Business Statistics

Coefficient of Quartile Deviation:

Given:

• Lower Quartile ($Q_1$) = $40$
• Upper Quartile ($Q_3$) = $70$

Formula:

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

Calculation:

$$\text{Coefficient of } QD = \frac{70 - 40}{70 + 40} = \frac{30}{110} = \frac{3}{11} \approx \mathbf{0.2727} \quad (\text{or } 27.27\%)$$

Final Answer: The coefficient of quartile deviation is 0.2727.

Calculation of Standard Deviation ($\sigma$):

Given:

• $\text{Mean } (\bar{X}) = 20$
• $\text{Coefficient of Variation } (CV) = 30\%$

Formula:

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

Calculation:

$$30 = \frac{\sigma}{20} \times 100$$

$$30 = 5\sigma \implies \sigma = \frac{30}{5} = \mathbf{6}$$

Final Answer: The standard deviation is 6.

Calculation of Coefficient of Kurtosis ($K$):

Given:

• Quartile Deviation ($QD$) = $20$
• $10^{\text{th}}$ Percentile ($P_{10}$) = $30$
• $90^{\text{th}}$ Percentile ($P_{90}$) = $90$

Formula:

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

Calculation:

$$K = \frac{20}{90 - 30} = \frac{20}{60} = \frac{1}{3} \approx \mathbf{0.3333}$$

Final Answer: The coefficient of kurtosis is 0.3333 (since $0.3333 > 0.263$, the distribution is leptokurtic).

Probability of Drawing Two Red Balls:

Bag Contents:

• Red balls = $3$
• Black balls = $2$
• White balls = $5$
• Total balls = $3 + 2 + 5 = 10$

Number of ways to draw 2 balls from 10:

$$n(S) = \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45$$

Number of ways to draw 2 red balls from 3:

$$n(E) = \binom{3}{2} = 3$$

$$P(\text{both red}) = \frac{\binom{3}{2}}{\binom{10}{2}} = \frac{3}{45} = \frac{1}{15} \approx \mathbf{0.0667} \quad (\text{or } 6.67\%)$$

Final Answer: The probability of getting both red balls is $\frac{1}{15}$ (or 0.0667).

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

Given:

• $\text{Cov}(X, Y) = 18$
• $\text{Var}(X) = \sigma_X^2 = 16 \implies \sigma_X = 4$
• $\text{Var}(Y) = \sigma_Y^2 = 81 \implies \sigma_Y = 9$

Formula:

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

Calculation:

$$r = \frac{18}{4 \times 9} = \frac{18}{36} = \mathbf{+0.50}$$

Final Answer: The correlation coefficient between $X$ and $Y$ is +0.50 (moderate positive correlation).

Regression Equation of $Y$ on $X$:

Given:

• $\bar{X} = 20$, $\bar{Y} = 15$
• $\sigma_X = 4$, $\sigma_Y = 3$
• $r = 0.7$

Step 1: Calculate Regression Coefficient $b_{YX}$

$$b_{YX} = r \times \frac{\sigma_Y}{\sigma_X} = 0.7 \times \frac{3}{4} = 0.7 \times 0.75 = \mathbf{0.525}$$

Step 2: Formulate Regression Equation

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

$$Y - 15 = 0.525(X - 20)$$

$$Y - 15 = 0.525X - 10.5$$

$$\mathbf{Y = 4.5 + 0.525X}$$

Final Answer: The regression equation of $Y$ on $X$ is $Y = 4.5 + 0.525X$.

Standard Error of Proportion of Bad Oranges:

Given:

• Sample size ($n$) = $400$
• Bad oranges ($x$) = $70$
• Sample proportion of bad oranges ($p$) = $\frac{70}{400} = 0.175$
• $q = 1 - p = 1 - 0.175 = 0.825$

Formula:

$$SE_p = \sqrt{\frac{p \cdot q}{n}}$$

Calculation:

$$SE_p = \sqrt{\frac{0.175 \times 0.825}{400}} = \sqrt{\frac{0.144375}{400}} = \sqrt{0.00036094} \approx \mathbf{0.0190}$$

Final Answer: The standard error of the proportion is 0.0190 (or 1.90%).

Calculation of Mode from Empirical Relationship:

Given:

• $\text{Mean} = 30$
• $\text{Median} = 28$

Empirical Formula:

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

Calculation:

$$\text{Mode} = 3(28) - 2(30) = 84 - 60 = \mathbf{24}$$

Final Answer: The value of the mode is 24.

