Business Statistics 2079 Board Question Paper

Step 1: Given values

• Mean ($\bar{X}$) = $35$
• Mode ($M_o$) = $40$
• Standard Deviation ($\sigma$) = $10$

Step 2: Karl Pearson’s formula based on Mode

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

Step 3: Calculation

$$S_k = \frac{35 - 40}{10} = \frac{-5}{10} = -0.5$$

Conclusion: Karl Pearson’s coefficient of skewness is -0.5, indicating that the distribution is moderately negatively skewed.

Step 1: State given summary statistics $N = 10, \quad \sum X = 125, \quad \sum Y = 50$ $\sum X^2 = 2080, \quad \sum Y^2 = 2085, \quad \sum XY = 1050$

Step 2: Apply Karl Pearson’s Product-Moment formula

$$r = \frac{N\sum XY - (\sum X)(\sum Y)}{\sqrt{\left[N\sum X^2 - (\sum X)^2\right] \left[N\sum Y^2 - (\sum Y)^2\right]}}$$

Step 3: Evaluate Numerator and Denominator

• Numerator:

  $$10(1050) - (125)(50) = 10,500 - 6,250 = 4,250$$

• Denominator:

  $$\sqrt{[10(2080) - (125)^2] [10(2085) - (50)^2]}$$
  $$= \sqrt{[20,800 - 15,625] [20,850 - 2,500]} = \sqrt{5,175 \times 18,350} = \sqrt{94,961,250} \approx 9744.806$$

Step 4: Compute $r$

$$r = \frac{4,250}{9744.806} \approx \mathbf{0.436}$$

Conclusion: The correlation coefficient is +0.436, indicating a moderate positive linear correlation between $X$ and $Y$.

Step 1: Identify given properties

• Standard Deviation ($\sigma$) = $7 \implies$ Variance ($\mu_2$) = $\sigma^2 = 7^2 = 49$.
• The distribution is mesokurtic (Normal).

Step 2: Property of a Mesokurtic Distribution For a mesokurtic distribution, the kurtosis coefficient $\beta_2 = 3$:

$$\beta_2 = \frac{\mu_4}{\mu_2^2} = 3$$

Step 3: Solve for the Fourth Central Moment ($\mu_4$)

$$\mu_4 = 3 \times \mu_2^2 = 3 \times (49)^2 = 3 \times 2,401 = \mathbf{7,203}$$

Conclusion: The fourth moment about the mean is 7,203.

Step 1: Define events and probabilities Let:

• $P(R) = \frac{3}{4} \implies P(R') = 1 - \frac{3}{4} = \frac{1}{4}$ (Probability Ram is not selected)
• $P(S) = \frac{1}{5} \implies P(S') = 1 - \frac{1}{5} = \frac{4}{5}$ (Probability Sita is not selected)

Since the interviews are for two different posts, events $R$ and $S$ are statistically independent.

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Part (a): Probability both will be selected

$$P(R \cap S) = P(R) \times P(S) = \frac{3}{4} \times \frac{1}{5} = \frac{3}{20} = \mathbf{0.15}$$

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Part (b): Probability none of them will be selected

$$P(R' \cap S') = P(R') \times P(S') = \frac{1}{4} \times \frac{4}{5} = \frac{1}{5} = \mathbf{0.20}$$

Step 1: Identify given equation and parameters

• Trend equation: $Y = 50 + 2.5X$ (Sales in million Rs.)
• Origin: Year $2015$ ($X = 0$ in 2015, unit of $X = 1$ year)

Step 2: Find $X$ for the year 2022

$$X = 2022 - 2015 = 7$$

Step 3: Estimate Sales ($Y$)

$$Y = 50 + 2.5(7) = 50 + 17.5 = 67.5$$

Conclusion: The estimated sales for the year 2022 is Rs. 67.5 million (Rs. 67,500,000).

