Dean's Office Official Model Question Paper

Solution:

We are given:

• $N_1 = 40, \quad \bar{X}_1 = 25$
• $N_2 = 60, \quad \bar{X}_2 = 35$

The formula for the combined arithmetic mean ($\bar{X}_{12}$) is:

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

Substituting the given values:

$$\bar{X}_{12} = \frac{(40 \times 25) + (60 \times 35)}{40 + 60} = \frac{1{,}000 + 2{,}100}{100} = \frac{3{,}100}{100} = \mathbf{31}$$

Thus, the combined mean of the combined group is 31.

Solution:

Let $Y = aX + b$, where $a = 3$ and $b = 15$. Given: $\sigma_X = 6$.

By the properties of standard deviation under linear transformation:

• Standard deviation is independent of change of origin ($b$), but dependent on change of scale ($|a|$):

  $$\sigma_Y = |a| \cdot \sigma_X = |3| \times 6 = \mathbf{18}$$

• Variance is the square of standard deviation:

  $$\text{Var}(Y) = \sigma_Y^2 = 18^2 = \mathbf{324}$$

Thus, $\sigma_Y = \mathbf{18}$ and $\text{Var}(Y) = \mathbf{324}$.

Solution:

Given:

• $Q_1 = 24, \quad M_d = 30, \quad Q_3 = 40$

Bowley’s coefficient of skewness ($S_b$) is given by:

$$S_b = \frac{Q_3 + Q_1 - 2M_d}{Q_3 - Q_1}$$

Substituting the values:

$$S_b = \frac{40 + 24 - 2(30)}{40 - 24} = \frac{64 - 60}{16} = \frac{4}{16} = \mathbf{+0.25}$$

Interpretation: Since $S_b = +0.25 > 0$, the distribution is positively skewed (skewed to the right).

Solution:

Given:

• $b_{yx} = -0.8$
• $b_{xy} = -0.45$

The correlation coefficient ($r$) is the geometric mean of the two regression coefficients, carrying their common sign:

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

Since both regression coefficients are negative, $r$ must also be negative:

$$r = -\sqrt{(-0.8) \times (-0.45)} = -\sqrt{0.36} = \mathbf{-0.60}$$

Thus, the correlation coefficient between $X$ and $Y$ is $-0.60$ (moderate negative correlation).

Answer:

Fisher’s Ideal Index Number is the geometric mean of Laspeyres’ index ($L$) and Paasche’s index ($P$):

$$P_{01}^F = \sqrt{L \times P} = \sqrt{\frac{\sum p_1 q_0}{\sum p_0 q_0} \times \frac{\sum p_1 q_1}{\sum p_0 q_1}} \times 100$$

Why it is called ‘Ideal’:

1. Geometric Mean: It utilizes the geometric mean, which is mathematically the most suitable average for ratios and index numbers.
2. Dual Weighting: It incorporates both base year quantities ($q_0$) and current year quantities ($q_1$), avoiding the upward bias of Laspeyres and downward bias of Paasche.
3. Fulfills Consistency Tests: It satisfies both the Time Reversal Test ($P_{01} \times P_{10} = 1$) and the Factor Reversal Test ($P_{01} \times Q_{01} = V_{01}$).

Answer:

Kurtosis refers to the degree of peakedness or flatness of a unimodal frequency distribution relative to a normal distribution.

Based on Pearson’s coefficient of kurtosis $\beta_2 = \frac{\mu_4}{\mu_2^2}$:

1. Leptokurtic ($\beta_2 > 3$ or $\gamma_2 > 0$): A distribution that is more peaked with fatter tails than the normal curve.
2. Mesokurtic ($\beta_2 = 3$ or $\gamma_2 = 0$): A distribution having normal peakedness (identical to the standard Gaussian/normal distribution).
3. Platykurtic ($\beta_2 < 3$ or $\gamma_2 < 0$): A distribution that is flatter-topped with thinner tails than the normal curve.

Solution:

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

• Let $A$ be the event of drawing a King: $n(A) = 4 \implies P(A) = \frac{4}{52}$.
• Let $B$ be the event of drawing a Heart: $n(B) = 13 \implies P(B) = \frac{13}{52}$.
• Event $(A \cap B)$ is drawing the King of Hearts: $n(A \cap B) = 1 \implies P(A \cap B) = \frac{1}{52}$.

