Business Statistics 2080 Board Question Paper

The two principal objectives of statistical classification are:

1. To Condense Mass Data into Manageable Form: Classification reduces vast, unorganized raw figures into systematic, homogenous groups, making them easy to comprehend, assimilate, and interpret.
2. To Facilitate Comparison and Analysis: By grouping similar items together and highlighting points of resemblance and difference, classification enables meaningful comparative analysis between different groups or time periods.

Step 1: Given values

• $\text{Mean } (\bar{X}) = 25$
• $\text{Mode } (M_o) = 28$
• $\text{Standard Deviation } (\sigma) = 5$

Step 2: Karl Pearson’s formula based on Mode

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

Step 3: Calculation

$$S_k = \frac{25 - 28}{5} = \frac{-3}{5} = \mathbf{-0.6}$$

Conclusion: Karl Pearson’s coefficient of skewness is -0.6, indicating that the distribution is negatively skewed.

Step 1: Given values

• Third Quartile ($Q_3$) = $30$
• Median ($M_d$) = $22$
• The distribution is symmetrical.

Step 2: Symmetry Condition In a perfectly symmetrical distribution, Bowley’s coefficient of skewness is zero, meaning the median is equidistant from the quartiles:

$$Q_3 - M_d = M_d - Q_1 \iff Q_3 + Q_1 - 2M_d = 0$$

Step 3: Solve for First Quartile ($Q_1$)

$$Q_1 = 2M_d - Q_3$$

$$Q_1 = 2(22) - 30 = 44 - 30 = \mathbf{14}$$

Conclusion: The value of the first quartile ($Q_1$) is 14.

Step 1: Given values

• Bowley’s Coefficient of Skewness ($S_{k(B)}$) = $0.6$
• Quartile Deviation ($Q.D.$) = $5$
• Third Quartile ($Q_3$) = $28$

Step 2: Find the First Quartile ($Q_1$) from $Q.D.$

$$Q.D. = \frac{Q_3 - Q_1}{2} = 5 \implies Q_3 - Q_1 = 10$$

$$28 - Q_1 = 10 \implies Q_1 = 28 - 10 = 18$$

Step 3: Apply Bowley’s Skewness Formula to find Median ($M_d$)

$$S_{k(B)} = \frac{Q_3 + Q_1 - 2M_d}{Q_3 - Q_1}$$

$$0.6 = \frac{28 + 18 - 2M_d}{10}$$

$$6 = 46 - 2M_d \implies 2M_d = 46 - 6 = 40 \implies M_d = \mathbf{20}$$

Conclusion: The median of the distribution is 20.

Step 1: Theoretical Property Both regression lines always intersect at the point of their means, $(\bar{X}, \bar{Y})$. Therefore, replacing $X$ and $Y$ with $\bar{X}$ and $\bar{Y}$ gives a simultaneous system:

1. $4\bar{X} - 5\bar{Y} = -33$ --- (Equation 1)
2. $20\bar{X} - 9\bar{Y} = 107$ --- (Equation 2)

Step 2: Solve the Simultaneous Equations Multiply Equation 1 by 5:

$$20\bar{X} - 25\bar{Y} = -165$$

Subtract this from Equation 2:

$$(20\bar{X} - 9\bar{Y}) - (20\bar{X} - 25\bar{Y}) = 107 - (-165)$$

$$16\bar{Y} = 272 \implies \bar{Y} = \frac{272}{16} = \mathbf{17}$$

Substitute $\bar{Y} = 17$ into Equation 1:

$$4\bar{X} - 5(17) = -33$$

$$4\bar{X} - 85 = -33 \implies 4\bar{X} = 85 - 33 = 52 \implies \bar{X} = \frac{52}{4} = \mathbf{13}$$

Conclusion: The mean values are $\bar{X} = 13$ and $\bar{Y} = 17$.

Method 1: Direct Expansion along Row 1 ($R_1$)

$$|A| = 1 \begin{vmatrix} 5 & 6 \\ 8 & 9 \end{vmatrix} - 2 \begin{vmatrix} 4 & 6 \\ 7 & 9 \end{vmatrix} + 3 \begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix}$$

$$|A| = 1(45 - 48) - 2(36 - 42) + 3(32 - 35)$$

$$|A| = 1(-3) - 2(-6) + 3(-3) = -3 + 12 - 9 = \mathbf{0}$$

Method 2: By Properties of Determinants Perform row operations: $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_2$:

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

Since Row 2 and Row 3 are identical, $|A| = 0$.