Step 1: Identify given values

• Company A: $N_1 = 100, \quad \bar{X}_1 = \text{Rs. } 2,550$
• Company B: $N_2 = 110, \quad \bar{X}_2 = \text{Rs. } 2,800$

Step 2: Formula for combined arithmetic mean

$$\bar{X}_{12} = \frac{N_1\bar{X}_1 + N_2\bar{X}_2}{N_1 + N_2}$$

Step 3: Calculation

$$\bar{X}_{12} = \frac{100(2,550) + 110(2,800)}{100 + 110} = \frac{255,000 + 308,000}{210} = \frac{563,000}{210} \approx \mathbf{2680.95}$$

Conclusion: The combined mean monthly income of workers across both companies is Rs. 2,680.95.

Step 1: Calculate total prices for Base Year (2020) and Current Year (2021)

• Base year prices:
  $$\sum P_0 = 160 + 140 + 150 + 130 + 155 + 120 = 855$$
• Current year prices:
  $$\sum P_1 = 170 + 135 + 155 + 140 + 150 + 30 = 780$$

Step 2: Formula for Simple Aggregative Price Index

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

Step 3: Calculation

$$P_{01} = \frac{780}{855} \times 100 \approx \mathbf{91.23}$$

Conclusion: The simple aggregative price index number for 2021 is 91.23, indicating an overall price decline of 8.77% ($100 - 91.23$) relative to 2020.

Step 1: Compute matrix sum $(A + B)$

$$A + B = \begin{bmatrix} 1+3 & 2+1 & 5+5 \\ 2+2 & 1+5 & 3+1 \\ 3+1 & 3+3 & 2+2 \end{bmatrix} = \begin{bmatrix} 4 & 3 & 10 \\ 4 & 6 & 4 \\ 4 & 6 & 4 \end{bmatrix}$$

Step 2: Scalar multiplication by 7

$$7(A + B) = 7 \begin{bmatrix} 4 & 3 & 10 \\ 4 & 6 & 4 \\ 4 & 6 & 4 \end{bmatrix} = \begin{bmatrix} 7(4) & 7(3) & 7(10) \\ 7(4) & 7(6) & 7(4) \\ 7(4) & 7(6) & 7(4) \end{bmatrix} = \mathbf{\begin{bmatrix} 28 & 21 & 70 \\ 28 & 42 & 28 \\ 28 & 42 & 28 \end{bmatrix}}$$

Let the determinant be:

$$|A| = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 5 & 1 \\ 9 & 3 & 2 \end{vmatrix}$$

Expansion along Row 1 ($R_1$):

$$|A| = 1 \begin{vmatrix} 5 & 1 \\ 3 & 2 \end{vmatrix} - 2 \begin{vmatrix} 2 & 1 \\ 9 & 2 \end{vmatrix} + 3 \begin{vmatrix} 2 & 5 \\ 9 & 3 \end{vmatrix}$$

Evaluate $2 \times 2$ determinants:

1. $1 \times [(5 \times 2) - (1 \times 3)] = 1 \times [10 - 3] = 1 \times 7 = 7$
2. $-2 \times [(2 \times 2) - (1 \times 9)] = -2 \times [4 - 9] = -2 \times (-5) = +10$
3. $+3 \times [(2 \times 3) - (5 \times 9)] = 3 \times [6 - 45] = 3 \times (-39) = -117$

Summing the terms:

$$|A| = 7 + 10 - 117 = \mathbf{-100}$$

Conclusion: The value of the determinant is -100.

Definition: Chronological (or Temporal) Classification is the systematic arrangement and categorization of statistical data according to the time of occurrence, such as years, quarters, months, weeks, or days. It displays how a variable evolves or fluctuates across time periods.

