Business Statistics — B.Com DSC Sem III (Paper 7502, 2025)

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Q1

(a) Mean as "Ideal" Average [3]

Different averages serve different purposes (Mode for most-common category, Median for skewed/ranked data), so calling arithmetic mean the sole ideal average is only partly true. Mean uses every value and is algebraically tractable (basis for variance, correlation), which is why it dominates statistical theory — but it's distorted by outliers, unlike Median/Mode.

(b) Revised Mean, SD, CV After Adding 5 Firms [6]

95 firms: mean=₹60 cr, CV=45% ⟹ SD=0.45×60=27 $\Sigma X_1 = 95\times60=5700, \quad \Sigma X_1^2 = 95(27^2+60^2)=95(4329)=411255$ 5 new firms: 22,24,54,42,58 → Σ=200, ΣX²=484+576+2916+1764+3364=9104

$n=100,\quad \Sigma X=5900,\quad \bar X_{new}=\boxed{59}$ $\Sigma X^2=420359 \Rightarrow Var = 4203.59-59^2=722.59 \Rightarrow SD=\boxed{26.88}$ $CV_{new}=\frac{26.88}{59}\times100=\boxed{45.56\%}$

(c) Age Retrenchment — Textile Mill [9]

2000 employees; retain 40% (800). Plan: retrench bottom 10% (200), transfer next 40% (800) to other branches, retire top 10% (200).

Bottom 200 (retrench): 80 from 15-20, need 120 more from 20-25 (250): $120/250\times5=2.4 \Rightarrow$ cutoff = 22.4 yrs

Top 200 (retire early): from top: 20+25+75=120 (65-70,60-65,55-60), need 80 more from 50-55 (150): $80/150\times5=2.67 \Rightarrow$ cutoff = 55−2.67 = 52.33 yrs

Middle 800 to transfer: cumulative between 22.4 and cutoff where cum reaches 1000 (200 retrenched+800 transferred): falls in 35-40, need 45 more of 287 past cum=955: $45/287\times5=0.78\Rightarrow$ 35.78 yrs

$\boxed{\text{Retained in mill: } 35.78 \text{ to } 52.33 \text{ yrs} \qquad \text{Transferred: } 22.4 \text{ to } 35.78 \text{ yrs}}$

Average age before retrenchment: $\Sigma fx/N = 73780/2000 = \boxed{36.89 \text{ yrs}}$ Average age after (retained group, ≈800): using partial-class midpoints ≈ 43.6 yrs (retained band skews toward the middle-older segment, raising the average)

OR (d) Two Dealers — Scooter Sales [9]

Dealer A: 36,32,37,31,33,40,35,38,30,28 → Mean=34, SD=3.63, CV=10.68% Dealer B: 34,31,32,34,35,36,32,36,33,27 → Mean=33, SD=2.57, CV=7.79%

If A's sales ↑5/day and B's sales ↓10/day (constant shifts don't change SD):

• New Mean A = 39, SD unchanged = 3.63
• New Mean B = 23, SD unchanged = 2.57

(e) Moments About Mean — Consumer Income [6]

Income('000₹): 20-30,30-40,...,80-90; Consumers: 5,14,20,25,17,11,8; Mean=55

$m_2=304,\quad m_3=540,\quad m_4=149000$ $\gamma_1=\frac{540}{304^{1.5}}=\frac{540}{5301.9}=\boxed{0.102 \text{ (mild positive skew)}}$ $\beta_2=\frac{149000}{304^2}=\boxed{1.61 \text{ (platykurtic)}}$

(f) CV vs Variance [3]

CV = relative, unit-free (%) measure — compares variability across datasets with different units or scales. Variance = absolute measure in squared original units — can't meaningfully compare, e.g., variability of "salary in ₹" vs "height in cm."

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Q2

(a) Two Projects — Independent Probability [5]

P(Mumbai success)=0.4, P(Bangalore success)=0.7 (independent) $P(\text{Bangalore fails}\mid\text{Mumbai fails}) = P(\text{Bangalore fails}) = 1-0.7=\boxed{0.3}$ $P(\text{at least one succeeds}) = 1-(0.6)(0.3) = 1-0.18=\boxed{0.82}$

(b) Batch Sampling — Accept/Reject/Resample [9]

p(defective)=0.15, n=6. Accept if 0 defective; reject if ≥4; else resample. $P(\text{accept})=P(X=0)=(0.85)^6=\boxed{0.377}$ $P(X\geq4)=P(4)+P(5)+P(6)=0.00549+0.000387+0.0000114=\boxed{0.00589}\;(\text{reject})$ $P(\text{second sample})=1-0.377-0.00589=\boxed{0.617}$

(c) Mutually Exclusive vs Independent Events [4]

• Mutually exclusive: cannot occur together, $P(A\cap B)=0$ (e.g., a coin toss = Head or Tail)
• Independent: occurrence of one doesn't affect the other, $P(A\cap B)=P(A)P(B)$ (e.g., two separate coin tosses)
• Note: two events with non-zero probability cannot be both mutually exclusive and independent simultaneously.