By the Addition Theorem of Probability:

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

$$P(A \cup B) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{\mathbf{4}}{\mathbf{13}} \approx \mathbf{0.3077}$$

The probability of drawing a King or a Heart is $\frac{4}{13}$.

Answer:

A business and economic time series comprises four classical components:

1. Secular Trend ($T$): The long-term, smooth, and persistent underlying direction (growth or decline) over an extended period.
2. Seasonal Variations ($S$): Periodic, repetitive fluctuations that occur within a single year at regular intervals (e.g., quarterly, monthly, festive peaks).
3. Cyclical Fluctuations ($C$): Wave-like oscillatory movements recurring over multi-year cycles, commonly linked to macroeconomic business cycles (prosperity, recession, depression, recovery).
4. Irregular or Random Fluctuations ($I$): Unpredictable, unsystematic residual variations resulting from extraordinary events (e.g., natural disasters, wars, strikes, pandemics).

Answer:

Key characteristics of the Normal Probability Distribution:

1. Bell-shaped and Symmetrical: The curve is perfectly symmetrical around the mean ($\mu$), with Skewness = 0.
2. Equality of Averages: The Mean, Median, and Mode coincide at the center: $\text{Mean} = \text{Median} = \text{Mode} = \mu$.
3. Asymptotic Tails: The curve approaches the horizontal axis asymptotically on both sides, extending to $\pm \infty$ without touching the axis.
4. Total Area: The total area under the normal curve is equal to 1 (or 100%), with 50% lying on either side of $\mu$.
5. Empirical Rule: Approximately 68.27% of data falls within $\mu \pm 1\sigma$, 95.45% within $\mu \pm 2\sigma$, and 99.73% within $\mu \pm 3\sigma$.

Answer:

Solution:

To compute Karl Pearson’s coefficient of skewness:

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

Let assumed mean $A = 170$, common class interval $h = 20$. Let $d' = \frac{m - A}{h} = \frac{m - 170}{20}$.

Calculation Table:

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Step 1: Calculation of Arithmetic Mean ($\bar{X}$)

$$\bar{X} = A + \left(\frac{\sum f d'}{N}\right) \times h = 170 + \left(\frac{-3}{100}\right) \times 20 = 170 - 0.6 = \mathbf{169.40 \text{ Rs.}}$$

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Step 2: Calculation of Mode ($M_o$)

The maximum frequency is $f_1 = 25$, corresponding to the modal class 160–180. Here:

• Lower limit $L = 160$
• Frequency of modal class $f_1 = 25$
• Pre-modal frequency $f_0 = 18$
• Post-modal frequency $f_2 = 16$
• Class width $h = 20$$M_o = L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$$
  $$M_o = 160 + \frac{25 - 18}{2(25) - 18 - 16} \times 20 = 160 + \frac{7}{50 - 34} \times 20 = 160 + \frac{140}{16} = 160 + 8.75 = \mathbf{168.75 \text{ Rs.}}$$

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Step 3: Calculation of Standard Deviation ($\sigma$)

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

$$\sigma = 20 \times \sqrt{\frac{273}{100} - \left(\frac{-3}{100}\right)^2} = 20 \times \sqrt{2.73 - 0.0009} = 20 \times \sqrt{2.7291} = 20 \times 1.6520 = \mathbf{33.04 \text{ Rs.}}$$

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Step 4: Karl Pearson’s Coefficient of Skewness ($S_k$)

$$S_k = \frac{\bar{X} - M_o}{\sigma} = \frac{169.40 - 168.75}{33.04} = \frac{0.65}{33.04} \approx \mathbf{+0.0197}$$

Interpretation: The coefficient of skewness is $+0.02$, indicating that the wage distribution is very slightly positively skewed (almost symmetrical), with a minor concentration of frequencies toward the lower wage brackets and an extremely mild tail extending toward higher wages.

Solution:

Let $X$ = Advertising Expenditure (Lakh Rs.) and $Y$ = Sales Volume (Crore Rs.). Number of observations $n = 6$.

Calculation Table:

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Part (a): Karl Pearson’s Correlation Coefficient ($r$)

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

Numerator:

$$n \sum XY - (\sum X)(\sum Y) = 6(4001) - (100)(222) = 24{,}006 - 22{,}200 = 1{,}806$$

Denominator:

$$n \sum X^2 - (\sum X)^2 = 6(1818) - (100)^2 = 10{,}908 - 10{,}000 = 908$$

$$n \sum Y^2 - (\sum Y)^2 = 6(8818) - (222)^2 = 52{,}908 - 49{,}284 = 3{,}624$$

$$\sqrt{908 \times 3624} = \sqrt{3{,}290{,}592} \approx 1{,}814.00$$

$$r = \frac{1{,}806}{1{,}814.00} \approx \mathbf{+0.9956}$$

Interpretation: There exists a very high degree of positive linear correlation ($r \approx 0.996$) between advertising expenditure and sales volume.

