{2560}{200} \times 10000 = 128,000$

Central Moments for Distribution X:

• $\mu_2 = \mu_2' - (\mu_1')^2 = 224 - (-3.5)^2 = 224 - 12.25 = \mathbf{211.75}$
• $\mu_3 = \mu_3' - 3\mu_2'\mu_1' + 2(\mu_1')^3 = -1790 - 3(224)(-3.5) + 2(-3.5)^3 = -1790 + 2352 - 85.75 = \mathbf{476.25}$
• $\mu_4 = \mu_4' - 4\mu_3'\mu_1' + 6\mu_2'(\mu_1')^2 - 3(\mu_1')^4$ $= 128000 - 4(-1790)(-3.5) + 6(224)(12.25) - 3(150.0625)$ $= 128000 - 25060 + 16464 - 450.19 = \mathbf{118,953.81}$

Coefficients for Distribution X:

• $\beta_1 = \frac{\mu_3^2}{\mu_2^3} = \frac{(476.25)^2}{(211.75)^3} = \frac{226,814.06}{9,494,402.73} \approx \mathbf{0.0239} \approx 0$
• $\beta_2 = \frac{\mu_4}{\mu_2^2} = \frac{118,953.81}{(211.75)^2} = \frac{118,953.81}{44,838.06} \approx \mathbf{2.653}$

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Step 2: Computations for Distribution Y

Assumed mean $A = 45$, step deviation $d = \frac{m - 45}{10}$.

Raw Moments for Distribution Y ($A=45$):

• $\mu_1' = \frac{10}{200} \times 10 = 0.5$
• $\mu_2' = \frac{470}{200} \times 100 = 235$
• $\mu_3' = \frac{680}{200} \times 1000 = 3400$
• $\mu_4' = \frac{3290}{200} \times 10000 = 164,500$

Central Moments for Distribution Y:

• $\mu_2 = 235 - (0.5)^2 = 235 - 0.25 = \mathbf{234.75}$
• $\mu_3 = 3400 - 3(235)(0.5) + 2(0.5)^3 = 3400 - 352.5 + 0.25 = \mathbf{3047.75}$
• $\mu_4 = 164500 - 4(3400)(0.5) + 6(235)(0.25) - 3(0.0625) = 164500 - 6800 + 352.5 - 0.19 = \mathbf{158,052.31}$

Coefficients for Distribution Y:

• $\beta_1 = \frac{(3047.75)^2}{(234.75)^3} = \frac{9,288,780.06}{12,936,547.45} \approx \mathbf{0.718}$
• $\beta_2 = \frac{158,052.31}{(234.75)^2} = \frac{158,052.31}{55,107.56} \approx \mathbf{2.868}$

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Comparison & Interpretation:

Conclusion:

1. Distribution X has $\beta_1 \approx 0$, showing near-perfect symmetry, with slight platykurtosis ($\beta_2 = 2.65$). It is reasonably close to normal.
2. Distribution Y exhibits significant positive skewness ($\beta_1 = 0.718 > 0$), departing substantially from normality.
3. Therefore, Distribution X is closer to a normal distribution than Distribution Y.