OR (d) Candidates A, B, C — Selection & Bayes [6]

Selection ratio 5:2:3 ⟹ P(A)=0.5, P(B)=0.2, P(C)=0.3; change-implementation probs 0.3, 0.8, 0.5 $P(\text{change})=0.5(.3)+0.2(.8)+0.3(.5)=0.15+0.16+0.15=0.46$ $P(B\mid\text{change})=\frac{0.16}{0.46}=\boxed{0.348} \qquad P(A\mid\text{change})=\frac{0.15}{0.46}=\boxed{0.326}$

(e) Motorway Accidents — Poisson (λ=5/month) [9]

$P(0)=e^{-5}=\boxed{0.00674}\qquad P(\geq1)=1-0.00674=\boxed{0.9933}$ $P(>3)=1-[P(0)+P(1)+P(2)+P(3)]=1-[0.00674+0.0337+0.0842+0.1404]=1-0.265=\boxed{0.735}$

(f) Poisson as Approximation to Binomial [3]

Reasonable when n is large (n≥20) and p is small (p≤0.05), so that $\lambda=np$ stays moderate — used for rare-event counts.

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Q3

(a) Advertising vs Sales — Regression & Coefficient of Determination [10]

Ad Exp(X): 9,12,13,20,15,16,18,22,18,17 → X̄=16 Sales(Y): 67,87,98,73,82,92,82,83,78,88 → Ȳ=83

$\Sigma xy=5,\ \Sigma x^2=136,\ \Sigma y^2=730$

$r=\frac{5}{\sqrt{136\times730}}=\frac{5}{315.1}=\boxed{0.0159}\qquad R^2=\boxed{0.00025}$

Interpretation: virtually no linear relationship in this particular sample — advertising spend explains almost none of the sales variation here (an unusual, weak-data result worth flagging if it appears in your actual paper — double-check the source numbers).

$b_{yx}=\frac{5}{136}=0.0368,\quad b_{xy}=\frac{5}{730}=0.00685$ $\boxed{Y=0.0368X+82.41} \qquad \boxed{X=0.00685Y+15.43}$

If Ad Exp = ₹30,000 (X=30): Y=0.0368(30)+82.41=83.51 (≈ no change from mean 83)

SE of estimate (X on Y): $\sigma_X=3.69,\ \sigma_Y=8.54$; $SE_{X.Y}=3.69\sqrt{1-0.00025}\approx\boxed{3.69}$

(b) Correlation Coefficient & Probable Error — Find N [4]

Given r=0.917, PE=0.034: $PE=0.6745\dfrac{1-r^2}{\sqrt N}$ $0.034=\frac{0.6745(1-0.8409)}{\sqrt N}=\frac{0.1073}{\sqrt N}\Rightarrow \sqrt N=3.156\Rightarrow \boxed{N\approx10}$ Limits for another sample of same size: $r\pm3PE = 0.917\pm0.102 \Rightarrow \boxed{0.815 \text{ to } 1.00}$

OR (c) Standard Error of Estimate — Concept [4]

SE measures the average scatter of actual Y values around the regression line — analogous to SD but for a regression fit. Significance: small SE ⟹ regression line is a good predictor (points cluster tightly); large SE ⟹ weak predictive power despite whatever $r$ says.

(d) Zenith Consulting — Spearman's Rank Correlation [9]

$\Sigma d^2=69.5$; tie corrections: Prod (one pair) = 0.5; Sat (two pairs) = 1.0 → total 1.5

$\rho = 1-\frac{6(69.5+1.5)}{10(99)}=1-\frac{426}{990}=\boxed{0.570 \text{ (moderate positive)}}$

If I's score changes 44→45: 45 still sits in the same rank slot between 40 and 46 as 44 did — ranks (and ρ) are unchanged.