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Part (b): Regression Equation of $Y$ on $X$

The regression line of $Y$ on $X$ is:

$$Y - \bar{Y} = b_{yx} (X - \bar{X})$$

Where:

$$\bar{X} = \frac{\sum X}{n} = \frac{100}{6} = 16.67$$

$$\bar{Y} = \frac{\sum Y}{n} = \frac{222}{6} = 37.00$$

$$b_{yx} = \frac{n \sum XY - (\sum X)(\sum Y)}{n \sum X^2 - (\sum X)^2} = \frac{1{,}806}{908} \approx \mathbf{1.989}$$

Regression equation:

$$Y - 37 = 1.989(X - 16.67)$$

$$Y = 1.989X - (1.989 \times 16.67) + 37 = 1.989X - 33.15 + 37$$

$$\mathbf{Y = 3.85 + 1.989X}$$

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Part (c): Estimation of Sales for $X = 30$ Lakh Rs.

Substituting $X = 30$ into the regression model:

$$\hat{Y} = 3.85 + 1.989(30) = 3.85 + 59.67 = \mathbf{63.52 \text{ Crore Rs.}}$$

The estimated sales volume is Rs. 63.52 Crore.

Solution:

Calculation Table:

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Step 1: Fisher’s Ideal Index Number ($P_{01}^F$)

$$P_{01}^F = \sqrt{\frac{\sum p_1 q_0}{\sum p_0 q_0} \times \frac{\sum p_1 q_1}{\sum p_0 q_1}} \times 100$$

$$P_{01}^F = \sqrt{\frac{1600}{1165} \times \frac{1830}{1324}} \times 100$$

$$P_{01}^F = \sqrt{1.37339 \times 1.38217} \times 100 = \sqrt{1.89826} \times 100 = 1.37777 \times 100 \approx \mathbf{137.78}$$

Interpretation: The price level of commodities in 2080 increased by 37.78% compared to the base year 2075.

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Step 2: Verification of Time Reversal Test

The Time Reversal Test requires:

$$P_{01} \times P_{10} = 1$$

Here:

$$P_{01} = \sqrt{\frac{\sum p_1 q_0}{\sum p_0 q_0} \times \frac{\sum p_1 q_1}{\sum p_0 q_1}} = \sqrt{\frac{1600}{1165} \times \frac{1830}{1324}}$$

Interchanging 0 and 1:

$$P_{10} = \sqrt{\frac{\sum p_0 q_1}{\sum p_1 q_1} \times \frac{\sum p_0 q_0}{\sum p_1 q_0}} = \sqrt{\frac{1324}{1830} \times \frac{1165}{1600}}$$

Now:

$$P_{01} \times P_{10} = \sqrt{\left(\frac{1600}{1165} \times \frac{1830}{1324}\right) \times \left(\frac{1324}{1830} \times \frac{1165}{1600}\right)} = \sqrt{1} = \mathbf{1}$$

Since $P_{01} \times P_{10} = 1$, Fisher’s Ideal Index satisfies the Time Reversal Test.

Solution:

Number of years $N = 7$ (odd). Let middle year $2077$ be origin ($x = \text{Year} - 2077$). Straight-line trend equation:

$$Y = a + bX$$

Since $\sum X = 0$:

$$a = \frac{\sum Y}{N}, \quad b = \frac{\sum XY}{\sum X^2}$$

Calculation Table:

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Step 1: Compute Parameters $a$ and $b$$a = \frac{462}{7} = \mathbf{66.00}$$

$$b = \frac{207}{28} \approx \mathbf{7.4286}$$

Straight-Line Trend Equation:

$$\mathbf{Y = 66.00 + 7.43 X} \quad (\text{Origin: 2077, } X \text{ unit = 1 year})$$

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Step 2: Forecast for 2081 and 2082

• For Year 2081: $X = 2081 - 2077 = +4$$\hat{Y}_{2081} = 66.00 + 7.4286(4) = 66.00 + 29.71 = \mathbf{95.71 \text{ Million Rs.}}$$

• For Year 2082: $X = 2082 - 2077 = +5$$\hat{Y}_{2082} = 66.00 + 7.4286(5) = 66.00 + 37.14 = \mathbf{103.14 \text{ Million Rs.}}$$

The projected turnovers are Rs. 95.71 Million for 2081 and Rs. 103.14 Million for 2082.

