Business Statistics 2023 Board Question Paper

Karl Pearson’s Coefficient of Correlation ($r$):

Given:

• $\text{Cov}(X, Y) = 30$
• $\text{Var}(X) = \sigma_X^2 = 25 \implies \sigma_X = \sqrt{25} = 5$
• $\text{Var}(Y) = \sigma_Y^2 = 64 \implies \sigma_Y = \sqrt{64} = 8$

Formula:

$$r = \frac{\text{Cov}(X, Y)}{\sigma_X \cdot \sigma_Y}$$

Calculation:

$$r = \frac{30}{5 \times 8} = \frac{30}{40} = \mathbf{+0.75}$$

Final Answer: Karl Pearson’s coefficient of correlation between $X$ and $Y$ is +0.75 (moderate to high positive correlation).

Coefficient of Quartile Deviation:

Given:

• Upper Quartile ($Q_3$) = $63$
• Lower Quartile ($Q_1$) = $59$

Formula:

$$\text{Coefficient of } QD = \frac{Q_3 - Q_1}{Q_3 + Q_1}$$

Calculation:

$$\text{Coefficient of } QD = \frac{63 - 59}{63 + 59} = \frac{4}{122} \approx \mathbf{0.0328} \quad (\text{or } 3.28\%)$$

Final Answer: The coefficient of quartile deviation is 0.0328.

Mean Values of $X$ and $Y$ from Regression Equations:

Property:

Both regression lines intersect at the point of their arithmetic means $(\bar{X}, \bar{Y})$.

Given Equations:

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

Solution:

Multiply Equation (1) by 5:

$$20\bar{X} - 25\bar{Y} = -165 \quad \text{--- (3)}$$

Subtract Equation (3) 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):

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

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

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

Calculation of Probability $P(A \cup B)$:

Given:

• $P(A) = 0.4$
• $P(B) = 0.5$
• $P(A \cap B) = 0.7$ (Note: In formal probability theory, $P(A \cap B)$ cannot exceed $P(A)$ or $P(B)$. This is an examination typo where $P(A \cup B) = 0.7$ was intended).

Method 1: Direct Formula Application

By the Addition Theorem of Probability:

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

$$P(A \cup B) = 0.4 + 0.5 - 0.7 = \mathbf{0.20}$$

Method 2: Typo-Corrected Interpretation

If the problem intended $P(A \cup B) = 0.7$ and asked to find $P(A \cap B)$:

$$P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.4 + 0.5 - 0.7 = \mathbf{0.20}$$

Final Answer: Using the standard addition rule formula, $P(A \cup B) = 0.20$.

Poisson Probability $P(X = 4)$:

Given:

• Mean parameter $\lambda = 1.5$
• Number of occurrences $x = 4$

Poisson PMF Formula:

$$P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!}$$

Step-by-Step Calculation:

• $e^{-1.5} \approx 0.22313$
• $(1.5)^4 = 5.0625$
• $4! = 4 \times 3 \times 2 \times 1 = 24$$P(X = 4) = \frac{0.22313 \times 5.0625}{24} = \frac{1.1296}{24} \approx \mathbf{0.0471}$$

Final Answer: $P(X = 4) =$ 0.0471 (or 4.71%).

Standard Error of the Mean with Finite Population Correction (FPC):

Given:

• Population size ($N$) = $1000$
• Sample size ($n$) = $100$
• Standard deviation ($\sigma$) = $10$

Sampling Fraction:

$$\frac{n}{N} = \frac{100}{1000} = 0.10 \quad (10\% > 5\%), \text{ so FPC is applied.}$$

Formula:

$$SE_{\bar{X}} = \frac{\sigma}{\sqrt{n}} \times \sqrt{\frac{N - n}{N - 1}}$$

Calculation:

$$\frac{\sigma}{\sqrt{n}} = \frac{10}{\sqrt{100}} = \frac{10}{10} = 1.0$$

$$\text{FPC} = \sqrt{\frac{1000 - 100}{1000 - 1}} = \sqrt{\frac{900}{999}} = \sqrt{0.9009} \approx 0.94916$$

$$SE_{\bar{X}} = 1.0 \times 0.94916 \approx \mathbf{0.9492}$$

(Without FPC: $SE = \frac{\sigma}{\sqrt{n}} = 1.0$).

Final Answer: The Standard Error of the mean is 0.9492.

