Business Statistics

Quartile Deviation ($QD$):

Given:

• First Quartile ($Q_1$) = $18$
• Third Quartile ($Q_3$) = $30$

Formula:

Calculation:

Final Answer: The Quartile Deviation is 6.

Calculation of Median from Mode and Mean:

Given:

• $\text{Mode} = 32$
• $\text{Mean} = 35$

Empirical Formula:

Calculation:

Final Answer: The median of the distribution is 34.

Pearson’s Coefficient of Skewness:

Given:

• $\text{Mean} = 25$
• $\text{Mode} = 20$
• $\text{Standard Deviation } (\sigma) = 10$

Formula:

Calculation:

Final Answer: Pearson’s coefficient of skewness is +0.5 (moderately positively skewed).

Calculation of Percentile Coefficient of Kurtosis:

Given:

• $Q_1 = 28$, $Q_3 = 52$
• $P_{90} = 61$, $P_{10} = 17$

Formula:

Calculation:

Final Answer: The coefficient of kurtosis is 0.2727 (since $0.2727 > 0.263$, it is leptokurtic).

Finding Missing Frequencies When Mean = 48:

Given Data:

• Total families ($N$) = $100$
• Mean expenditure ($\bar{X}$) = $48$

Let the missing frequency of class 20–40 be $f_1$ and class 60–80 be $f_2$.

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Step 1: Total Frequency Equation

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Step 2: Mean Equation

Divide by 10:

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Step 3: Solve Simultaneously

Multiply Equation (1) by 3:

Subtract Equation (3) from Equation (2):

Substitute $f_2 = 16$ into Equation (1):

Final Answer: The missing frequencies are $f_1 = 28$ (for class 20–40) and $f_2 = 16$ (for class 60–80).

Coefficient of Quartile Deviation of Household Expenditure:

Cumulative Frequency Table:

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Step 1: Lower Quartile ($Q_1$)

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Step 2: Upper Quartile ($Q_3$)

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Step 3: Coefficient of Quartile Deviation

Final Answer: The coefficient of quartile deviation is 0.1923 (or 19.23%).

Lowest Marks of the Top 10% of Students ($P_{90}$):

Concept:

The cutoff score separating the top 10% of students is the $90^{\text{th}}$ Percentile ($P_{90}$).

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Cumulative Frequency Table:

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Step 1: Position of $P_{90}$$\text{Position} = \frac{90 \times N}{100} = \frac{90 \times 104}{100} = 93.6^{\text{th}} \text{ item}$$

The cumulative frequency just exceeding 93.6 is 101, corresponding to class 50 – 60.

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Step 2: Percentile Interpolation Formula

Where $L = 50$, $cf = 92$, $f = 9$, $h = 10$:

Final Answer: The lowest marks of the top 10% of students is 51.78 marks.

Percentile Coefficient of Kurtosis for Nepal Bank Limited Employees:

Cumulative Frequency Table:

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Step 1: Calculate $Q_1$ and $Q_3$

• $Q_1$ position: $\frac{N}{4} = \frac{62}{4} = 15.5^{\text{th}} \implies$ Class 30 – 40

  Swipe to see the full formula↔

  $$Q_1 = 30 + \left(\frac{15.5 - 7}{10}\right) \times 10 = 30 + 8.5 = \mathbf{38.5}$$
• $Q_3$ position: $\frac{3N}{4} = \frac{186}{4} = 46.5^{\text{th}} \implies$ Class 50 – 60

  Swipe to see the full formula↔

  $$Q_3 = 50 + \left(\frac{46.5 - 37}{18}\right) \times 10 = 50 + \left(\frac{9.5}{18}\right) \times 10 = 50 + 5.28 = \mathbf{55.28}$$
• Semi-Interquartile Range ($QD$):

  Swipe to see the full formula↔

  $$QD = \frac{Q_3 - Q_1}{2} = \frac{55.28 - 38.5}{2} = \frac{16.78}{2} = \mathbf{8.39}$$

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

• $P_{10}$ position: $\frac{10 \times 62}{100} = 6.2^{\text{th}} \implies$ Class 20 – 30

  Swipe to see the full formula↔

  $$P_{10} = 20 + \left(\frac{6.2 - 0}{7}\right) \times 10 = 20 + 8.86 = \mathbf{28.86}$$
• $P_{90}$ position: $\frac{90 \times 62}{100} = 55.8^{\text{th}} \implies$ Class 60 – 70

  Swipe to see the full formula↔

  $$P_{90} = 60 + \left(\frac{55.8 - 55}{7}\right) \times 10 = 60 + 1.14 = \mathbf{61.14}$$

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Step 3: Compute Kurtosis ($K$)

Interpretation:

• The standard normal (mesokurtic) value is $K = 0.263$.
• Since $K = 0.2599 < 0.263$, the distribution is slightly platykurtic, indicating a slightly flatter peak with thinner tails compared to a normal distribution.