Binomial Probability $P(X = 2)$:

Given:

• $n = 4$
• $p = 0.45 \implies q = 1 - 0.45 = 0.55$
• $x = 2$

Formula:

$$P(X = x) = \binom{n}{x} p^x q^{n-x}$$

Calculation:

$$\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$$

$$(0.45)^2 = 0.2025$$

$$(0.55)^2 = 0.3025$$

$$P(X = 2) = 6 \times 0.2025 \times 0.3025 = 6 \times 0.061256 = \mathbf{0.3675}$$

Final Answer: $P(X = 2) =$ 0.3675 (or 36.75%).

Types of Random (Probability) Sampling Techniques:

1. Simple Random Sampling (SRS): Lottery method or random number table.
2. Stratified Random Sampling: Proportionate or disproportionate sampling across distinct strata.
3. Systematic Random Sampling: Selecting every $k^{\text{th}}$ item from a sequential list.
4. Cluster / Area Sampling: Sampling naturally clustered units or geographical zones.
5. Multi-Stage Sampling: Successive random sampling conducted in multiple stages.

95% Confidence Interval for Mean Article Diameter:

Given Data:

• Sample size ($n$) = $100$ ($n \ge 30$, large sample)
• Sample mean ($\bar{X}$) = $0.354$
• Sample standard deviation ($s$) = $0.048$
• Confidence level = $95\% \implies Z_{\alpha/2} = 1.96$

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

$$SE_{\bar{X}} = \frac{s}{\sqrt{n}} = \frac{0.048}{\sqrt{100}} = \frac{0.048}{10} = 0.0048$$

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Step 2: Margin of Error ($E$)

$$E = Z_{\alpha/2} \times SE_{\bar{X}} = 1.96 \times 0.0048 = 0.009408$$

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Step 3: Compute Confidence Limits

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

• Lower Limit: $0.354 - 0.009408 = \mathbf{0.3446}$
• Upper Limit: $0.354 + 0.009408 = \mathbf{0.3634}$

Final Answer: The 95% confidence interval for the true population mean diameter is [0.3446, 0.3634].

Calculation of Corrected Mean:

Given Data:

• Number of observations ($n$) = $100$
• Incorrect mean ($\bar{X}$) = $40$
• Incorrect values included: $30$ and $72$
• Correct values to include: $3$ and $27$

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Step 1: Compute Incorrect Sum ($\sum X$)

$$\text{Incorrect } \sum X = n \times \bar{X} = 100 \times 40 = 4,000$$

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Step 2: Compute Corrected Sum

$$\text{Correct } \sum X = \text{Incorrect } \sum X - (\text{Wrong Values}) + (\text{Correct Values})$$

$$\text{Correct } \sum X = 4000 - (30 + 72) + (3 + 27)$$

$$\text{Correct } \sum X = 4000 - 102 + 30 = \mathbf{3,928}$$

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Step 3: Calculate Corrected Mean

$$\text{Correct Mean } (\bar{X}_{\text{corr}}) = \frac{\text{Correct } \sum X}{n} = \frac{3928}{100} = \mathbf{39.28}$$

Final Answer: The corrected mean of the observations is 39.28.

Poisson Probabilities for $\lambda = 3$:

Given:

• Poisson mean parameter $\lambda = 3$

Poisson PMF Formula:

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

Given $e^{-3} \approx 0.049787$:

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a. Calculation of $P(X = 0)$:

$$P(X = 0) = \frac{e^{-3} \cdot 3^0}{0!} = \frac{0.049787 \times 1}{1} = \mathbf{0.0498} \quad (\text{or } 4.98\%)$$

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b. Calculation of $P(X = 1)$:

$$P(X = 1) = \frac{e^{-3} \cdot 3^1}{1!} = \frac{0.049787 \times 3}{1} = \mathbf{0.1494} \quad (\text{or } 14.94\%)$$

Final Answer:

• $P(X = 0) =$ 0.0498
• $P(X = 1) =$ 0.1494

Comparison of Uniformity in Running Life: Samsung vs. Apple:

Given Data ($n = 50$ for both brands):

• Samsung:
  • Mean life ($\bar{X}_1$) = $11 \text{ years}$
  • Variance ($\sigma_1^2$) = $25 \implies \sigma_1 = \sqrt{25} = 5 \text{ years}$
• Apple:
  • Mean life ($\bar{X}_2$) = $12 \text{ years}$
  • Variance ($\sigma_2^2$) = $36 \implies \sigma_2 = \sqrt{36} = 6 \text{ years}$

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Step 1: Compute Coefficient of Variation (CV)

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

• For Samsung:

  $$CV_1 = \frac{5}{11} \times 100 = \mathbf{45.45\%}$$

• For Apple:

  $$CV_2 = \frac{6}{12} \times 100 = \mathbf{50.00\%}$$

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Conclusion:

• A lower Coefficient of Variation indicates greater consistency, stability, and uniformity.
• Since $CV_{\text{Samsung}} (45.45\%) < CV_{\text{Apple}} (50.00\%)$, Samsung shows greater uniformity in running life compared to Apple.