Example: Annual Foreign Direct Investment (FDI) inflows into Nepal:

Analytical Interpretation:

The middle 80% of students lies symmetrically between the lower 10% and the upper 10% of the distribution. Therefore, we must find:

1. Lower Limit: $10^{\text{th}}$ Percentile ($P_{10}$)
2. Upper Limit: $90^{\text{th}}$ Percentile ($P_{90}$)

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Step 1: Cumulative Frequency Table

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Step 2: Calculate Lower Limit ($P_{10}$)

• Position of $P_{10} = \frac{10 \times N}{100} = \frac{10 \times 100}{100} = 10^{\text{th}}$ item.
• Looking at the $c.f.$ column, the $10^{\text{th}}$ item falls in class 20 - 30.
  • $L = 20, \quad c.f. = 8, \quad f = 16, \quad h = 10$$P_{10} = L + \left(\frac{\frac{10N}{100} - c.f.}{f}\right) \times h$$
    $$P_{10} = 20 + \left(\frac{10 - 8}{16}\right) \times 10 = 20 + \frac{20}{16} = 20 + 1.25 = \mathbf{21.25}$$

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Step 3: Calculate Upper Limit ($P_{90}$)

• Position of $P_{90} = \frac{90 \times N}{100} = \frac{90 \times 100}{100} = 90^{\text{th}}$ item.
• Looking at the $c.f.$ column, the $90^{\text{th}}$ item falls in class 60 - 70.
  • $L = 60, \quad c.f. = 86, \quad f = 9, \quad h = 10$$P_{90} = L + \left(\frac{\frac{90N}{100} - c.f.}{f}\right) \times h$$
    $$P_{90} = 60 + \left(\frac{90 - 86}{9}\right) \times 10 = 60 + \frac{40}{9} = 60 + 4.44 = \mathbf{64.44}$$

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Conclusion:

The limits of marks obtained by the middle 80% of the students are 21.25 marks to 64.44 marks.

Step 1: Set up the Cost of Living Index (Family Budget Method)

Let:

• Base year (2021) price $= P_0$
• Current year (2022) price $= P_1$
• Weight (percentage expense) $= W$
• Price Relative $I = \frac{P_1}{P_0} \times 100$

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Step 2: Compute Cost of Living Index Number ($CLI$)

$$CLI = \frac{\sum IW}{\sum W} = \frac{12,612.95}{100} = \mathbf{126.13}$$

The cost of living in 2022 increased by 26.13% compared to 2021.

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Step 3: Compute Adjusted Salary in Current Period

To maintain the same standard of living as in the base period:

$$\text{Required Salary in Current Period} = \text{Base Period Salary} \times \left(\frac{CLI}{100}\right)$$

$$\text{Required Salary} = 40,500 \times \left(\frac{126.13}{100}\right) = 405 \times 126.13 = \mathbf{\text{Rs. } 51,082.65}$$

Conclusion:

• The Cost of Living Index for 2022 is 126.13.
• The employee’s monthly salary in the current period should be raised to Rs. 51,083 to maintain their real purchasing power.

Step 1: Formulate Boundary Equations for the Constraints

1. Line 1: $5x + y = 10$

   • $(x=0 \implies y=10) \implies (0, 10)$
   • $(y=0 \implies x=2) \implies (2, 0)$
   • Testing $(0, 0)$: $0 \ge 10$ is False (region is away from the origin).

2. Line 2: $2x + 2y = 12 \iff x + y = 6$

   • $(x=0 \implies y=6) \implies (0, 6)$
   • $(y=0 \implies x=6) \implies (6, 0)$
   • Testing $(0, 0)$: $0 \ge 6$ is False (region is away from the origin).

3. Line 3: $x + 4y = 12$

   • $(x=0 \implies y=3) \implies (0, 3)$
   • $(y=0 \implies x=12) \implies (12, 0)$
   • Testing $(0, 0)$: $0 \ge 12$ is False (region is away from the origin).