Calculation of Median from Empirical Relationship:

Given:

• $\text{Mean} = 20$
• $\text{Mode} = 23$

Empirical Formula for Moderately Asymmetric Distributions:

$$\text{Mode} = 3\text{Median} - 2\text{Mean}$$

Calculation:

$$23 = 3\text{Median} - 2(20)$$

$$23 = 3\text{Median} - 40$$

$$3\text{Median} = 23 + 40$$

$$3\text{Median} = 63 \implies \text{Median} = \frac{63}{3} = \mathbf{21}$$

Final Answer: The value of the median is 21.

Calculation of Coefficient of Variation (CV):

Given:

• $\text{Mean } (\bar{X}) = 32$
• $\text{Standard Deviation } (\sigma) = 17$

Formula:

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

Calculation:

$$CV = \frac{17}{32} \times 100 = \mathbf{53.125\%}$$

Final Answer: The Coefficient of Variation is 53.125%.

Coefficient of Determination and Interpretation:

Given:

• Correlation coefficient ($r$) = $0.84$

Calculation of Coefficient of Determination ($r^2$):

$$r^2 = (0.84)^2 = \mathbf{0.7056} \quad (\text{or } 70.56\%)$$

Interpretation:

• 70.56% of the total variation in the dependent variable is explained by or accounted for by the variation in the independent variable through the linear relationship.
• The remaining 29.44% of the total variation remains unexplained, attributable to other omitted variables or random error.

Four Types of Random (Probability) Sampling Techniques:

1. Simple Random Sampling (SRS): Every element in the population has an equal, known, and independent chance of selection (e.g., lottery method or random number generator).
2. Stratified Random Sampling: The heterogeneous population is partitioned into mutually exclusive homogeneous strata, and random samples are drawn proportionally from each stratum.
3. Systematic Random Sampling: Selecting every $k^{\text{th}}$ item from a sampling frame after picking a random starting point between $1$ and $k$.
4. Cluster Sampling: Dividing the population into geographically or naturally occurring heterogeneous clusters, and randomly selecting entire clusters for survey.

Lowest Marks of the Top 20% of Students ($P_{80}$):

Concept:

The lowest marks obtained by the top 20% of students corresponds to the $80^{\text{th}}$ Percentile ($P_{80}$) of the distribution.

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Cumulative Frequency Table:

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Step 1: Locate $P_{80}$ Class Position

$$\text{Position of } P_{80} = \frac{80 \times N}{100} = \frac{80 \times 130}{100} = 104^{\text{th}} \text{ item}$$

The cumulative frequency just greater than 104 is 118, which corresponds to the class 40 – 50.

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Step 2: Apply the Percentile Interpolation Formula

$$P_{80} = L + \left(\frac{\frac{80N}{100} - cf}{f}\right) \times h$$

Where:

• $L = 40$ (Lower limit of $P_{80}$ class)
• $cf = 100$ (Cumulative frequency of the preceding class)
• $f = 18$ (Frequency of $P_{80}$ class)
• $h = 10$ (Class width)

$$P_{80} = 40 + \left(\frac{104 - 100}{18}\right) \times 10 = 40 + \left(\frac{4}{18}\right) \times 10 = 40 + 2.22 = \mathbf{42.22}$$

Final Answer: The lowest marks scored by the top 20% of the students is 42.22 marks.

95% Confidence Limits for Population Mean Weight:

Given Data:

• Sample size ($n$) = $50$ ($n \ge 30$, large sample)
• Sample mean ($\bar{X}$) = $7.5 \text{ kg}$
• Sample standard deviation ($s$) = $1.5 \text{ kg}$
• Confidence level = $95\% \implies \alpha = 0.05$

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Step 1: Critical Value ($Z_{\alpha/2}$)

For a 95% confidence interval, the two-tailed standard normal critical value is:

$$Z_{0.025} = 1.96$$

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Step 2: Standard Error of the Mean ($SE$)

$$SE_{\bar{X}} = \frac{s}{\sqrt{n}} = \frac{1.5}{\sqrt{50}} = \frac{1.5}{7.0711} \approx 0.2121 \text{ kg}$$

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Step 3: Compute Confidence Limits

$$\text{Confidence Limits} = \bar{X} \pm Z_{\alpha/2} \cdot SE_{\bar{X}}$$

$$\text{Margin of Error} = 1.96 \times 0.2121 = 0.4157 \text{ kg}$$

• Lower Limit: $7.5 - 0.4157 = \mathbf{7.0843 \text{ kg}}$
• Upper Limit: $7.5 + 0.4157 = \mathbf{7.9157 \text{ kg}}$

Final Answer: We are 95% confident that the true population mean weight lies between 7.08 kg and 7.92 kg ($[7.084, 7.916]$).