Conclusion: The value of the determinant is 0.

Total number of cards in a standard deck: $n(S) = 52$.

Part (a): Probability of getting a black card There are 26 black cards (13 Spades + 13 Clubs):

$$P(\text{Black}) = \frac{n(\text{Black})}{n(S)} = \frac{26}{52} = \frac{1}{2} = \mathbf{0.5}$$

Part (b): Probability of getting a King There are 4 Kings in a standard deck (1 of each suit):

$$P(\text{King}) = \frac{n(\text{King})}{n(S)} = \frac{4}{52} = \frac{1}{13} \approx \mathbf{0.0769}$$

Step 1: Compute sums of prices Base year (2014) price $= P_0$; Current year (2015) price $= P_1$.

• Base year prices sum:
  $$\sum P_0 = 5 + 4 + 10 + 7 = 26$$
• Current year prices sum:
  $$\sum P_1 = 8 + 6 + 12 + 5 = 31$$

Step 2: Formula for Simple Aggregative Price Index

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

Step 3: Calculation

$$P_{01} = \frac{31}{26} \times 100 = \mathbf{119.23}$$

Conclusion: The simple aggregative price index number is 119.23, representing an increase of 19.23% in price level from 2014 to 2015.

Definition: The transpose of a matrix $A$ (denoted $A^T$ or $A'$) is obtained by interchanging its rows into columns.

Calculation:

$$A = \begin{bmatrix} 1 & -2 \\ 4 & 3 \end{bmatrix} \implies A^T = \mathbf{\begin{bmatrix} 1 & 4 \\ -2 & 3 \end{bmatrix}}$$

Step 1: State the parameters

• Trend equation: $y = 40 + 1.5x$
• Origin: Year $2017$ ($x = 0$ in 2017, unit of $x = 1$ year)

Step 2: Calculate $x$ for the year 2033

$$x = 2033 - 2017 = 16$$

Step 3: Compute Estimated Production ($y$)

$$y = 40 + 1.5(16) = 40 + 24 = \mathbf{64} \text{ tonnes}$$

Conclusion: The estimated production for the year 2033 is 64 tonnes.

Part (i): Correction of Rank Correlation Coefficient (5 Marks)

Step 1: Given data $n = 10, \quad r_s = 0.5$

Step 2: Apply Spearman’s formula to find incorrect $\sum d^2$

$$r_s = 1 - \frac{6 \sum d^2}{n(n^2 - 1)}$$

$$0.5 = 1 - \frac{6 \sum d^2}{10(100 - 1)} = 1 - \frac{6 \sum d^2}{990}$$

$$\frac{6 \sum d^2}{990} = 1 - 0.5 = 0.5$$

$$\sum d^2 = \frac{0.5 \times 990}{6} = \frac{495}{6} = 82.5$$

Step 3: Adjust for the incorrect observation

• Incorrect difference: $3 \implies 3^2 = 9$
• Correct difference: $7 \implies 7^2 = 49$$\text{Correct } \sum d^2 = 82.5 - 9 + 49 = \mathbf{122.5}$$

Step 4: Compute correct $r_s$

$$\text{Correct } r_s = 1 - \frac{6(122.5)}{990} = 1 - \frac{735}{990} = 1 - 0.7424 = \mathbf{+0.2576}$$

The corrected rank correlation coefficient is +0.258.

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Part (ii): Comparison of Wage Bill and Variability (5 Marks)

Given data:

• Firm A: $N_1 = 500, \quad \bar{X}_1 = \text{Rs. } 480, \quad \sigma_1^2 = 400 \implies \sigma_1 = \sqrt{400} = 20$
• Firm B: $N_2 = 600, \quad \bar{X}_2 = \text{Rs. } 475, \quad \sigma_2^2 = 625 \implies \sigma_2 = \sqrt{625} = 25$

(a) Which firm pays larger wage bill?

$$\text{Total Wage Bill} = N \times \bar{X}$$

• Firm A: $500 \times 480 = \mathbf{\text{Rs. } 240,000}$
• Firm B: $600 \times 475 = \mathbf{\text{Rs. } 285,000}$

Conclusion: Firm B pays a larger total monthly wage bill (Rs. 285,000 > Rs. 240,000).

(b) In which firm is there greater variability in individual wages?

