Business Statistics

Quartile Deviation ($QD$):

Given:

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

Formula:

$$QD = \frac{Q_3 - Q_1}{2}$$

Calculation:

$$QD = \frac{30 - 18}{2} = \frac{12}{2} = \mathbf{6}$$

Final Answer: The Quartile Deviation is 6.

Combined Mean Calculation:

Given:

• Group A: $n_1 = 40$, $\bar{X}_1 = 65$
• Group B: $n_2 = 60$, $\bar{X}_2 = 60$

Formula:

$$\bar{X}_{12} = \frac{n_1 \bar{X}_1 + n_2 \bar{X}_2}{n_1 + n_2}$$

Calculation:

$$\bar{X}_{12} = \frac{40(65) + 60(60)}{40 + 60} = \frac{2600 + 3600}{100} = \frac{6200}{100} = \mathbf{62}$$

Final Answer: The combined mean of the two groups is 62.

Calculation of Standard Deviation of $Y$ ($\sigma_Y$):

Given:

• Correlation coefficient ($r$) = $0.48$
• $\text{Cov}(X, Y) = 36$
• $\text{Var}(X) = \sigma_X^2 = 16 \implies \sigma_X = 4$

Formula:

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

Calculation:

$$0.48 = \frac{36}{4 \cdot \sigma_Y}$$

$$0.48 = \frac{9}{\sigma_Y} \implies \sigma_Y = \frac{9}{0.48} = \mathbf{18.75}$$

Final Answer: The standard deviation of $Y$ is 18.75.

Calculation of Median from Mode and Mean:

Given:

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

Empirical Formula:

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

Calculation:

$$32 = 3\text{Median} - 2(35)$$

$$32 = 3\text{Median} - 70$$

$$3\text{Median} = 32 + 70 = 102$$

$$\text{Median} = \frac{102}{3} = \mathbf{34}$$

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:

$$Sk_p = \frac{\text{Mean} - \text{Mode}}{\sigma}$$

Calculation:

$$Sk_p = \frac{25 - 20}{10} = \frac{5}{10} = \mathbf{+0.5}$$

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

Poisson Probability $P(X = 4)$:

Given:

• $\lambda = 1.2$
• $x = 4$

Formula:

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

Calculation:

• $e^{-1.2} \approx 0.301194$
• $(1.2)^4 = 2.0736$
• $4! = 24$

$$P(X = 4) = \frac{0.301194 \times 2.0736}{24} = \frac{0.62456}{24} \approx \mathbf{0.0260}$$

Final Answer: $P(X = 4) =$ 0.0260 (or 2.60%).

Standard Error with Finite Population Correction (FPC):

Given:

• Sample size ($n$) = $36$
• Population size ($N$) = $400$
• Population standard deviation ($\sigma$) = $12$

Sampling Fraction:

$$\frac{n}{N} = \frac{36}{400} = 0.09 > 0.05 \implies \text{FPC must be applied.}$$

Formula:

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

Calculation:

$$\frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{36}} = \frac{12}{6} = 2.0$$

$$\text{FPC} = \sqrt{\frac{400 - 36}{400 - 1}} = \sqrt{\frac{364}{399}} = \sqrt{0.91228} \approx 0.95513$$

$$SE_{\bar{X}} = 2.0 \times 0.95513 = \mathbf{1.9103}$$

Final Answer: The standard error of the sample mean is 1.9103.

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

Given:

• $P(A) = \frac{1}{3}$
• $P(B) = \frac{3}{5}$
• $P(A \cap B) = \frac{1}{15}$

Formula:

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

Calculation:

$$P(A \cup B) = \frac{1}{3} + \frac{3}{5} - \frac{1}{15}$$

Common denominator is $15$:

$$P(A \cup B) = \frac{5 + 9 - 1}{15} = \frac{13}{15} \approx \mathbf{0.8667}$$

Final Answer: $P(A \cup B) = \frac{13}{15}$ (or 0.8667).

Calculation of Percentile Coefficient of Kurtosis:

Given:

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

Formula:

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

Calculation:

$$QD = \frac{52 - 28}{2} = \frac{24}{2} = 12$$

$$P_{90} - P_{10} = 61 - 17 = 44$$

$$K = \frac{12}{44} = \frac{3}{11} \approx \mathbf{0.2727}$$

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

Non-Random (Non-Probability) Sampling Methods:

1. Convenience Sampling: Selecting units that are easiest to reach and immediately accessible to the researcher.
2. Purposive / Judgmental Sampling: Selecting sample members deliberately based on the expert judgment of the researcher.
3. Quota Sampling: Segmenting the population into categories and gathering non-random data until pre-assigned category quotas are filled.
4. Snowball (Referral) Sampling: Existing participants recruit future subjects from among their acquaintances, useful for hidden or rare populations.