Appropriate Measure of Central Tendency for Open-Ended Income:

Choice of Measure and Support:

• The given income distribution features open-ended terminal classes (“Below 1000” and “5000 & above”).
• The arithmetic mean cannot be determined without arbitrary assumptions about class limits. The Median is the most appropriate measure because it is a positional average independent of extreme or undefined boundary values.

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Cumulative Frequency Table:

(Using continuous class boundaries with correction factor $0.5$)

Monthly Income (Rs)	Families ($f$)	Cumulative Frequency ($cf$)
Below 999.5	50	50
999.5 – 1999.5	500	550
1999.5 – 2999.5	555	1,105
2999.5 – 3999.5	100	1,205
3999.5 – 4999.5	30	1,235
4999.5 & above	15	1,250
Total	$N = 1,250$	—

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Step 1: Locate Median Class

$$\text{Position of Median} = \frac{N}{2} = \frac{1250}{2} = 625^{\text{th}} \text{ item}$$

The cumulative frequency just greater than 625 is 1,105, which falls in class 1999.5 – 2999.5.

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Step 2: Compute Median

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

Where $L = 1999.5$, $cf = 550$, $f = 555$, $h = 1000$:

$$M_d = 1999.5 + \left(\frac{625 - 550}{555}\right) \times 1000 = 1999.5 + \left(\frac{75}{555}\right) \times 1000 = 1999.5 + 135.14 = \mathbf{Rs \; 2,134.64}$$

Final Answer: The appropriate measure is the Median, and its value is Rs 2,134.64.

Five-Number Summary and Distribution Shape:

Step 1: Arrange Data in Ascending Order

Given data ($n = 11$): 39, 26, 15, 8, 70, 11, 45, 60, 20, 32, 52

Sorted order: 8, 11, 15, 20, 26, 32, 39, 45, 52, 60, 70

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Step 2: Calculate Five-Number Summary

1. Minimum ($Min$): 8
2. First Quartile ($Q_1$):
   $$\text{Position} = \left(\frac{11 + 1}{4}\right)^{\text{th}} = 3^{\text{rd}} \text{ value} = \mathbf{15}$$
3. Median ($Q_2$):
   $$\text{Position} = \left(\frac{11 + 1}{2}\right)^{\text{th}} = 6^{\text{th}} \text{ value} = \mathbf{32}$$
4. Third Quartile ($Q_3$):
   $$\text{Position} = \left(\frac{3 \times 12}{4}\right)^{\text{th}} = 9^{\text{th}} \text{ value} = \mathbf{52}$$
5. Maximum ($Max$): 70

Five-Number Summary: $[8, 15, 32, 52, 70]$

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Step 3: Comment on the Shape of the Distribution

• Left quartile distance: $Q_2 - Q_1 = 32 - 15 = 17$
• Right quartile distance: $Q_3 - Q_2 = 52 - 32 = 20$
• Bowley’s Skewness:
  $$Sk_B = \frac{Q_3 + Q_1 - 2Q_2}{Q_3 - Q_1} = \frac{52 + 15 - 64}{52 - 15} = \frac{3}{37} \approx \mathbf{+0.081}$$

Conclusion: Since $(Q_3 - Q_2) > (Q_2 - Q_1)$ ($20 > 17$) and $Sk_B > 0$, the distribution exhibits a slight positive skewness (skewed to the right).

Lowest Income of the Richest 10% of People ($P_{90}$):

Concept:

The cutoff separating the lowest income of the richest 10% corresponds to the $90^{\text{th}}$ Percentile ($P_{90}$).

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Cumulative Frequency Table:

(Converted to continuous class boundaries)

Income (Rs '000)	True Boundaries	Persons ($f$)	Cumulative Frequency ($cf$)
0 – 4	0.0 – 4.5	160	160
5 – 9	4.5 – 9.5	200	360
10 – 14	9.5 – 14.5	430	790
15 – 19	14.5 – 19.5	140	930
20 – 24	19.5 – 24.5	70	1000
Total	—	$N = 1000$	—

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Step 1: Locate $P_{90}$ Class

$$\text{Position} = \frac{90 \times N}{100} = \frac{90 \times 1000}{100} = 900^{\text{th}} \text{ item}$$

The cumulative frequency exceeding 900 is 930, so the $P_{90}$ Class is 14.5 – 19.5.

