Business Statistics 2022 Board Question Paper

Calculation of Lower ($Q_1$) and Upper ($Q_3$) Quartiles:

Step 1: Arrange the Data in Ascending Order

Given sample data ($n = 14$): 78, 121, 143, 88, 110, 107, 62, 122, 130, 95, 78, 139, 89, 125

Arranging in ascending order: 62, 78, 78, 88, 89, 95, 107, 110, 121, 122, 125, 130, 139, 143

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Step 2: Calculate Lower Quartile ($Q_1$)

$$\text{Position of } Q_1 = \left(\frac{n + 1}{4}\right)^{\text{th}} \text{ item} = \left(\frac{14 + 1}{4}\right)^{\text{th}} = 3.75^{\text{th}} \text{ item}$$

$$Q_1 = \text{Value of } 3^{\text{rd}} \text{ item} + 0.75 \times (\text{Value of } 4^{\text{th}} \text{ item} - \text{Value of } 3^{\text{rd}} \text{ item})$$

$$Q_1 = 78 + 0.75 \times (88 - 78) = 78 + 0.75 \times 10 = 78 + 7.5 = \mathbf{85.5}$$

(Note: Under the alternative integer-rank formula $n/4 = 3.5^{\text{th}}$ item, $Q_1 = 88$).

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Step 3: Calculate Upper Quartile ($Q_3$)

$$\text{Position of } Q_3 = \left(\frac{3(n + 1)}{4}\right)^{\text{th}} \text{ item} = \left(\frac{3 \times 15}{4}\right)^{\text{th}} = 11.25^{\text{th}} \text{ item}$$

$$Q_3 = \text{Value of } 11^{\text{th}} \text{ item} + 0.25 \times (\text{Value of } 12^{\text{th}} \text{ item} - \text{Value of } 11^{\text{th}} \text{ item})$$

$$Q_3 = 125 + 0.25 \times (130 - 125) = 125 + 0.25 \times 5 = 125 + 1.25 = \mathbf{126.25}$$

(Note: Under the alternative integer-rank formula $3n/4 = 10.5^{\text{th}}$ item, $Q_3 = 125$).

Final Answer:

• Lower Quartile ($Q_1$): 85.5
• Upper Quartile ($Q_3$): 126.25

Calculation of Median from Bowley’s Skewness:

Given Data:

• First Quartile ($Q_1$) = $8$
• Third Quartile ($Q_3$) = $16$
• Bowley’s Coefficient of Skewness ($Sk_B$) = $0.5$

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Formula:

Bowley’s coefficient of skewness based on quartiles is given by:

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

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Step-by-Step Substitution:

$$0.5 = \frac{16 + 8 - 2M_d}{16 - 8}$$

$$0.5 = \frac{24 - 2M_d}{8}$$

Multiplying both sides by $8$:

$$0.5 \times 8 = 24 - 2M_d$$

$$4 = 24 - 2M_d$$

$$2M_d = 24 - 4$$

$$2M_d = 20$$

$$M_d = \frac{20}{2} = \mathbf{10}$$

Final Answer: The median ($M_d$) of the distribution is 10.

Probability Calculations for Numbered Balls (1 to 20):

Total number of balls in the sample space:

$$S = \{1, 2, 3, \dots, 20\} \implies n(S) = 20$$

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(a) Probability of Selecting a Multiple of 3 or 7:

• Multiples of 3: $A = \{3, 6, 9, 12, 15, 18\} \implies n(A) = 6$
• Multiples of 7: $B = \{7, 14\} \implies n(B) = 2$
• Common multiples (multiples of 21): $A \cap B = \emptyset \implies n(A \cap B) = 0$ (Mutually exclusive events)

By the Addition Rule:

$$P(A \cup B) = P(A) + P(B) = \frac{6}{20} + \frac{2}{20} = \frac{8}{20} = \frac{2}{5} = \mathbf{0.40}$$