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 $f_2$.

Expenditure	Mid-point ($m$)	Frequency ($f$)	$fm$
0 – 20	10	14	140
20 – 40	30	$f_1$	$30f_1$
40 – 60	50	27	1,350
60 – 80	70	$f_2$	$70f_2$
80 – 100	90	15	1,350
Total	—	$N = 100$	$\sum fm = 2,840 + 30f_1 + 70f_2$

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Step 1: Total Frequency Equation

$$14 + f_1 + 27 + f_2 + 15 = 100$$

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

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Step 2: Mean Equation

$$\bar{X} = \frac{\sum fm}{N} \implies 48 = \frac{2840 + 30f_1 + 70f_2}{100}$$

$$4,800 = 2,840 + 30f_1 + 70f_2$$

$$30f_1 + 70f_2 = 1,960$$

Divide by 10:

$$3f_1 + 7f_2 = 196 \quad \text{--- (2)}$$

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Step 3: Solve Simultaneously

Multiply Equation (1) by 3:

$$3f_1 + 3f_2 = 132 \quad \text{--- (3)}$$

Subtract Equation (3) from Equation (2):

$$(3f_1 + 7f_2) - (3f_1 + 3f_2) = 196 - 132$$

$$4f_2 = 64 \implies f_2 = \frac{64}{4} = \mathbf{16}$$

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

$$f_1 + 16 = 44 \implies f_1 = 44 - 16 = \mathbf{28}$$

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

95% Confidence Limits for Population Mean Bulb Life:

Given Data:

• Sample size ($n$) = $500$ (Large sample, $n \ge 30$)
• Sample mean ($\bar{X}$) = $1400 \text{ hours}$
• Sample standard deviation ($s$) = $30 \text{ hours}$
• Confidence level = $95\% \implies Z_{\alpha/2} = 1.96$

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Step 1: Standard Error of Mean ($SE$)

$$SE_{\bar{X}} = \frac{s}{\sqrt{n}} = \frac{30}{\sqrt{500}} = \frac{30}{22.3607} \approx 1.3416 \text{ hours}$$

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Step 2: Calculate Confidence Limits

$$\text{Margin of Error } (E) = Z_{\alpha/2} \times SE_{\bar{X}} = 1.96 \times 1.3416 = 2.6295 \text{ hours}$$

$$\text{Confidence Limits} = \bar{X} \pm E$$

• Lower Limit: $1400 - 2.6295 = \mathbf{1397.37 \text{ hours}}$
• Upper Limit: $1400 + 2.6295 = \mathbf{1402.63 \text{ hours}}$

Final Answer: The 95% confidence interval for the true population mean life is 1397.37 hours to 1402.63 hours ($[1397.37, 1402.63]$).

Coefficient of Quartile Deviation of Household Expenditure:

Cumulative Frequency Table:

Expenditure (Rs '00)	Households ($f$)	Cumulative Frequency ($cf$)
30 – 50	54	54
50 – 70	100	154
70 – 90	140	294
90 – 110	300	594
110 – 130	230	824
130 – 150	125	949
Above 150	51	1000
Total	$N = 1000$	—

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Step 1: Lower Quartile ($Q_1$)

$$\text{Position of } Q_1 = \frac{N}{4} = \frac{1000}{4} = 250^{\text{th}} \implies \text{Class: 70 – 90}$$

$$Q_1 = 70 + \left(\frac{250 - 154}{140}\right) \times 20 = 70 + \left(\frac{96}{140}\right) \times 20 = 70 + 13.71 = \mathbf{83.71}$$

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Step 2: Upper Quartile ($Q_3$)

$$\text{Position of } Q_3 = \frac{3N}{4} = \frac{3000}{4} = 750^{\text{th}} \implies \text{Class: 110 – 130}$$

$$Q_3 = 110 + \left(\frac{750 - 594}{230}\right) \times 20 = 110 + \left(\frac{156}{230}\right) \times 20 = 110 + 13.57 = \mathbf{123.57}$$

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Step 3: Coefficient of Quartile Deviation

$$\text{Coefficient of } QD = \frac{Q_3 - Q_1}{Q_3 + Q_1} = \frac{123.57 - 83.71}{123.57 + 83.71} = \frac{39.86}{207.28} \approx \mathbf{0.1923}$$