Compare their Coefficients of Variation ($C.V.$):

$$C.V._A = \frac{\sigma_1}{\bar{X}_1} \times 100 = \frac{20}{480} \times 100 = \mathbf{4.17\%}$$

$$C.V._B = \frac{\sigma_2}{\bar{X}_2} \times 100 = \frac{25}{475} \times 100 = \mathbf{5.26\%}$$

Conclusion: Since $C.V._B > C.V._A$, there is greater variability in individual wages in Firm B.

Step 1: Convert Inclusive Classes to Exclusive Boundaries

Subtract $0.5$ from lower limits and add $0.5$ to upper limits:

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Step 2: Calculate Required Partition Values

Class width $h = 10$.

1. Lower Quartile ($Q_1$):

   • Position: $\frac{N}{4} = \frac{240}{4} = 60^{\text{th}}$ item $\implies$ Class 19.5 - 29.5 ($c.f.=67, L=19.5, f=32, \text{prec } c.f.=35$).
     $$Q_1 = 19.5 + \left(\frac{60 - 35}{32}\right) \times 10 = 19.5 + \frac{250}{32} = 19.5 + 7.81 = \mathbf{27.31}$$

2. Upper Quartile ($Q_3$):

   • Position: $\frac{3N}{4} = \frac{3(240)}{4} = 180^{\text{th}}$ item $\implies$ Class 49.5 - 59.5 ($c.f.=213, L=49.5, f=43, \text{prec } c.f.=170$).

     $$Q_3 = 49.5 + \left(\frac{180 - 170}{43}\right) \times 10 = 49.5 + \frac{100}{43} = 49.5 + 2.33 = \mathbf{51.83}$$

   • Quartile Deviation ($Q.D.$):

     $$Q.D. = \frac{Q_3 - Q_1}{2} = \frac{51.83 - 27.31}{2} = \frac{24.52}{2} = \mathbf{12.26}$$

3. $10^{\text{th}}$ Percentile ($P_{10}$):

   • Position: $\frac{10 \times 240}{100} = 24^{\text{th}}$ item $\implies$ Class 9.5 - 19.5 ($c.f.=35, L=9.5, f=35, \text{prec } c.f.=0$).
     $$P_{10} = 9.5 + \left(\frac{24 - 0}{35}\right) \times 10 = 9.5 + \frac{240}{35} = 9.5 + 6.86 = \mathbf{16.36}$$

4. $90^{\text{th}}$ Percentile ($P_{90}$):

   • Position: $\frac{90 \times 240}{100} = 216^{\text{th}}$ item $\implies$ Class 59.5 - 69.5 ($c.f.=230, L=59.5, f=17, \text{prec } c.f.=213$).
     $$P_{90} = 59.5 + \left(\frac{216 - 213}{17}\right) \times 10 = 59.5 + \frac{30}{17} = 59.5 + 1.76 = \mathbf{61.26}$$

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Step 3: Compute Percentile Coefficient of Kurtosis ($\kappa$)

$$\kappa = \frac{Q.D.}{P_{90} - P_{10}} = \frac{12.26}{61.26 - 16.36} = \frac{12.26}{44.90} = \mathbf{0.273}$$

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Interpretation:

• Normal (mesokurtic) distribution standard: $\kappa = 0.263$.
• Since $\kappa = 0.273 > 0.263$, the distribution is Leptokurtic (more peaked than a normal distribution with relatively heavier tails).

Step 1: Set up the Regression Calculation Table

Let $X =$ Age (years) and $Y =$ Cholesterol level (mg/dL). $N = 10$.

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Step 2: Compute Means and Regression Slope ($b_{yx}$)

• $\bar{X} = \frac{\sum X}{N} = \frac{512}{10} = 51.2$
• $\bar{Y} = \frac{\sum Y}{N} = \frac{1896}{10} = 189.6$$b_{yx} = \frac{N \sum XY - (\sum X)(\sum Y)}{N \sum X^2 - (\sum X)^2}$$
  $$b_{yx} = \frac{10(98,307) - (512)(1,896)}{10(28,110) - (512)^2} = \frac{983,070 - 970,752}{281,100 - 262,144} = \frac{12,318}{18,956} \approx \mathbf{0.6498}$$

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Step 3: Develop the Regression Line of $Y$ on $X$$Y - \bar{Y} = b_{yx}(X - \bar{X})$$

$$Y - 189.6 = 0.6498(X - 51.2)$$

$$Y = 189.6 + 0.6498X - 33.27$$

$$Y = \mathbf{156.33 + 0.650X}$$

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Step 4: Predict Cholesterol Level for Age $X = 60$$Y_{X=60} = 156.33 + 0.6498(60) = 156.33 + 38.99 = \mathbf{195.32}$$

Conclusion:

• The fitted regression equation of cholesterol level on age is $Y = 156.33 + 0.650X$.
• The predicted cholesterol level of a 60-year-old man is 195.32 mg/dL.