Solution:

Let $E_1, E_2, E_3$ denote the events that a component is manufactured by Machine $M_1, M_2,$ and $M_3$ respectively. Let $D$ denote the event that a component is defective.

Given prior probabilities:

• $P(E_1) = 0.30$
• $P(E_2) = 0.45$
• $P(E_3) = 0.25$

Given conditional probabilities of defectiveness:

• $P(D|E_1) = 0.02$
• $P(D|E_2) = 0.03$
• $P(D|E_3) = 0.04$

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Part (a): Total Probability of a Defective Component $P(D)$

By the Theorem of Total Probability:

$$P(D) = P(E_1)P(D|E_1) + P(E_2)P(D|E_2) + P(E_3)P(D|E_3)$$

$$P(D) = (0.30 \times 0.02) + (0.45 \times 0.03) + (0.25 \times 0.04)$$

$$P(D) = 0.0060 + 0.0135 + 0.0100 = \mathbf{0.0295} \quad (\text{or } \mathbf{2.95\%})$$

The probability that a randomly selected component is defective is 0.0295 (2.95%).

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Part (b): Posterior Probability $P(E_2|D)$ using Bayes’ Theorem

By Bayes’ Theorem:

$$P(E_2|D) = \frac{P(E_2) \cdot P(D|E_2)}{P(D)}$$

$$P(E_2|D) = \frac{0.45 \times 0.03}{0.0295} = \frac{0.0135}{0.0295} = \frac{135}{295} \approx \mathbf{0.4576} \quad (\text{or } \mathbf{45.76\%})$$

Given that the component is defective, the probability that it was manufactured by Machine $M_2$ is 0.4576 (45.76%).

Solution:

Given:

• Team Alpha: $N_1 = 50, \quad \bar{X}_1 = 120, \quad \sigma_1 = 15$
• Team Beta: $N_2 = 60, \quad \bar{X}_2 = 115, \quad \sigma_2 = 18$

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Part (a): Comparison of Total Monthly Sales Revenue

$$\text{Total Sales} = N \times \bar{X}$$

• Team Alpha: $50 \times 120 = \mathbf{6{,}000 \text{ Lakh Rs.}}$ (Rs. 60 Crore)
• Team Beta: $60 \times 115 = \mathbf{6{,}900 \text{ Lakh Rs.}}$ (Rs. 69 Crore)

Conclusion: Team Beta generates greater total monthly sales revenue (Rs. 6,900 Lakh vs Rs. 6,000 Lakh).

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Part (b): Comparison of Consistency (Coefficient of Variation)

Consistency/uniformity is measured by the Coefficient of Variation ($CV$):

$$CV = \frac{\sigma}{\bar{X}} \times 100\%$$

• For Team Alpha:

  $$CV_1 = \frac{15}{120} \times 100\% = 0.125 \times 100\% = \mathbf{12.50\%}$$

• For Team Beta:

  $$CV_2 = \frac{18}{115} \times 100\% = 0.15652 \times 100\% = \mathbf{15.65\%}$$

Interpretation: A lower $CV$ indicates greater consistency and less relative variability. Since $CV_1 (12.50\%) < CV_2 (15.65\%)$, Team Alpha exhibits greater consistency and uniformity in sales performance compared to Team Beta.

Solution:

Let $X$ denote the scholarship test scores. $X \sim N(\mu = 65, \sigma = 12)$, Total students $N = 1{,}000$. Standard normal variate: $Z = \frac{X - \mu}{\sigma} = \frac{X - 65}{12}$.

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Part (a): Percentage of students scoring above 80 marks ($X > 80$)

For $X = 80$:

$$Z = \frac{80 - 65}{12} = \frac{15}{12} = +1.25$$

$$P(X > 80) = P(Z > 1.25) = 0.5 - P(0 < Z < 1.25)$$

From standard normal area tables, area between $Z = 0$ and $Z = 1.25$ is $0.3944$.

$$P(Z > 1.25) = 0.5000 - 0.3944 = \mathbf{0.1056} \quad (\text{or } \mathbf{10.56\%})$$

Result: 10.56% of students scored above 80 marks (approximately 106 students).