Calculation of Missing Frequencies Given Mode = 48:

Given Data:

• Total students ($N$) = $100$
• $\text{Mode} = 48$

Let the missing frequency for class $20-40$ be $f_1$ and for $60-80$ be $f_2$.

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Step 1: Formulate Total Frequency Equation

$$\sum f = 14 + f_1 + 27 + f_2 + 15 = 100$$

$$f_1 + f_2 + 56 = 100 \implies f_1 + f_2 = 44 \quad \text{--- (1)}$$

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Step 2: Use Mode Formula

Since $\text{Mode} = 48$, it lies in the modal class 40 – 60.

• Lower limit ($L$) = $40$
• Frequency of modal class ($f_m$) = $27$
• Frequency of pre-modal class ($f_0$) = $f_1$
• Frequency of post-modal class ($f_2'$) = $f_2$
• Class height ($h$) = $20$$\text{Mode} = L + \left[\frac{f_m - f_0}{2f_m - f_0 - f_2'}\right] \times h$$
  $$48 = 40 + \left[\frac{27 - f_1}{2(27) - f_1 - f_2}\right] \times 20$$
  $$8 = \left[\frac{27 - f_1}{54 - (f_1 + f_2)}\right] \times 20$$

Substitute $f_1 + f_2 = 44$ from Equation (1):

$$8 = \left[\frac{27 - f_1}{54 - 44}\right] \times 20$$

$$8 = \left[\frac{27 - f_1}{10}\right] \times 20$$

$$8 = 2(27 - f_1)$$

$$4 = 27 - f_1 \implies f_1 = 27 - 4 = \mathbf{23}$$

Now find $f_2$ from Equation (1):

$$f_2 = 44 - 23 = \mathbf{21}$$

Final Answer: The missing frequencies are $f_1 = 23$ (for 20–40) and $f_2 = 21$ (for 60–80).

Calculation of Median Income of 500 Workers:

Cumulative Frequency Table:

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Step 1: Identify Median Class

$$\text{Position of Median} = \frac{N}{2} = \frac{500}{2} = 250^{\text{th}} \text{ item}$$

The cumulative frequency just greater than 250 is 290, so the Median Class is 40 – 60.

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Step 2: Apply Median Formula

$$M_d = L + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h$$

Where:

• $L = 40$
• $cf = 140$
• $f = 150$
• $h = 20$$M_d = 40 + \left(\frac{250 - 140}{150}\right) \times 20 = 40 + \left(\frac{110}{150}\right) \times 20 = 40 + 14.67 = \mathbf{54.67 \text{ (in '000 Rs)}}$$

Final Answer: The median income of the workers is Rs 54,667 (or 54.67 thousand Rs).

Appropriate Measure of Dispersion for Open-Ended Distribution:

Justification:

The given distribution has open-ended classes (“Below 50” and “250 and above”). For open-ended distributions, neither the Range nor the Standard Deviation can be calculated without arbitrary class boundaries. Therefore, the Quartile Deviation (Semi-Interquartile Range) is the uniquely appropriate measure of dispersion.

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Cumulative Frequency with True Class Boundaries:

(Correction factor $= (51 - 50)/2 = 0.5$)

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Step 1: Lower Quartile ($Q_1$)

$$\text{Position of } Q_1 = \frac{N}{4} = \frac{100}{4} = 25^{\text{th}} \text{ item} \implies \text{Class: 50.5 – 100.5}$$

$$Q_1 = 50.5 + \left(\frac{25 - 10}{18}\right) \times 50 = 50.5 + \left(\frac{15}{18}\right) \times 50 = 50.5 + 41.67 = \mathbf{92.17}$$