Hypothesis Testing: Difference in Average Marks Between Class A and Class B:

Step 1: Summary of Sample Statistics

• Class A: $n_1 = 150$, $\bar{X}_1 = 45$, $s_1 = 5$
• Class B: $n_2 = 100$, $\bar{X}_2 = 43$, $s_2 = 4$
• 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 average marks between the two classes).
• Alternative Hypothesis ($H_1$): $\mu_1 \ne \mu_2$ (There is a significant difference in average marks — Two-tailed test).

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Step 3: Compute Test Statistic ($Z$)

Calculate Standard Error ($SE$):

Calculate $Z_{\text{cal}}$:

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Step 4: Decision and Conclusion

• For a two-tailed test at $\alpha = 0.05$, the tabulated critical value is $Z_{\text{tab}} = \pm 1.96$.
• Since $|Z_{\text{cal}}| = 3.50 > 1.96$, we reject the null hypothesis $H_0$.

Conclusion: There is a statistically significant difference in the average marks of students between Class A and Class B at the 5% level of significance (Class A students perform significantly better).

Consistency Comparison of Two Horses Using Coefficient of Variation (CV):

Data ($n = 6$ observations each):

• Horse A: $[250, 255, 280, 290, 295, 300]$
• Horse B: $[280, 282, 290, 295, 298, 295]$

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Step 1: Horse A Calculations

Deviations squared $\sum (X_A - \bar{X}_A)^2$:

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Step 2: Horse B Calculations

Deviations squared $\sum (X_B - \bar{X}_B)^2$:

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Conclusion:

• A lower Coefficient of Variation signifies greater stability and consistency.
• Since $CV_B (2.35\%) < CV_A (6.93\%)$, Horse B is significantly more consistent in its running performance than Horse A.

Coefficient of Quartile Deviation:

Given:

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

Formula:

Calculation:

Final Answer: The coefficient of quartile deviation is 0.0328.

Calculation of Median from Empirical Relationship:

Given:

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

Empirical Formula for Moderately Asymmetric Distributions:

Calculation:

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:

Calculation:

Final Answer: The Coefficient of Variation is 53.125%.

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

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

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)

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

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

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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$$\text{Mode} = L + \left[\frac{f_m - f_0}{2f_m - f_0 - f_2'}\right] \times h$$

  Swipe to see the full formula↔

  $$48 = 40 + \left[\frac{27 - f_1}{2(27) - f_1 - f_2}\right] \times 20$$

  Swipe to see the full formula↔

  $$8 = \left[\frac{27 - f_1}{54 - (f_1 + f_2)}\right] \times 20$$

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

Now find $f_2$ from Equation (1):

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

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

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 ($Q_1$)

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Step 2: Upper Quartile ($Q_3$)

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Step 3: Quartile Deviation ($QD$)

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 ($\bar{X}_{12}$)

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Step 2: Deviations of Individual Means from Combined Mean

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

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

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

Calculation of Karl Pearson’s Coefficient of Skewness:

Calculation Table:

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

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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 ($\sigma$)

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

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:

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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

  Swipe to see the full formula↔

  $$Q_1 = 10 + \left(\frac{125 - 50}{100}\right) \times 10 = 10 + 7.5 = \mathbf{17.5}$$
• $Q_3$ position: $\frac{3N}{4} = 375^{\text{th}} \implies$ Class 30 – 40

  Swipe to see the full formula↔

  $$Q_3 = 30 + \left(\frac{375 - 300}{90}\right) \times 10 = 30 + 8.33 = \mathbf{38.33}$$
• Semi-Interquartile Range ($QD$):

  Swipe to see the full formula↔

  $$QD = \frac{38.33 - 17.5}{2} = \frac{20.83}{2} = \mathbf{10.415}$$

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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:

  Swipe to see the full formula↔

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

  Swipe to see the full formula↔

  $$P_{90} = \mathbf{50.0}$$

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Step 3: Compute Kurtosis ($K$)

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):

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 ($r$):

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 $X$ ($b_{YX}$):

   Swipe to see the full formula↔

   $$b_{YX} = \frac{5}{132} \approx 0.0379$$
   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 $Y$ ($b_{XY}$):

   Swipe to see the full formula↔

   $$b_{XY} = \frac{5}{84} \approx 0.0595$$
   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 ($X = 70$):