4. Non-negativity: $x \ge 0, y \ge 0$ restricts the region to the first quadrant.

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Step 2: Find Intersection Points between Constraints

1. Intersection of Line 1 ($5x+y=10$) and Line 2 ($x+y=6$): Subtract equations: $4x = 4 \implies x = 1$. Then $y = 6 - 1 = 5$. Point $B = (1, 5)$. Check Line 3: $1 + 4(5) = 21 \ge 12$ (Feasible).

2. Intersection of Line 2 ($x+y=6$) and Line 3 ($x+4y=12$): Subtract equations: $3y = 6 \implies y = 2$. Then $x = 6 - 2 = 4$. Point $C = (4, 2)$. Check Line 1: $5(4) + 2 = 22 \ge 10$ (Feasible).

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Step 3: Identify the Feasible Region Corner Points The feasible region is unbounded above, with extreme boundary vertices:

• $A = (0, 10)$ on the Y-axis (since $\max(10, 6, 3) = 10$).
• $B = (1, 5)$
• $C = (4, 2)$
• $D = (12, 0)$ on the X-axis (since $\max(2, 6, 12) = 12$).

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Step 4: Evaluate Objective Function $Z = 2x + 3y$

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Step 5: Verification for Unbounded Region Since the feasible region is unbounded, we must verify if the open half-plane $2x + 3y < 14$ shares any points with the feasible region. Graphing $2x + 3y < 14$:

• At $(0, 10)$: $30 > 14$
• At $(1, 5)$: $17 > 14$
• At $(12, 0)$: $24 > 14$ The half-plane $2x + 3y < 14$ lies entirely below the boundary and shares no points with the feasible region.

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Conclusion: The minimum cost is $Z_{min} = 14$, obtained at $x = 4$ and $y = 2$.

Part (a): Decision Making Under Uncertainty (5 Marks)

Given Payoff Matrix:

Transpose to strategies as rows (actions under control):

(i) Maximax Approach (Optimistic):

• Max payoffs: $S_1 = 500, S_2 = 600, S_3 = 700$.
• $\max(500, 600, 700) = \mathbf{700}$ for $S_3$.
• Decision: Select strategy $S_3$.

(ii) Maximin Approach (Pessimistic):

• Min payoffs: $S_1 = 100, S_2 = 200, S_3 = 100$.
• $\max(100, 200, 100) = \mathbf{200}$ for $S_2$.
• Decision: Select strategy $S_2$.

(iii) Minimax Regret Approach:

Find maximum payoff in each state of nature:

• Max for $N_1 = 700$, Max for $N_2 = 600$, Max for $N_3 = 500$.

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

• $\min(500, 500, 400) = \mathbf{400}$ for $S_3$.
• Decision: Select strategy $S_3$.

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Part (b): Dice Probability Problem (5 Marks)

Total possible outcomes when two fair dice are rolled: $n(S) = 6 \times 6 = 36$.

(i) Probability of getting a sum of 8 or 9:

• Outcomes for Sum = 8: $(2,6), (3,5), (4,4), (5,3), (6,2) \implies 5$ outcomes.
• Outcomes for Sum = 9: $(3,6), (4,5), (5,4), (6,3) \implies 4$ outcomes.
• Total favorable outcomes $= 5 + 4 = 9$.
  $$P(\text{Sum is 8 or 9}) = \frac{9}{36} = \frac{1}{4} = \mathbf{0.25}$$

(ii) Probability of getting a sum less than 5:

• Sum can be 2, 3, or 4:
  • Sum = 2: $(1, 1) \implies 1$ outcome
  • Sum = 3: $(1, 2), (2, 1) \implies 2$ outcomes
  • Sum = 4: $(1, 3), (2, 2), (3, 1) \implies 3$ outcomes
• Total favorable outcomes $= 1 + 2 + 3 = 6$.
  $$P(\text{Sum } < 5) = \frac{6}{36} = \frac{1}{6} \approx \mathbf{0.1667}$$

Step 1: Set up the Least Squares Calculation Table

Number of years $n = 5$ (odd). Let middle year $2018$ be the origin ($A = 2018$). Step deviation $X = \text{Year} - 2018$, so $\sum X = 0$.