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Step 2: Upper Quartile ($Q_3$)

$$\text{Position of } Q_3 = \frac{3N}{4} = \frac{300}{4} = 75^{\text{th}} \text{ item} \implies \text{Class: 200.5 – 250.5}$$

$$Q_3 = 200.5 + \left(\frac{75 - 73}{16}\right) \times 50 = 200.5 + \left(\frac{2}{16}\right) \times 50 = 200.5 + 6.25 = \mathbf{206.75}$$

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Step 3: Quartile Deviation ($QD$)

$$QD = \frac{Q_3 - Q_1}{2} = \frac{206.75 - 92.17}{2} = \frac{114.58}{2} = \mathbf{57.29}$$

$$\text{Coefficient of } QD = \frac{Q_3 - Q_1}{Q_3 + Q_1} = \frac{114.58}{298.92} = \mathbf{0.3833}$$

Final Answer:

• Quartile Deviation: Rs 57.29
• Coefficient of QD: 0.3833

Calculation of Combined Standard Deviation:

Given Data:

• Company A: $n_1 = 100$, $\bar{X}_1 = 28$, $\sigma_1^2 = 25$
• Company B: $n_2 = 100$, $\bar{X}_2 = 37$, $\sigma_2^2 = 36$

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

$$\bar{X}_{12} = \frac{n_1 \bar{X}_1 + n_2 \bar{X}_2}{n_1 + n_2} = \frac{100(28) + 100(37)}{100 + 100} = \frac{2800 + 3700}{200} = \frac{6500}{200} = \mathbf{32.5}$$

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Step 2: Deviations of Individual Means from Combined Mean

$$d_1 = \bar{X}_1 - \bar{X}_{12} = 28 - 32.5 = -4.5 \implies d_1^2 = 20.25$$

$$d_2 = \bar{X}_2 - \bar{X}_{12} = 37 - 32.5 = +4.5 \implies d_2^2 = 20.25$$

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Step 3: Combined Variance ($\sigma_{12}^2$)

$$\sigma_{12}^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}$$

$$\sigma_{12}^2 = \frac{100(25 + 20.25) + 100(36 + 20.25)}{200}$$

$$\sigma_{12}^2 = \frac{100(45.25) + 100(56.25)}{200} = \frac{4525 + 5625}{200} = \frac{10150}{200} = \mathbf{50.75}$$

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

$$\sigma_{12} = \sqrt{50.75} \approx \mathbf{7.1239 \text{ thousand Rs}}$$

Final Answer: The combined standard deviation is Rs 7,124 (or 7.12 thousand Rs).

Hypothesis Testing: Difference in Mean Life of Electric Bulbs:

Step 1: Summary of Sample Statistics

• Company X: $n_1 = 100$, $\bar{X}_1 = 1300 \text{ hours}$, $s_1 = 80 \text{ hours}$
• Company Y: $n_2 = 100$, $\bar{X}_2 = 1250 \text{ hours}$, $s_2 = 90 \text{ hours}$
• Level of significance ($\alpha$) = $0.05$

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Step 2: Formulate Hypotheses

• Null Hypothesis ($H_0$): $\mu_1 = \mu_2$ (There is no significant difference in the mean life of bulbs between Company X and Company Y).
• Alternative Hypothesis ($H_1$): $\mu_1 \ne \mu_2$ (There is a significant difference in mean life — Two-tailed test).

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Step 3: Test Statistic ($Z$)

$$Z = \frac{\bar{X}_1 - \bar{X}_2}{SE_{\bar{X}_1 - \bar{X}_2}} = \frac{\bar{X}_1 - \bar{X}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$$

Calculate Standard Error:

$$\frac{s_1^2}{n_1} = \frac{80^2}{100} = \frac{6400}{100} = 64$$

$$\frac{s_2^2}{n_2} = \frac{90^2}{100} = \frac{8100}{100} = 81$$

$$SE = \sqrt{64 + 81} = \sqrt{145} \approx 12.0416$$

Calculate Test Statistic:

$$Z_{\text{cal}} = \frac{1300 - 1250}{12.0416} = \frac{50}{12.0416} = \mathbf{4.152}$$

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Step 4: Decision and Interpretation

• For a two-tailed test at $\alpha = 0.05$, the tabulated critical value is $Z_{\text{tab}} = \pm 1.96$.
• Decision Rule: Reject $H_0$ if $|Z_{\text{cal}}| > 1.96$.
• Since $|Z_{\text{cal}}| = 4.152 > 1.96$, we reject the null hypothesis $H_0$.