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Part (b): Number of students scoring between 50 and 75 marks ($50 \le X \le 75$)

• For $X_1 = 50$:
  $$Z_1 = \frac{50 - 65}{12} = \frac{-15}{12} = -1.25$$
• For $X_2 = 75$:
  $$Z_2 = \frac{75 - 65}{12} = \frac{10}{12} = +0.833 \approx +0.83$$

$$P(50 \le X \le 75) = P(-1.25 \le Z \le 0.83) = P(-1.25 \le Z \le 0) + P(0 \le Z \le 0.83)$$

From normal tables:

• Area for $Z = 1.25$ is $0.3944$
• Area for $Z = 0.83$ is $0.2967$$P(50 \le X \le 75) = 0.3944 + 0.2967 = \mathbf{0.6911}$$

Expected Number of Students:

$$\text{Number} = N \times P(50 \le X \le 75) = 1{,}000 \times 0.6911 \approx \mathbf{691 \text{ students}}$$

Approximately 691 students scored between 50 and 75 marks.

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Part (c): Minimum score for Top 5% Merit Scholarship

Top 5% implies area in the right tail is $0.05$. Therefore, area between $Z = 0$ and critical value $Z_c$ is:

$$\text{Area} = 0.5000 - 0.0500 = 0.4500$$

From standard normal tables, the $Z$-value corresponding to an area of $0.4500$ is:

$$Z_c = +1.645$$

Using the transformation $X = \mu + Z \sigma$:

$$X_c = 65 + (1.645 \times 12) = 65 + 19.74 = \mathbf{84.74 \text{ marks}}$$

Result: A candidate must score a minimum of 84.74 marks (or at least 85 marks) to qualify for the top 5% merit scholarship awards.

Solution:

Let $X$ = Output units and $Y$ = Total Cost (Thousand Rs.). $n = 8$.

Calculation Table:

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Step 1: Means and Regression Coefficients

$$\bar{X} = \frac{\sum X}{n} = \frac{440}{8} = 55.00$$

$$\bar{Y} = \frac{\sum Y}{n} = \frac{1560}{8} = 195.00$$

$$b_{yx} = \frac{n \sum XY - (\sum X)(\sum Y)}{n \sum X^2 - (\sum X)^2} = \frac{8(98100) - (440)(1560)}{8(28400) - (440)^2}$$

$$b_{yx} = \frac{784{,}800 - 686{,}400}{227{,}200 - 193{,}600} = \frac{98{,}400}{33{,}600} \approx \mathbf{2.9286}$$

$$b_{xy} = \frac{n \sum XY - (\sum X)(\sum Y)}{n \sum Y^2 - (\sum Y)^2} = \frac{98{,}400}{8(340600) - (1560)^2} = \frac{98{,}400}{2{,}724{,}800 - 2{,}433{,}600} = \frac{98{,}400}{291{,}200} \approx \mathbf{0.3379}$$

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Step 2: Part (a) Regression Equations

• Regression Equation of $Y$ on $X$:

  $$Y - \bar{Y} = b_{yx}(X - \bar{X}) \implies Y - 195 = 2.9286(X - 55)$$
  $$Y = 2.9286X - 161.07 + 195 \implies \mathbf{Y = 33.93 + 2.9286 X}$$

• Regression Equation of $X$ on $Y$:

  $$X - \bar{X} = b_{xy}(Y - \bar{Y}) \implies X - 55 = 0.3379(Y - 195)$$
  $$X = 0.3379Y - 65.89 + 55 \implies \mathbf{X = -10.89 + 0.3379 Y}$$

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Step 3: Part (b) Coefficient of Determination ($r^2$)

$$r^2 = b_{yx} \times b_{xy} = 2.9286 \times 0.3379 \approx \mathbf{0.9896} \quad (\text{or } \mathbf{98.96\%})$$

Significance: $r^2 = 0.9896$ signifies that 98.96% of the total variation in production costs is explained by variations in output volume, indicating an exceptionally robust, near-perfect predictive fit. Only 1.04% of variation is attributable to random unexplained factors.