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(b) Probability of Selecting a Multiple of 3 or 4:

• Multiples of 3: $A = \{3, 6, 9, 12, 15, 18\} \implies n(A) = 6$
• Multiples of 4: $C = \{4, 8, 12, 16, 20\} \implies n(C) = 5$
• Common multiples (multiples of 12): $A \cap C = \{12\} \implies n(A \cap C) = 1$

By the General Addition Theorem:

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

$$P(A \cup C) = \frac{6}{20} + \frac{5}{20} - \frac{1}{20} = \frac{10}{20} = \frac{1}{2} = \mathbf{0.50}$$

Final Answer:

• (a) $P(\text{multiple of 3 or 7}) =$ 0.40
• (b) $P(\text{multiple of 3 or 4}) =$ 0.50

Expected Value and Variance of Discrete Random Variable $X$:

Calculation Table:

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Step 1: Expected Value $E(X)$$E(X) = \mu = \sum [X \cdot P(X)] = 0.18 + 0.70 + 0.45 + 0.40 = \mathbf{1.73}$$

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Step 2: Variance $Var(X)$$\text{Var}(X) = \sigma^2 = E(X^2) - [E(X)]^2$$

$$\text{Var}(X) = 4.53 - (1.73)^2 = 4.53 - 2.9929 = \mathbf{1.5371}$$

Final Answer:

• Expected Value $E(X)$: 1.73
• Variance $\text{Var}(X)$: 1.5371

Five-Number Summary and Distribution Shape:

Step 1: Arrange Data in Ascending Order

Given data ($n = 11$): 7, 11, 25, 23, 19, 34, 29, 31, 9, 15, 30

Ordered data: 7, 9, 11, 15, 19, 23, 25, 29, 30, 31, 34

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Step 2: Compute the Five-Number Summary

1. Minimum ($Min$): 7
2. First Quartile ($Q_1$):
   $$\text{Position of } Q_1 = \left(\frac{n + 1}{4}\right)^{\text{th}} = \left(\frac{11 + 1}{4}\right)^{\text{th}} = 3^{\text{rd}} \text{ value} = \mathbf{11}$$
3. Median ($Q_2$):
   $$\text{Position of } Q_2 = \left(\frac{n + 1}{2}\right)^{\text{th}} = \left(\frac{11 + 1}{2}\right)^{\text{th}} = 6^{\text{th}} \text{ value} = \mathbf{23}$$
4. Third Quartile ($Q_3$):
   $$\text{Position of } Q_3 = \left(\frac{3(n + 1)}{4}\right)^{\text{th}} = 9^{\text{th}} \text{ value} = \mathbf{30}$$
5. Maximum ($Max$): 34

Summary Vector: $[Min, Q_1, Q_2, Q_3, Max] = [7, 11, 23, 30, 34]$

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Step 3: Comment on the Shape of the Distribution

Compare quartile distances:

• Distance from $Q_1$ to Median: $Q_2 - Q_1 = 23 - 11 = 12$
• Distance from Median to $Q_3$: $Q_3 - Q_2 = 30 - 23 = 7$$Sk_B = \frac{Q_3 + Q_1 - 2Q_2}{Q_3 - Q_1} = \frac{30 + 11 - 2(23)}{30 - 11} = \frac{-5}{19} \approx -0.263$$

Conclusion on Shape: Since $(Q_2 - Q_1) > (Q_3 - Q_2)$ ($12 > 7$) and Bowley’s skewness is negative ($-0.263$), the distribution has a longer tail to the left and is negatively skewed (skewed to the left).