Conclusion: There is a statistically significant difference in the average life of electric bulbs produced by Company X and Company Y at the 5% level of significance (Company X bulbs have a significantly longer lifespan).

Normal Distribution Probabilities for Bank Employee Income:

Given Parameters:

• Mean ($\mu$) = $\text{Rs } 1200$
• Standard Deviation ($\sigma$) = $\text{Rs } 120$
• Variable $X \sim N(1200, 120^2)$

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(a) Probability of Income Between Rs 1,050 and Rs 1,400:

Convert to standard normal $Z$-scores:

$$Z_1 = \frac{1050 - 1200}{120} = \frac{-150}{120} = -1.25$$

$$Z_2 = \frac{1400 - 1200}{120} = \frac{200}{120} \approx +1.67$$

$$P(1050 \le X \le 1400) = P(-1.25 \le Z \le 1.67) = P(0 \le Z \le 1.25) + P(0 \le Z \le 1.67)$$

From standard normal area tables:

• $\text{Area}(0 \le Z \le 1.25) = 0.3944$
• $\text{Area}(0 \le Z \le 1.67) = 0.4525$$P = 0.3944 + 0.4525 = \mathbf{0.8469} \quad (\text{or } 84.69\%)$$

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(b) Probability of Income Between Rs 1,250 and Rs 1,500:

Convert to standard normal $Z$-scores:

$$Z_1 = \frac{1250 - 1200}{120} = \frac{50}{120} \approx +0.42$$

$$Z_2 = \frac{1500 - 1200}{120} = \frac{300}{120} = +2.50$$

$$P(1250 \le X \le 1500) = P(0.42 \le Z \le 2.50) = P(0 \le Z \le 2.50) - P(0 \le Z \le 0.42)$$

From standard normal area tables:

• $\text{Area}(0 \le Z \le 2.50) = 0.4938$
• $\text{Area}(0 \le Z \le 0.42) = 0.1628$$P = 0.4938 - 0.1628 = \mathbf{0.3310} \quad (\text{or } 33.10\%)$$

Final Answer:

• (a) Probability between Rs 1,050 and Rs 1,400: 0.8469 (approx. 42 out of 50 employees)
• (b) Probability between Rs 1,250 and Rs 1,500: 0.3310 (approx. 17 out of 50 employees)

Calculation of Karl Pearson’s Coefficient of Skewness:

Calculation Table:

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

$$\bar{X} = A + \left(\frac{\sum fd}{N}\right) \times h = 250 + \left(\frac{50}{220}\right) \times 100 = 250 + 22.73 = \mathbf{272.73 \text{ thousand Rs}}$$

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Step 2: Mode ($M_o$)

Modal class is 200 – 300 (highest frequency $f_1 = 75$).

• $L = 200$, $f_0 = 50$, $f_2 = 40$, $h = 100$$M_o = L + \left[\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right] \times h = 200 + \left[\frac{75 - 50}{2(75) - 50 - 40}\right] \times 100 = 200 + \left(\frac{25}{60}\right) \times 100 = \mathbf{241.67}$$

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Step 3: Standard Deviation ($\sigma$)

$$\sigma = h \times \sqrt{\frac{\sum fd^2}{N} - \left(\frac{\sum fd}{N}\right)^2} = 100 \times \sqrt{\frac{360}{220} - \left(\frac{50}{220}\right)^2} = 100 \times \sqrt{1.6364 - 0.05165} = 100 \times 1.2589 = \mathbf{125.89}$$

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

$$Sk_p = \frac{\bar{X} - M_o}{\sigma} = \frac{272.73 - 241.67}{125.89} = \frac{31.06}{125.89} = \mathbf{+0.2467}$$

Interpretation: The distribution exhibits a positive skewness ($Sk_p = +0.2467$), indicating that the distribution has a longer tail to the right and a concentration of lower to middle-income workers.