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Step 4: Part (c) Predictions

1. Cost for Output $X = 110$ units:

   $$\hat{Y} = 33.93 + 2.9286(110) = 33.93 + 322.15 = \mathbf{356.08 \text{ Thousand Rs.}}$$

2. Output for Budget $Y = 220$ Thousand Rs.:

   $$\hat{X} = -10.89 + 0.3379(220) = -10.89 + 74.34 = \mathbf{63.45 \approx 63 \text{ units}}$$

Solution:

Let mid-point $m$:

• 0–10: $m = 5$
• 10–20: $m = 15$
• 20–30: $m = 25$
• 30–40: $m = 35$
• 40–50: $m = 45$
• 50–60: $m = 55$

Let assumed mean $A = 25$, step length $h = 10$. $d' = \frac{m - 25}{10}$.

Calculation Table:

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Step 1: Mean and Standard Deviation for Plant A

$$\bar{X}_A = A + \left(\frac{\sum f_A d'}{N_A}\right) \times h = 25 + \left(\frac{13}{100}\right) \times 10 = 25 + 1.3 = \mathbf{26.30 \text{ hours}}$$

$$\sigma_A = h \times \sqrt{\frac{\sum f_A (d')^2}{N_A} - \left(\frac{\sum f_A d'}{N_A}\right)^2} = 10 \times \sqrt{\frac{171}{100} - (0.13)^2}$$

$$\sigma_A = 10 \times \sqrt{1.71 - 0.0169} = 10 \times \sqrt{1.6931} = 10 \times 1.3012 = \mathbf{13.01 \text{ hours}}$$

$$CV_A = \frac{\sigma_A}{\bar{X}_A} \times 100\% = \frac{13.01}{26.30} \times 100\% = \mathbf{49.47\%}$$

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Step 2: Mean and Standard Deviation for Plant B

$$\bar{X}_B = A + \left(\frac{\sum f_B d'}{N_B}\right) \times h = 25 + \left(\frac{49}{100}\right) \times 10 = 25 + 4.9 = \mathbf{29.90 \text{ hours}}$$

$$\sigma_B = h \times \sqrt{\frac{\sum f_B (d')^2}{N_B} - \left(\frac{\sum f_B d'}{N_B}\right)^2} = 10 \times \sqrt{\frac{205}{100} - (0.49)^2}$$

$$\sigma_B = 10 \times \sqrt{2.05 - 0.2401} = 10 \times \sqrt{1.8099} = 10 \times 1.3453 = \mathbf{13.45 \text{ hours}}$$

$$CV_B = \frac{\sigma_B}{\bar{X}_B} \times 100\% = \frac{13.45}{29.90} \times 100\% = \mathbf{44.98\%}$$

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Step 3: Evaluations

• Part (a): Higher Average Overtime: Since $\bar{X}_B (29.90 \text{ hrs}) > \bar{X}_A (26.30 \text{ hrs})$, Plant B has a higher average overtime workload.
• Part (b): Greater Variability: Since $CV_A (49.47\%) > CV_B (44.98\%)$, Plant A exhibits greater relative variability (less uniformity) in its overtime distribution.

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Step 4: Part (c) Combined Standard Deviation ($\sigma_{12}$)

Combined Mean:

$$\bar{X}_{12} = \frac{N_A \bar{X}_A + N_B \bar{X}_B}{N_A + N_B} = \frac{(100 \times 26.30) + (100 \times 29.90)}{200} = \frac{5{,}620}{200} = \mathbf{28.10 \text{ hours}}$$

Deviations:

• $d_1 = \bar{X}_A - \bar{X}_{12} = 26.30 - 28.10 = -1.80 \implies d_1^2 = 3.24$
• $d_2 = \bar{X}_B - \bar{X}_{12} = 29.90 - 28.10 = +1.80 \implies d_2^2 = 3.24$

Combined Variance ($\sigma_{12}^2$):

$$\sigma_{12}^2 = \frac{N_A (\sigma_A^2 + d_1^2) + N_B (\sigma_B^2 + d_2^2)}{N_A + N_B}$$

$$\sigma_A^2 = (13.01)^2 = 169.26 \implies \sigma_A^2 + d_1^2 = 169.26 + 3.24 = 172.50$$

$$\sigma_B^2 = (13.45)^2 = 180.90 \implies \sigma_B^2 + d_2^2 = 180.90 + 3.24 = 184.14$$

$$\sigma_{12}^2 = \frac{100(172.50) + 100(184.14)}{200} = \frac{17{,}250 + 18{,}414}{200} = \frac{35{,}664}{200} = 178.32$$

$$\sigma_{12} = \sqrt{178.32} \approx \mathbf{13.35 \text{ hours}}$$

The combined standard deviation across both plants is 13.35 hours.