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Step 2: Percentile Interpolation Formula

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

Where:

• $L = 14.5$
• $cf = 790$
• $f = 140$
• $h = 5$

$$P_{90} = 14.5 + \left(\frac{900 - 790}{140}\right) \times 5 = 14.5 + \left(\frac{110}{140}\right) \times 5 = 14.5 + 3.9286 = \mathbf{18.43 \text{ (in '000 Rs)}}$$

Final Answer: The lowest income of the richest 10% of the people is Rs 18,429 (or 18.43 thousand Rs).

Hypothesis Testing: Difference in Sales Between Two Salesmen:

Step 1: Summary of Sample Statistics

• Salesman A: $n_1 = 20$, $\bar{X}_1 = 170$, $s_1 = 20$
• Salesman B: $n_2 = 18$, $\bar{X}_2 = 205$, $s_2 = 25$
• Level of significance ($\alpha$) = $0.05$ (Two-tailed test)

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Step 2: Formulate Hypotheses

• Null Hypothesis ($H_0$): $\mu_1 = \mu_2$ (There is no significant difference in average sales between salesmen A and B).
• Alternative Hypothesis ($H_1$): $\mu_1 \ne \mu_2$ (There is a significant difference in average sales).

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Step 3: Compute Pooled Variance and Test Statistic ($t$)

Degrees of freedom ($df$) = $n_1 + n_2 - 2 = 20 + 18 - 2 = 36$.

Pooled sample variance ($s_p^2$):

$$s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2} = \frac{19(400) + 17(625)}{36} = \frac{7600 + 10625}{36} = \frac{18225}{36} = 506.25$$

$$s_p = \sqrt{506.25} = 22.5$$

Standard Error ($SE$):

$$SE = s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}} = 22.5 \sqrt{\frac{1}{20} + \frac{1}{18}} = 22.5 \sqrt{0.05 + 0.05556} = 22.5 \times 0.3249 = 7.310$$

Calculate $t$:

$$t_{\text{cal}} = \frac{\bar{X}_1 - \bar{X}_2}{SE} = \frac{170 - 205}{7.310} = \frac{-35}{7.310} = -4.788 \implies |t_{\text{cal}}| = \mathbf{4.79}$$

(Using large-sample approximation $Z$: $Z = \frac{-35}{\sqrt{20 + 34.72}} = \frac{-35}{7.397} = -4.73$).

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Step 4: Decision and Conclusion

• The critical value for two-tailed test at $\alpha = 0.05$ ($df = 36$) is $t_{0.05} \approx 2.028$ (or $Z_{0.05} = 1.96$).
• Since $|t_{\text{cal}}| = 4.79 > 2.028$, we reject the null hypothesis $H_0$.

Conclusion: There is a statistically significant difference in the average sales of the two salesmen at the 5% level of significance (Salesman B achieves significantly higher average sales).

Karl Pearson’s Coefficient of Skewness for 100 Families:

Calculation Table:

Income (Rs '00)	Mid-point ($m$)	Families ($f$)	$d = \frac{m - 50}{20}$	$fd$	$fd^2$
0 – 20	10	13	-2	-26	52
20 – 40	30	21	-1	-21	21
40 – 60	50	27	0	0	0
60 – 80	70	23	1	23	23
80 – 100	90	16	2	32	64
Total	—	$N = 100$	—	$\sum fd = 8$	$\sum fd^2 = 160$

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

$$\bar{X} = A + \left(\frac{\sum fd}{N}\right) \times h = 50 + \left(\frac{8}{100}\right) \times 20 = 50 + 1.6 = \mathbf{51.60 \text{ (in '00 Rs)}}$$

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Step 2: Mode ($M_o$)

Modal class is 40 – 60 ($f_1 = 27, f_0 = 21, f_2 = 23, L = 40, h = 20$):

$$M_o = 40 + \left[\frac{27 - 21}{2(27) - 21 - 23}\right] \times 20 = 40 + \left(\frac{6}{54 - 44}\right) \times 20 = 40 + \left(\frac{6}{10}\right) \times 20 = 40 + 12 = \mathbf{52.00}$$