Step 1: Cost of Living Index Calculation (Family Budget Method)

Base year (2015) price $= P_0$; Current year (2016) price $= P_1$; Weight $= W$.

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

$$CLI = \frac{\sum IW}{\sum W} = \frac{10,765.35}{100} = \mathbf{107.65}$$

The cost of living in 2016 rose by 7.65% compared to 2015.

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Step 3: Evaluate Salary Adequacy

• Base Salary (2015): Rs. 20,000 per month
• Required Salary in 2016:
  $$\text{Required Salary} = 20,000 \times \left(\frac{107.65}{100}\right) = \mathbf{\text{Rs. } 21,530}$$
• Actual Salary Offered in 2016: Rs. 21,000

Adequacy Assessment: Since the offered salary of Rs. 21,000 is less than the required salary of Rs. 21,530, the increment is NOT adequate.

Required Additional Increment:

$$\text{Required Increment} = 21,530 - 20,000 = \text{Rs. } 1,530$$

Since the employee was only given Rs. 1,000 ($21,000 - 20,000$), they need an additional increment of Rs. 530 per month ($21,530 - 21,000$) to fully preserve their real standard of living.

Step 1: Boundary Equations of Constraints

1. Line 1: $3x + 5y = 45$

   • When $x = 0 \implies y = 9 \implies (0, 9)$
   • When $y = 0 \implies x = 15 \implies (15, 0)$
   • Test $(0, 0)$: $0 \ge 45$ is False (region is away from the origin).

2. Line 2: $2x + y = 20$

   • When $x = 0 \implies y = 20 \implies (0, 20)$
   • When $y = 0 \implies x = 10 \implies (10, 0)$
   • Test $(0, 0)$: $0 \ge 20$ is False (region is away from the origin).

3. Non-negativity: $x \ge 0, y \ge 0$ (First quadrant).

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Step 2: Find the Point of Intersection between Line 1 and Line 2 From Line 2: $y = 20 - 2x$. Substitute into Line 1:

$$3x + 5(20 - 2x) = 45$$

$$3x + 100 - 10x = 45 \implies -7x = -55 \implies x = \frac{55}{7} \approx 7.857$$

Then:

$$y = 20 - 2\left(\frac{55}{7}\right) = 20 - \frac{110}{7} = \frac{30}{7} \approx 4.286$$

Intersection point $B = \left(\frac{55}{7}, \frac{30}{7}\right)$.

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Step 3: Corner Points of the Feasible Region The feasible region is unbounded above, with extreme boundary corner points:

• $A = (0, 20)$ on the Y-axis (since $\max(20, 9) = 20$).
• $B = \left(\frac{55}{7}, \frac{30}{7}\right)$
• $C = (15, 0)$ on the X-axis (since $\max(10, 15) = 15$).

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Step 4: Evaluate Objective Function $Z = 20x + 30y$ at Corner Points

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Step 5: Verification of Minimum in Unbounded Region Since the region is unbounded, we check if the open half-plane $20x + 30y < 285.71$ has any points in common with the feasible region. Dividing by 10: $2x + 3y < 28.57$.

• At $A(0, 20)$: $2(0) + 3(20) = 60 > 28.57$
• At $C(15, 0)$: $2(15) + 3(0) = 30 > 28.57$ The line $20x + 30y = 285.71$ touches the boundary only at point $B$ and has no common points with the interior of the feasible region.

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Conclusion: The minimum cost is $Z_{min} = \frac{2000}{7} \approx \mathbf{\text{Rs. } 285.71}$, which occurs at $x = \frac{55}{7} \approx 7.86$ and $y = \frac{30}{7} \approx 4.29$.

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

Step 1: Calculate Coefficient Determinant ($D$)

$$D = \begin{vmatrix} 3 & 2 & 5 \\ 2 & -3 & 7 \\ 1 & 1 & 1 \end{vmatrix}$$

Expanding along Row 3 ($R_3$):

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

$$D = 1(14 - (-15)) - 1(21 - 10) + 1(-9 - 4)$$

$$D = 1(29) - 1(11) + 1(-13) = 29 - 11 - 13 = \mathbf{5}$$

Since $D = 5 \ne 0$, a unique solution exists.