Mean and Standard Deviation of Age Distribution:

Calculation Table:

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

$$\bar{X} = \frac{\sum fX}{N} = \frac{874}{50} = \mathbf{17.48 \text{ years}}$$

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Step 2: Calculation of Standard Deviation ($\sigma$)

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

$$\sigma = \sqrt{\frac{15372}{50} - (17.48)^2} = \sqrt{307.44 - 305.5504} = \sqrt{1.8896} \approx \mathbf{1.3746 \text{ years}}$$

Final Answer:

• Mean Age: 17.48 years
• Standard Deviation: 1.37 years

Range of Height for Middle 80% of Students:

Given Data:

• Total students ($N$) = $500$
• Mean height ($\mu$) = $165 \text{ cm}$
• Variance ($\sigma^2$) = $25 \text{ cm}^2 \implies \sigma = \sqrt{25} = 5 \text{ cm}$

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Step 1: Determine the Standard Normal $Z$-Value

For the middle 80% of a normal distribution:

• Area in the middle = $0.80$
• Area in each tail = $\frac{1 - 0.80}{2} = 0.10$
• Area from the mean $\mu$ to each boundary = $\frac{0.80}{2} = 0.4000$

From the standard normal distribution table:

$$P(0 \le Z \le z) = 0.4000 \implies z = 1.28 \quad (\text{more precisely, } 1.2816)$$

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Step 2: Calculate Lower and Upper Height Boundaries

$$X = \mu \pm z \cdot \sigma$$

• Lower Limit ($X_1$):

  $$X_1 = \mu - z \cdot \sigma = 165 - (1.28 \times 5) = 165 - 6.4 = \mathbf{158.60 \text{ cm}}$$

• Upper Limit ($X_2$):

  $$X_2 = \mu + z \cdot \sigma = 165 + (1.28 \times 5) = 165 + 6.4 = \mathbf{171.40 \text{ cm}}$$

Final Answer: The height range for the middle 80% of students is from 158.60 cm to 171.40 cm (Range = $171.40 - 158.60 = 12.80 \text{ cm}$).

Hypothesis Testing: Difference Between Two Means (British vs. Indian Heights):

Step 1: Summary of Sample Statistics

• Sample 1 (Indians): $n_1 = 6,400$, $\bar{X}_1 = 67.85 \text{ inches}$, $s_1 = 2.56 \text{ inches}$
• Sample 2 (British): $n_2 = 1,600$, $\bar{X}_2 = 68.55 \text{ inches}$, $s_2 = 2.52 \text{ inches}$
• Level of significance ($\alpha$) = $0.05$

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Step 2: Formulate Hypotheses

• Null Hypothesis ($H_0$): $\mu_2 \le \mu_1$ (British are not taller on average than Indians).
• Alternative Hypothesis ($H_1$): $\mu_2 > \mu_1$ (British are on average taller than Indians — Right-tailed test).

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Step 3: Test Statistic ($Z$)

Under $H_0$, the test statistic for large independent samples is:

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

Calculate Standard Error ($SE$):

$$\frac{s_1^2}{n_1} = \frac{(2.56)^2}{6400} = \frac{6.5536}{6400} = 0.001024$$

$$\frac{s_2^2}{n_2} = \frac{(2.52)^2}{1600} = \frac{6.3504}{1600} = 0.003969$$

$$SE = \sqrt{0.001024 + 0.003969} = \sqrt{0.004993} \approx 0.07066$$

Calculate Calculated $Z$:

$$Z_{\text{cal}} = \frac{68.55 - 67.85}{0.07066} = \frac{0.70}{0.07066} = \mathbf{9.907}$$

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Step 4: Critical Value and Decision Rule

• For a right-tailed test at $\alpha = 0.05$, the tabulated critical value is $Z_{\text{tab}} = +1.645$.
• Decision: Since $Z_{\text{cal}} = 9.907 > 1.645$, it falls far into the critical rejection region. We reject the null hypothesis $H_0$.

Conclusion: There is overwhelming statistical evidence at the 5% significance level to conclude that British are on average taller than Indians.