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Step 2: Fit the Straight Line Trend Equation

Linear trend equation:

$$Y_c = a + bX$$

Since $\sum X = 0$:

$$a = \frac{\sum Y}{n} = \frac{336}{5} = \mathbf{67.2}$$

$$b = \frac{\sum XY}{\sum X^2} = \frac{124}{10} = \mathbf{12.4}$$

The fitted trend line equation is:

$$Y_c = 67.2 + 12.4X \quad (\text{Origin: 2018, } X \text{ in 1-year units, } Y \text{ in '000 Rs.})$$

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Step 3: Estimate Sales for the Year 2024

For Year $2024$:

$$X = 2024 - 2018 = 6$$

$$Y_{2024} = 67.2 + 12.4(6) = 67.2 + 74.4 = \mathbf{141.6}$$

Conclusion:

• The fitted trend equation is $Y_c = 67.2 + 12.4X$.
• Estimated sales for 2024 is Rs. 141,600 ($141.6 \times 1,000$).

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

Step 1: Compute Coefficient Determinant ($D$)

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

Expanding along Row 1 ($R_1$):

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

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

$$D = 2(-5) + 1(-11) + 2(9) = -10 - 11 + 18 = \mathbf{-3}$$

Since $D = -3 \ne 0$, a unique solution exists.

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Step 2: Calculate $D_x$ (replace 1st column with constant terms $[-1, 4, 4]^T$)

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

Expanding along Row 1:

$$D_x = -1(-2 - 3) - (-1)(4 - 12) + 2(4 - (-8))$$

$$D_x = -1(-5) + 1(-8) + 2(12) = 5 - 8 + 24 = \mathbf{21}$$

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Step 3: Calculate $D_y$ (replace 2nd column with $[-1, 4, 4]^T$)

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

Expanding along Row 1:

$$D_y = 2(4 - 12) - (-1)(1 - 12) + 2(4 - 16)$$

$$D_y = 2(-8) + 1(-11) + 2(-12) = -16 - 11 - 24 = \mathbf{-51}$$

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Step 4: Calculate $D_z$ (replace 3rd column with $[-1, 4, 4]^T$)

$$D_z = \begin{vmatrix} 2 & -1 & -1 \\ 1 & -2 & 4 \\ 4 & 1 & 4 \end{vmatrix}$$

Expanding along Row 1:

$$D_z = 2(-8 - 4) - (-1)(4 - 16) + (-1)(1 - (-8))$$

$$D_z = 2(-12) + 1(-12) - 1(9) = -24 - 12 - 9 = \mathbf{-45}$$

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Step 5: Apply Cramer’s Rule

$$x = \frac{D_x}{D} = \frac{21}{-3} = \mathbf{-7}$$

$$y = \frac{D_y}{D} = \frac{-51}{-3} = \mathbf{17}$$

$$z = \frac{D_z}{D} = \frac{-45}{-3} = \mathbf{15}$$

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Verification: Substitute into equation 1: $2(-7) - 17 + 2(15) = -14 - 17 + 30 = -31 + 30 = -1$ (Matches). Substitute into equation 3: $4(-7) + 17 + 15 = -28 + 32 = 4$ (Matches).

Step 1: Calculation Table

Mid-values $m$: $1, 3, 5, 7, 9$. Let assumed mean $A = 5$, class width $h = 2$. Step deviation $d = \frac{m - 5}{2} = -2, -1, 0, 1, 2$.