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Step 2: Calculate $D_x$ (replace 1st column with $[10, 9, 5]^T$)

$$D_x = \begin{vmatrix} 10 & 2 & 5 \\ 9 & -3 & 7 \\ 5 & 1 & 1 \end{vmatrix}$$

Expanding along Row 3:

$$D_x = 5(14 - (-15)) - 1(70 - 45) + 1(-30 - 18)$$

$$D_x = 5(29) - 1(25) + 1(-48) = 145 - 25 - 48 = \mathbf{72}$$

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Step 3: Calculate $D_y$ (replace 2nd column with $[10, 9, 5]^T$)

$$D_y = \begin{vmatrix} 3 & 10 & 5 \\ 2 & 9 & 7 \\ 1 & 5 & 1 \end{vmatrix}$$

Expanding along Row 3:

$$D_y = 1(70 - 45) - 5(21 - 10) + 1(27 - 20)$$

$$D_y = 1(25) - 5(11) + 1(7) = 25 - 55 + 7 = \mathbf{-23}$$

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Step 4: Calculate $D_z$ (replace 3rd column with $[10, 9, 5]^T$)

$$D_z = \begin{vmatrix} 3 & 2 & 10 \\ 2 & -3 & 9 \\ 1 & 1 & 5 \end{vmatrix}$$

Expanding along Row 3:

$$D_z = 1(18 - (-30)) - 1(27 - 20) + 5(-9 - 4)$$

$$D_z = 1(48) - 1(7) + 5(-13) = 48 - 7 - 65 = \mathbf{-24}$$

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Step 5: Apply Cramer’s Rule

$$x = \frac{D_x}{D} = \frac{72}{5} = \mathbf{14.4}$$

$$y = \frac{D_y}{D} = \frac{-23}{5} = \mathbf{-4.6}$$

$$z = \frac{D_z}{D} = \frac{-24}{5} = \mathbf{-4.8}$$

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Verification: Substitute into Equation 3:

$$x + y + z = 14.4 + (-4.6) + (-4.8) = 14.4 - 9.4 = \mathbf{5} \quad \text{(Matches perfectly)}$$

Step 1: Economic Parameters

• Cost Price ($C$) = Rs. 10 per dozen
• Selling Price ($S$) = Rs. 30 per dozen
• Salvage Value for unsold roses = Rs. 0 (donated to hospital)
• Profit per dozen sold = $S - C = 30 - 10 = \text{Rs. } 20$
• Loss per unsold dozen = $C = \text{Rs. } 10$

Payoff Formula: Let $S_i$ be the quantity stocked and $D_j$ be the quantity demanded:

$$\text{Payoff} = \begin{cases} 20 \times D_j - 10 \times (S_i - D_j) = 30D_j - 10S_i, & \text{if } D_j < S_i \\ 20 \times S_i, & \text{if } D_j \ge S_i \end{cases}$$

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Step 2: Conditional Payoff Table (in Rs.)

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Step 3: Compute Expected Monetary Value ($EMV$) for Each Strategy

$$EMV(S_i) = \sum [\text{Payoff}(S_i, D_j) \times P(D_j)]$$

1. $EMV(\text{Stock 70}):$

   $$1400(0.1) + 1400(0.2) + 1400(0.4) + 1400(0.3) = \mathbf{\text{Rs. } 1,400}$$

2. $EMV(\text{Stock 80}):$

   $$1300(0.1) + 1600(0.2) + 1600(0.4) + 1600(0.3) = 130 + 320 + 640 + 480 = \mathbf{\text{Rs. } 1,570}$$

3. $EMV(\text{Stock 90}):$

   $$1200(0.1) + 1500(0.2) + 1800(0.4) + 1800(0.3) = 120 + 300 + 720 + 540 = \mathbf{\text{Rs. } 1,680} \quad (\text{Maximum})$$

4. $EMV(\text{Stock 100}):$

   $$1100(0.1) + 1400(0.2) + 1700(0.4) + 2000(0.3) = 110 + 280 + 680 + 600 = \mathbf{\text{Rs. } 1,670}$$

Optimal Decision: Stock 90 dozens of roses with the highest expected profit of $EMV^* = \text{Rs. } 1,680$.