Comparison of Expenditure Uniformity Between Village A and Village B Using CV:

Calculation Table:

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Step 1: Village A Statistics

$$\bar{X}_A = \frac{\sum f_A X}{N_A} = \frac{1695}{103} = \mathbf{16.46 \text{ thousand Rs}}$$

$$\sigma_A = \sqrt{\frac{\sum f_A X^2}{N_A} - \bar{X}_A^2} = \sqrt{\frac{30825}{103} - (16.456)^2} = \sqrt{299.27 - 270.81} = \sqrt{28.46} = \mathbf{5.33}$$

$$CV_A = \frac{\sigma_A}{\bar{X}_A} \times 100 = \frac{5.33}{16.46} \times 100 = \mathbf{32.38\%}$$

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Step 2: Village B Statistics

$$\bar{X}_B = \frac{\sum f_B X}{N_B} = \frac{1750}{105} = \mathbf{16.67 \text{ thousand Rs}}$$

$$\sigma_B = \sqrt{\frac{\sum f_B X^2}{N_B} - \bar{X}_B^2} = \sqrt{\frac{33800}{105} - (16.667)^2} = \sqrt{321.90 - 277.78} = \sqrt{44.12} = \mathbf{6.64}$$

$$CV_B = \frac{\sigma_B}{\bar{X}_B} \times 100 = \frac{6.64}{16.67} \times 100 = \mathbf{39.83\%}$$

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Conclusion:

Since $CV_A (32.38\%) < CV_B (39.83\%)$, Village A has a lower coefficient of variation, meaning the people of Village A have more uniform (consistent) monthly expenditure compared to Village B.

Correlation and Linear Regression Analysis:

Data Summary ($n = 12$):

• $X$ = Number of machines serviced: $[11, 8, 9, 10, 7, 6, 8, 4, 10, 5, 5, 12]$
• $Y$ = Service time in minutes: $[115, 60, 80, 90, 55, 65, 70, 33, 95, 50, 40, 110]$

Summations:

• $\sum X = 95 \implies \bar{X} = \frac{95}{12} = 7.917$
• $\sum Y = 863 \implies \bar{Y} = \frac{863}{12} = 71.917$
• $\sum X^2 = 825$
• $\sum Y^2 = 69,789$
• $\sum XY = 7,552$

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a) Coefficient of Correlation ($r$):

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

$$n\sum XY - (\sum X)(\sum Y) = 12(7552) - (95)(863) = 90624 - 81985 = 8,639$$

$$n\sum X^2 - (\sum X)^2 = 12(825) - 95^2 = 9900 - 9025 = 875$$

$$n\sum Y^2 - (\sum Y)^2 = 12(69789) - 863^2 = 837468 - 744769 = 92,699$$

$$r = \frac{8639}{\sqrt{875 \times 92699}} = \frac{8639}{\sqrt{81111625}} = \frac{8639}{9006.20} \approx \mathbf{+0.9592}$$

Interpretation: There is a very strong positive linear correlation ($r = +0.9592$) between the number of machines serviced and the required service call time.

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b) Regression Equation of $Y$ on $X$ and Prediction for $X = 6$:

Regression coefficient $b_{YX}$:

$$b_{YX} = \frac{n\sum XY - (\sum X)(\sum Y)}{n\sum X^2 - (\sum X)^2} = \frac{8639}{875} \approx \mathbf{9.8731}$$

Intercept $a$:

$$a = \bar{Y} - b_{YX}\bar{X} = 71.9167 - (9.8731 \times 7.9167) = 71.9167 - 78.1623 = \mathbf{-6.2456}$$

The fitted regression line is:

$$\hat{Y} = -6.25 + 9.87X$$

Prediction for 6 machines ($X = 6$):

$$\hat{Y} = -6.2456 + 9.8731(6) = -6.2456 + 59.2386 = \mathbf{52.99 \text{ minutes}}$$

Final Answer:

• Correlation coefficient ($r$): +0.9592 (Very strong positive correlation)
• Regression equation: $\hat{Y} = -6.25 + 9.87X$
• Expected time for 6 machines: ~53 minutes