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Step 2: Statistical Measures for Model A

• Mean ($\bar{X}_A$):

  $$\bar{X}_A = A + \left(\frac{\sum f_A d}{N_A}\right) h = 5 + \left(\frac{5}{60}\right) 2 = 5 + \frac{10}{60} = 5 + 0.167 = \mathbf{5.167} \text{ years}$$

• Standard Deviation ($\sigma_A$):

  $$\sigma_A = h \times \sqrt{\frac{\sum f_A d^2}{N_A} - \left(\frac{\sum f_A d}{N_A}\right)^2} = 2 \times \sqrt{\frac{73}{60} - \left(\frac{5}{60}\right)^2}$$
  $$\sigma_A = 2 \times \sqrt{1.2167 - 0.0069} = 2 \times \sqrt{1.2098} = 2 \times 1.0999 = \mathbf{2.20} \text{ years}$$

• Coefficient of Variation ($C.V._A$):

  $$C.V._A = \frac{\sigma_A}{\bar{X}_A} \times 100 = \frac{2.20}{5.167} \times 100 = \mathbf{42.58\%}$$

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Step 3: Statistical Measures for Model B

• Mean ($\bar{X}_B$):

  $$\bar{X}_B = A + \left(\frac{\sum f_B d}{N_B}\right) h = 5 + \left(\frac{-18}{60}\right) 2 = 5 - 0.6 = \mathbf{4.40} \text{ years}$$

• Standard Deviation ($\sigma_B$):

  $$\sigma_B = h \times \sqrt{\frac{\sum f_B d^2}{N_B} - \left(\frac{\sum f_B d}{N_B}\right)^2} = 2 \times \sqrt{\frac{78}{60} - \left(\frac{-18}{60}\right)^2}$$
  $$\sigma_B = 2 \times \sqrt{1.30 - 0.09} = 2 \times \sqrt{1.21} = 2 \times 1.10 = \mathbf{2.20} \text{ years}$$

• Coefficient of Variation ($C.V._B$):

  $$C.V._B = \frac{\sigma_B}{\bar{X}_B} \times 100 = \frac{2.20}{4.40} \times 100 = \mathbf{50.00\%}$$

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Part (a): Uniformity Comparison

• $C.V._A = 42.58\%$
• $C.V._B = 50.00\%$

Reasoning: Since $C.V._A < C.V._B$, Model A has a smaller relative variation in life. Therefore, Model A has greater uniformity of life.

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Part (b): Combined Standard Deviation ($\sigma_{12}$)

1. Combined Mean ($\bar{X}_{12}$):

   $$\bar{X}_{12} = \frac{N_A \bar{X}_A + N_B \bar{X}_B}{N_A + N_B} = \frac{60(5.167) + 60(4.40)}{60 + 60} = \frac{310.02 + 264}{120} = \frac{574.02}{120} = \mathbf{4.7835} \text{ years}$$

2. Deviations from Combined Mean:

   • $d_A = \bar{X}_A - \bar{X}_{12} = 5.167 - 4.7835 = +0.3835 \implies d_A^2 = 0.1471$
   • $d_B = \bar{X}_B - \bar{X}_{12} = 4.40 - 4.7835 = -0.3835 \implies d_B^2 = 0.1471$

3. Combined Variance Formula:

   $$\sigma_{12} = \sqrt{\frac{N_A(\sigma_A^2 + d_A^2) + N_B(\sigma_B^2 + d_B^2)}{N_A + N_B}}$$
   $$\sigma_{12} = \sqrt{\frac{60(2.20^2 + 0.1471) + 60(2.20^2 + 0.1471)}{120}} = \sqrt{4.84 + 0.1471} = \sqrt{4.9871} = \mathbf{2.233} \text{ years}$$

Conclusion: The combined standard deviation of both models is 2.233 years.

Step 1: Set up the Calculation Table

Mid-values $m$: $15, 25, 35, 45, 55, 65, 75, 85$. Let assumed mean $A = 45$, class width $h = 10$. Step deviation $d = \frac{m - 45}{10} = -3, -2, -1, 0, 1, 2, 3, 4$.