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Step 3: Standard Deviation ($\sigma$)

$$\sigma = h \times \sqrt{\frac{\sum fd^2}{N} - \left(\frac{\sum fd}{N}\right)^2} = 20 \times \sqrt{\frac{160}{100} - \left(\frac{8}{100}\right)^2} = 20 \times \sqrt{1.60 - 0.0064} = 20 \times \sqrt{1.5936} = 20 \times 1.2624 = \mathbf{25.25}$$

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

$$Sk_p = \frac{\bar{X} - M_o}{\sigma} = \frac{51.60 - 52.00}{25.25} = \frac{-0.40}{25.25} \approx \mathbf{-0.0158}$$

Interpretation: The distribution has a very slight negative skewness ($-0.0158$), indicating it is nearly symmetrical.

Normal Probability of Electronic Tube Lifespan:

Given Parameters:

• Mean lifespan ($\mu$) = $155 \text{ hours}$
• Standard Deviation ($\sigma$) = $19 \text{ hours}$
• Variable $X \sim N(155, 19^2)$

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Step 1: Compute Standard Normal $Z$-Score for $X = 117$

$$Z = \frac{X - \mu}{\sigma} = \frac{117 - 155}{19} = \frac{-38}{19} = \mathbf{-2.00}$$

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Step 2: Probability Calculation

$$P(X < 117) = P(Z < -2.00) = 0.5000 - P(0 \le Z \le 2.00)$$

From standard normal area tables:

$$P(0 \le Z \le 2.00) = 0.4772$$

$$P(X < 117) = 0.5000 - 0.4772 = \mathbf{0.0228} \quad (\text{or } 2.28\%)$$

Final Answer: The probability that the life of a randomly selected electronic tube is less than 117 hours is 0.0228 (or 2.28%).

Promotional Expenses ($X$) and Sales ($Y$): Correlation & Regression Analysis

Data Summary ($n = 7$):

• $X$ = Expenses (Rs '000): $[4, 4, 6, 10, 10, 12, 15]$
• $Y$ = Sales (Rs '00,000): $[16, 20, 18, 24, 20, 22, 25]$

Summations:

• $\sum X = 61 \implies \bar{X} = \frac{61}{7} \approx 8.7143$
• $\sum Y = 145 \implies \bar{Y} = \frac{145}{7} \approx 20.7143$
• $\sum (X - \bar{X})^2 = 107.43$
• $\sum (Y - \bar{Y})^2 = 61.43$
• $\sum (X - \bar{X})(Y - \bar{Y}) = 68.71$

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a. Calculate Two Regression Coefficients:

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

   $$b_{YX} = \frac{\sum (X - \bar{X})(Y - \bar{Y})}{\sum (X - \bar{X})^2} = \frac{68.71}{107.43} = \mathbf{0.6396}$$

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

   $$b_{XY} = \frac{\sum (X - \bar{X})(Y - \bar{Y})}{\sum (Y - \bar{Y})^2} = \frac{68.71}{61.43} = \mathbf{1.1185}$$

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b. Correlation Coefficient ($r$) and Test of Significance:

$$r = \sqrt{b_{YX} \times b_{XY}} = \sqrt{0.6396 \times 1.1185} = \sqrt{0.7154} = \mathbf{+0.8458}$$

Probable Error ($PE_r$):

$$PE_r = 0.6745 \times \frac{1 - r^2}{\sqrt{n}} = 0.6745 \times \frac{1 - 0.7154}{\sqrt{7}} = 0.6745 \times \frac{0.2846}{2.6458} = \mathbf{0.0725}$$

$$6 \times PE_r = 6 \times 0.0725 = 0.435$$

Since $r (0.8458) > 6 \times PE_r (0.435)$, the correlation is statistically significant.

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c. Two Regression Lines & Estimated Sales for Expenses = Rs 20,000 ($X = 20$):

1. Line of $Y$ on $X$:

   $$Y - \bar{Y} = b_{YX}(X - \bar{X})$$
   $$Y - 20.714 = 0.6396(X - 8.714) \implies \mathbf{Y = 15.14 + 0.6396X}$$

2. Line of $X$ on $Y$:

   $$X - 8.714 = 1.1185(Y - 20.714) \implies \mathbf{X = -14.45 + 1.1185Y}$$

Prediction for Promotional Expenses of Rs 20,000 ($X = 20$):

$$\hat{Y} = 15.14 + 0.6396(20) = 15.14 + 12.79 = \mathbf{27.93 \text{ (in '00,000 Rs)}} = \mathbf{Rs \; 2,793,000}$$