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:

Marks	Frequency ($f$)	Cumulative Frequency ($cf$)
0 – 10	7	7
10 – 20	8	15
20 – 30	13	28
30 – 40	29	57
40 – 50	35	92
50 – 60	9	101
60 – 70	3	104
Total	$N = 104$	—

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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

$$P_{90} = L + \left(\frac{\frac{90N}{100} - cf}{f}\right) \times h$$

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

$$P_{90} = 50 + \left(\frac{93.6 - 92}{9}\right) \times 10 = 50 + \left(\frac{1.6}{9}\right) \times 10 = 50 + 1.78 = \mathbf{51.78}$$

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

Fitting a Binomial Distribution to Dice Experiment:

Problem Parameters:

• Number of dice thrown simultaneously ($n$) = $5$
• Total number of trials ($N$) = $96$
• Success: Getting face 4, 5, or 6 on a single die.
  $$p = \frac{3}{6} = \frac{1}{2} = 0.5, \quad q = 1 - p = \frac{1}{2} = 0.5$$

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Theoretical Binomial Probability:

$$P(X = x) = \binom{5}{x} p^x q^{5-x} = \binom{5}{x} \left(\frac{1}{2}\right)^5 = \frac{\binom{5}{x}}{32}$$

Expected frequency for each outcome:

$$f_e(x) = N \times P(X = x) = 96 \times \frac{\binom{5}{x}}{32} = 3 \times \binom{5}{x}$$

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Fitting Table:

Successes ($x$)	$\binom{5}{x}$	Observed Frequency ($f_o$)	Expected Frequency $f_e = 3 \binom{5}{x}$
0	1	2	$3 \times 1 = \mathbf{3}$
1	5	10	$3 \times 5 = \mathbf{15}$
2	10	26	$3 \times 10 = \mathbf{30}$
3	10	35	$3 \times 10 = \mathbf{30}$
4	5	15	$3 \times 5 = \mathbf{15}$
5	1	8	$3 \times 1 = \mathbf{3}$
Total	32	96	96

Conclusion: The fitted binomial distribution frequencies are: 3, 15, 30, 30, 15, 3.

Committee Selection Probability (Combinations):

Group Composition:

• Number of boys = $8$
• Number of girls = $7$
• Total persons = $8 + 7 = 15$
• Committee size to choose = $5$

Total possible committee combinations ($n(S)$):

$$n(S) = \binom{15}{5} = \frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} = 3,003$$

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a. Probability that All 5 Members are Boys:

Number of ways to choose 5 boys from 8:

$$n(A) = \binom{8}{5} = \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$$

$$P(\text{All Boys}) = \frac{\binom{8}{5}}{\binom{15}{5}} = \frac{56}{3003} = \frac{8}{429} \approx \mathbf{0.01865} \quad (\text{or } 1.87\%)$$

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b. Probability that All 5 Members are Girls:

Number of ways to choose 5 girls from 7:

$$n(B) = \binom{7}{5} = \binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21$$

$$P(\text{All Girls}) = \frac{\binom{7}{5}}{\binom{15}{5}} = \frac{21}{3003} = \frac{1}{143} \approx \mathbf{0.00699} \quad (\text{or } 0.70\%)$$

Final Answer:

• a. $P(\text{All Boys}) =$ 0.0187 (or $\frac{8}{429}$)
• b. $P(\text{All Girls}) =$ 0.0070 (or $\frac{1}{143}$)

Percentile Coefficient of Kurtosis for Nepal Bank Limited Employees:

Cumulative Frequency Table:

Income (Rs '000)	Employees ($f$)	Cumulative Frequency ($cf$)
20 – 30	7	7
30 – 40	10	17
40 – 50	20	37
50 – 60	18	55
60 – 70	7	62
Total	$N = 62$	—

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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
  $$Q_1 = 30 + \left(\frac{15.5 - 7}{10}\right) \times 10 = 30 + 8.5 = \mathbf{38.5}$$
• $Q_3$ position: $\frac{3N}{4} = \frac{186}{4} = 46.5^{\text{th}} \implies$ Class 50 – 60
  $$Q_3 = 50 + \left(\frac{46.5 - 37}{18}\right) \times 10 = 50 + \left(\frac{9.5}{18}\right) \times 10 = 50 + 5.28 = \mathbf{55.28}$$
• Semi-Interquartile Range ($QD$):
  $$QD = \frac{Q_3 - Q_1}{2} = \frac{55.28 - 38.5}{2} = \frac{16.78}{2} = \mathbf{8.39}$$