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Step 2: Calculate Mean, Mode, and Standard Deviation

1. Mean ($\bar{X}$):

   $$\bar{X} = A + \left(\frac{\sum fd}{N}\right) h = 45 + \left(\frac{26}{100}\right) 10 = 45 + 2.6 = \mathbf{47.60} \text{ million Rs.}$$

2. Mode ($M_o$): Highest frequency $f_1 = 26$ lies in modal class 40 - 50.

   • $L = 40, \quad f_1 = 26, \quad f_0 = 18, \quad f_2 = 20, \quad h = 10$$M_o = L + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) h = 40 + \left(\frac{26 - 18}{2(26) - 18 - 20}\right) 10$$
     $$M_o = 40 + \left(\frac{8}{52 - 38}\right) 10 = 40 + \frac{80}{14} = 40 + 5.71 = \mathbf{45.71} \text{ million Rs.}$$

3. Standard Deviation ($\sigma$):

   $$\sigma = h \times \sqrt{\frac{\sum fd^2}{N} - \left(\frac{\sum fd}{N}\right)^2} = 10 \times \sqrt{\frac{278}{100} - (0.26)^2}$$
   $$\sigma = 10 \times \sqrt{2.78 - 0.0676} = 10 \times \sqrt{2.7124} = 10 \times 1.6469 = \mathbf{16.47} \text{ million Rs.}$$

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Step 3: Karl Pearson’s Coefficient of Skewness ($S_k$)

$$S_k = \frac{\bar{X} - M_o}{\sigma} = \frac{47.60 - 45.71}{16.47} = \frac{1.89}{16.47} = \mathbf{+0.115}$$

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Step 4: Percentile Coefficient of Kurtosis ($\kappa$)

$$\kappa = \frac{Q.D.}{P_{90} - P_{10}} = \frac{0.5(Q_3 - Q_1)}{P_{90} - P_{10}}$$

1. $Q_1$ ($25^{\text{th}}$ item): Falls in class 30 - 40 ($c.f.=32, L=30, f=18, \text{prec } c.f.=14$).

   $$Q_1 = 30 + \left(\frac{25 - 14}{18}\right) 10 = 30 + \frac{110}{18} = 30 + 6.11 = \mathbf{36.11}$$

2. $Q_3$ ($75^{\text{th}}$ item): Falls in class 50 - 60 ($c.f.=78, L=50, f=20, \text{prec } c.f.=58$).

   $$Q_3 = 50 + \left(\frac{75 - 58}{20}\right) 10 = 50 + \frac{170}{20} = 50 + 8.50 = \mathbf{58.50}$$
   $$Q.D. = \frac{58.50 - 36.11}{2} = \frac{22.39}{2} = \mathbf{11.195}$$

3. $P_{10}$ ($10^{\text{th}}$ item): Falls in class 20 - 30 ($c.f.=14, L=20, f=9, \text{prec } c.f.=5$).

   $$P_{10} = 20 + \left(\frac{10 - 5}{9}\right) 10 = 20 + 5.56 = \mathbf{25.56}$$

4. $P_{90}$ ($90^{\text{th}}$ item): Falls in class 60 - 70 ($c.f.=90, L=60, f=12, \text{prec } c.f.=78$).

   $$P_{90} = 60 + \left(\frac{90 - 78}{12}\right) 10 = 60 + 10.00 = \mathbf{70.00}$$

5. Coefficient of Kurtosis ($\kappa$):

   $$\kappa = \frac{11.195}{70.00 - 25.56} = \frac{11.195}{44.44} = \mathbf{0.252}$$

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Comments on the Nature of Income Distribution:

1. Skewness ($S_k = +0.115$): The distribution is slightly positively skewed, indicating that the tail stretches slightly toward higher income levels, with a majority of families earning below the mean income of Rs. 47.60 million.
2. Kurtosis ($\kappa = 0.252$): Since $\kappa = 0.252 < 0.263$, the distribution is platykurtic, meaning it has a broader, flatter peak than a standard bell-shaped normal curve.