Calculation of Percentile Coefficient of Kurtosis ($K$):

Concept:

The Percentile Coefficient of Kurtosis is given by:

$$K = \frac{QD}{P_{90} - P_{10}} = \frac{\frac{Q_3 - Q_1}{2}}{P_{90} - P_{10}}$$

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Cumulative Frequency Table:

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Step 1: Calculate $Q_1$ and $Q_3$

• $Q_1$ position: $\frac{N}{4} = 125^{\text{th}} \implies$ Class 10 – 20
  $$Q_1 = 10 + \left(\frac{125 - 50}{100}\right) \times 10 = 10 + 7.5 = \mathbf{17.5}$$
• $Q_3$ position: $\frac{3N}{4} = 375^{\text{th}} \implies$ Class 30 – 40
  $$Q_3 = 30 + \left(\frac{375 - 300}{90}\right) \times 10 = 30 + 8.33 = \mathbf{38.33}$$
• Semi-Interquartile Range ($QD$):
  $$QD = \frac{38.33 - 17.5}{2} = \frac{20.83}{2} = \mathbf{10.415}$$

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Step 2: Calculate $P_{10}$ and $P_{90}$

• $P_{10}$ position: $\frac{10N}{100} = 50^{\text{th}} \implies$ Exactly the upper boundary of class 0–10:
  $$P_{10} = \mathbf{10.0}$$
• $P_{90}$ position: $\frac{90N}{100} = 450^{\text{th}} \implies$ Exactly the upper boundary of class 40–50:
  $$P_{90} = \mathbf{50.0}$$

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Step 3: Compute Kurtosis ($K$)

$$K = \frac{10.415}{50.0 - 10.0} = \frac{10.415}{40} = \mathbf{0.2604}$$

Interpretation:

• For a normal (mesokurtic) distribution, $K = 0.263$.
• Since $K = 0.2604 < 0.263$, the distribution is slightly platykurtic (flatter-topped than normal).

Income and Expenditure: Correlation and Regression Analysis

Data Summary ($n = 10$):

• $X$ = Income (000 Rs): $[60, 55, 54, 52, 48, 52, 53, 47, 49, 50]$
• $Y$ = Expenditure (000 Rs): $[22, 20, 15, 17, 25, 21, 16, 17, 18, 19]$

Summations:

• $\sum X = 520 \implies \bar{X} = 52.0$
• $\sum Y = 190 \implies \bar{Y} = 19.0$
• $\sum X^2 = 27,172 \implies \sum (X - \bar{X})^2 = 132$
• $\sum Y^2 = 3,694 \implies \sum (Y - \bar{Y})^2 = 84$
• $\sum XY = 9,885 \implies \sum (X - \bar{X})(Y - \bar{Y}) = 5$

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a. Coefficients of Variation (CV):

$$\sigma_X = \sqrt{\frac{132}{10}} = \sqrt{13.2} = 3.633 \implies CV_X = \frac{3.633}{52} \times 100 = \mathbf{6.99\%}$$

$$\sigma_Y = \sqrt{\frac{84}{10}} = \sqrt{8.4} = 2.898 \implies CV_Y = \frac{2.898}{19} \times 100 = \mathbf{15.25\%}$$

Interpretation: Since $CV_X (6.99\%) < CV_Y (15.25\%)$, income is significantly more consistent and uniform than expenditure.

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b. Correlation Coefficient ($r$):

$$r = \frac{\sum (X - \bar{X})(Y - \bar{Y})}{\sqrt{\sum (X - \bar{X})^2 \sum (Y - \bar{Y})^2}} = \frac{5}{\sqrt{132 \times 84}} = \frac{5}{\sqrt{11088}} = \frac{5}{105.30} = \mathbf{+0.0475}$$

There is a negligible, weak positive linear correlation between income and expenditure in this sample.

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c. Two Regression Lines:

1. Regression coefficient of $Y$ on $X$ ($b_{YX}$):

   $$b_{YX} = \frac{5}{132} \approx 0.0379$$
   Line: $Y - \bar{Y} = b_{YX}(X - \bar{X})$$Y - 19 = 0.0379(X - 52) \implies \mathbf{Y = 17.03 + 0.0379X}$$

2. Regression coefficient of $X$ on $Y$ ($b_{XY}$):

   $$b_{XY} = \frac{5}{84} \approx 0.0595$$
   Line: $X - \bar{X} = b_{XY}(Y - \bar{Y})$$X - 52 = 0.0595(Y - 19) \implies \mathbf{X = 50.87 + 0.0595Y}$$

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d. Estimated Expenditure for Income = Rs 70,000 ($X = 70$):

$$\hat{Y} = 17.03 + 0.0379(70) = 17.03 + 2.65 = \mathbf{19.68 \text{ thousand Rs (Rs 19,680)}}$$