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Step 4: Compute Expected Profit with Perfect Information ($EPPI$)

Under perfect information, the manager would always stock the exact quantity demanded:

$$\text{Payoff under certainty} = 20 \times D_j$$

$$EPPI = \mathbf{\text{Rs. } 1,780}$$

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Step 5: Compute Expected Value of Perfect Information ($EVPI$)

$$EVPI = EPPI - EMV^* = 1,780 - 1,680 = \mathbf{\text{Rs. } 100}$$

Interpretation: The manager should be willing to pay at most Rs. 100 per day for complete, 100% accurate market intelligence regarding rose demand.

Step 1: Set up Least Squares Calculation Table

Number of years $n = 7$ (odd). Middle year $2014$ is chosen as origin ($A = 2014$). Step deviation $X = \text{Year} - 2014$, so $\sum X = 0$.

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Part (i): Fit the Straight Line Trend Equation

Linear trend equation:

$$Y_c = a + bX$$

Since $\sum X = 0$:

$$a = \frac{\sum Y}{n} = \frac{623}{7} = \mathbf{89}$$

$$b = \frac{\sum XY}{\sum X^2} = \frac{56}{28} = \mathbf{2}$$

(Using exact integers: $b = 2.00$ exactly!)

The fitted trend line equation is:

$$Y_c = 89 + 2X \quad (\text{Origin: 2014, } X \text{ in 1-year units, } Y \text{ in '000 tons})$$

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Part (ii): Trend Values

Recalculated with exact $b = 2$:

• 2011: $89 + 2(-3) = \mathbf{83}$
• 2012: $89 + 2(-2) = \mathbf{85}$
• 2013: $89 + 2(-1) = \mathbf{87}$
• 2014: $89 + 2(0) = \mathbf{89}$
• 2015: $89 + 2(1) = \mathbf{91}$
• 2016: $89 + 2(2) = \mathbf{93}$
• 2017: $89 + 2(3) = \mathbf{95}$

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Part (iii): Description of Graph Plotting

• Horizontal Axis (X-axis): Represent years from 2011 to 2017 with uniform spacing.
• Vertical Axis (Y-axis): Represent sugar production in thousand tons (scale from 70 to 105).
• Actual Line: Plot the actual points $(2011, 77), (2012, 88), \dots, (2017, 90)$ and join them with straight line segments (zigzag actual production curve).
• Trend Line: Plot the points $(2011, 83)$ and $(2017, 95)$ and connect them with a single straight line through the center to illustrate the upward secular growth.

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Part (iv): Monthly Increase in Production

The annual rate of increase in production is $b = 2$ thousand tons per year.

$$\text{Monthly Increase} = \frac{b}{12} = \frac{2}{12} = \frac{1}{6} \approx \mathbf{0.1667} \text{ thousand tons} = \mathbf{166.67} \text{ tons per month}$$

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Part (v): Estimate Production for the Year 2020

For Year $2020$:

$$X = 2020 - 2014 = 6$$

$$Y_{2020} = 89 + 2(6) = 89 + 12 = \mathbf{101} \text{ thousand tons}$$

Conclusion: Estimated sugar production for 2020 is 101,000 tons.

Step 1: Identification of Extremes and Class Structure

• Total observations ($N$) = $50$
• Minimum value = $28$
• Maximum value = $69$
• Given class intervals: $25 - 34, \quad 35 - 44, \quad 45 - 54, \quad 55 - 64, \quad 65 - 74$
• This is an inclusive class interval system with class width $h = 10$.

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Step 2: Tally Marks and Frequency Count

1. Class 25 - 34:

   • Observations: $28, 29, 33, 31, 32$
   • Count = 5

2. Class 35 - 44:

   • Observations: $35, 36, 40, 42, 41, 37, 35, 40, 38, 39, 42, 38, 44, 39, 38, 44, 38, 44$
   • Count = 18

3. Class 45 - 54:

   • Observations: $48, 50, 48, 49, 51, 51, 54, 46, 45, 45, 47, 47, 47$
   • Count = 13

4. Class 55 - 64:

   • Observations: $61, 59, 62, 63, 60, 56, 61, 59, 58, 57, 64$
   • Count = 11

5. Class 65 - 74:

   • Observations: $67, 69, 69$
   • Count = 3

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Step 3: Complete Frequency Distribution Table

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Interpretation:

• The modal profit group is Rs. 35 to 44 lakhs, containing $36\%$ of the firms ($18$ firms).
• $72\%$ of the firms ($36$ out of $50$) earn an annual profit of less than Rs. 54.5 lakhs.