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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
  $$P_{10} = 20 + \left(\frac{6.2 - 0}{7}\right) \times 10 = 20 + 8.86 = \mathbf{28.86}$$
• $P_{90}$ position: $\frac{90 \times 62}{100} = 55.8^{\text{th}} \implies$ Class 60 – 70
  $$P_{90} = 60 + \left(\frac{55.8 - 55}{7}\right) \times 10 = 60 + 1.14 = \mathbf{61.14}$$

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Step 3: Compute Kurtosis ($K$)

$$K = \frac{QD}{P_{90} - P_{10}} = \frac{8.39}{61.14 - 28.86} = \frac{8.39}{32.28} \approx \mathbf{0.2599}$$

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 ($Z$)

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

Calculate Standard Error ($SE$):

$$\frac{s_1^2}{n_1} = \frac{5^2}{150} = \frac{25}{150} = 0.1667$$

$$\frac{s_2^2}{n_2} = \frac{4^2}{100} = \frac{16}{100} = 0.1600$$

$$SE = \sqrt{0.1667 + 0.1600} = \sqrt{0.3267} \approx 0.5715$$

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

$$Z_{\text{cal}} = \frac{45 - 43}{0.5715} = \frac{2}{0.5715} = \mathbf{3.4995}$$

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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

$$\sum X_A = 250 + 255 + 280 + 290 + 295 + 300 = 1,670$$

$$\bar{X}_A = \frac{1670}{6} = \mathbf{278.33}$$

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

$$(250 - 278.33)^2 + \dots + (300 - 278.33)^2 = 802.78 + 544.44 + 2.78 + 136.11 + 277.78 + 469.44 = 2,233.33$$

$$\sigma_A = \sqrt{\frac{2233.33}{6}} = \sqrt{372.22} = \mathbf{19.29}$$

$$CV_A = \frac{\sigma_A}{\bar{X}_A} \times 100 = \frac{19.29}{278.33} \times 100 = \mathbf{6.93\%}$$

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Step 2: Horse B Calculations

$$\sum X_B = 280 + 282 + 290 + 295 + 298 + 295 = 1,740$$

$$\bar{X}_B = \frac{1740}{6} = \mathbf{290.00}$$

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

$$(-10)^2 + (-8)^2 + (0)^2 + (5)^2 + (8)^2 + (5)^2 = 100 + 64 + 0 + 25 + 64 + 25 = 278.00$$

$$\sigma_B = \sqrt{\frac{278}{6}} = \sqrt{46.33} = \mathbf{6.81}$$

$$CV_B = \frac{\sigma_B}{\bar{X}_B} \times 100 = \frac{6.81}{290} \times 100 = \mathbf{2.35\%}$$

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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.

Normal Distribution Probabilities for Student Test Marks:

Given Parameters:

• Mean ($\mu$) = $60$
• Standard Deviation ($\sigma$) = $5$
• Variable $X \sim N(60, 25)$

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a. Percentage of Students Scoring More Than 65 Marks:

Convert to $Z$-score:

$$Z = \frac{X - \mu}{\sigma} = \frac{65 - 60}{5} = \frac{5}{5} = +1.00$$

$$P(X > 65) = P(Z > 1.00) = 0.5000 - P(0 \le Z \le 1.00)$$

From standard normal area tables:

$$P(0 \le Z \le 1.00) = 0.3413$$

$$P(X > 65) = 0.5000 - 0.3413 = \mathbf{0.1587} \implies \mathbf{15.87\%}$$

Answer: 15.87% of students scored more than 65 marks.

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b. Percentage of Students Scoring Between 50 and 65 Marks:

Convert both bounds to $Z$-scores:

$$Z_1 = \frac{50 - 60}{5} = \frac{-10}{5} = -2.00$$

$$Z_2 = \frac{65 - 60}{5} = \frac{5}{5} = +1.00$$

$$P(50 \le X \le 65) = P(-2.00 \le Z \le 1.00) = P(0 \le Z \le 2.00) + P(0 \le Z \le 1.00)$$

From standard normal area tables:

• $P(0 \le Z \le 2.00) = 0.4772$
• $P(0 \le Z \le 1.00) = 0.3413$

$$P(50 \le X \le 65) = 0.4772 + 0.3413 = \mathbf{0.8185} \implies \mathbf{81.85\%}$$

Answer: 81.85% of students scored between 50 and 65 marks.