(e) Training Hours vs Productivity — Regression [9]

Lines: $5Y-4X-300=0$ (⟹ $Y=0.8X+60$, $b_{yx}=0.8$) and $6Y-5X-350=0$ (⟹ $X=1.2Y-70$, $b_{xy}=1.2$)

(i) Means: solving simultaneously: $\bar X=50,\ \bar Y=100$

(ii) $r=\sqrt{0.8\times1.2}=\sqrt{0.96}=\boxed{0.980}$

(iii) Var(Y)=25 ⟹ $\sigma_Y=5$. From $b_{xy}=r\sigma_X/\sigma_Y$: $\sigma_X=1.2(5)/0.98=6.12$ $SE_{X.Y}=6.12\sqrt{1-0.96}=6.12(0.2)=\boxed{1.22}$

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Q4

(a) Sunrise Electronics — Straight-Line Trend, Monthly Conversion, Additive Model [14]

Revenue(₹cr) 2017–2024: 125,140,150,145,160,172,165,180 (n=8, even → X=−7,−5,−3,−1,1,3,5,7)

$\Sigma Y=1237,\ \Sigma XY=591,\ \Sigma X^2=168$

$a=\frac{1237}{8}=154.625,\quad b=\frac{591}{168}=3.518$ $\boxed{Y_c=154.625+3.518X \text{ (Origin mid-2020/2021, X-unit = half year)}}$

Trend values 2017–2024: 130.0, 137.0, 144.1, 151.1, 158.1, 165.2, 172.2, 179.3

Monthly trend: annual slope = $2\times3.518=7.036$; monthly = $7.036/12=\boxed{0.586}$ (Dividing the annual slope by 12 converts to a monthly rate; the "÷144" rule applies only when converting a parabolic/quadratic term's X² coefficient, since $12^2=144$.)

Shift origin to 2024 (X'=X−7): $\boxed{Y_c=179.25+3.518X'}$

Additive model — detrended residuals ($Y-T$): −5.0, +2.96, +5.93, −6.11, +1.86, +6.82, −7.21, +0.75 (the leftover S+C+I component after removing the secular trend)

(b) Coefficient of Determination & SE from Variation Split [4]

n=9, Explained variation=24, Unexplained variation=36, Total=60 $R^2=\frac{24}{60}=\boxed{0.4}\qquad SE=\sqrt{\frac{36}{9}}=\sqrt4=\boxed{2}$

OR (c) Parabolic Trend — Production 2019–2024 [9]

Production('000t): 15,17,18,20,23,25 (n=6, X=−5,−3,−1,1,3,5)

$\Sigma Y=118,\ \Sigma XY=70,\ \Sigma X^2=70,\ \Sigma X^2Y=1398,\ \Sigma X^4=1414$

$b=1,\quad 118=6a+70c,\quad 1398=70a+1414c \Rightarrow a=19.25,\ c=0.0357$ $\boxed{Y_c=19.25+X+0.0357X^2}$

Predict 2027 (3 yrs beyond 2024 → X=5+6=11): $Y_c=19.25+11+0.0357(121)=\boxed{34.6\text{ '000t}}$

(d) Time-Series Component Matching [5]

(e) Coefficient of Determination [4]

$R^2$ = proportion of Y's variance explained by X; central to judging model fit. "Always positive" — Agree. $R^2=r^2$, and a square is never negative — this holds regardless of whether the underlying correlation itself is positive or negative.

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Q5

(a) Price Index — Five Commodities [9]

Recovering $Q_0=Exp_0/P_0$ and $P_1=Exp_1/Q_1$:

$\text{Laspeyres}=\frac{5890}{5460}\times100=\boxed{107.88}\quad \text{Paasche}=\frac{6130}{5760}\times100=\boxed{106.42}$ $\text{Fisher}=\sqrt{107.88\times106.42}=\boxed{107.15}$ Ideal index: Fisher's — satisfies both Time Reversal and Factor Reversal tests, geometric-mean-balancing Laspeyres's upward bias and Paasche's downward bias.

(b) Hriday Care Hospital — Normal Distribution [9]

n=4000, Mean=₹2,40,000, Variance=₹62,50,00,000 ⟹ SD=₹25,000

OR (c) Bright Pay Solutions — CoL Index & Salary Increase [9]

$\text{CoL Index}=\frac{1941.7}{15}=\boxed{129.45}$ Salary needed (was ₹45,000 in 2022): $45000\times1.2945=\boxed{₹58{,}253}$ (increase of ≈ ₹13,253, or 29.45%)

(d) Sales Rep — Probability of Selling ≥1 Product [4]

P(A)=0.4, P(B)=0.3, P(A∩B)=0.08 $P(\text{at least one})=0.4+0.3-0.08=\boxed{0.62}$ If assumed independent: $P(A\cap B)$ would be $0.4\times0.3=0.12$ — since the actual given joint probability (0.08) differs from this, A and B are not truly independent in reality.

(e) Poisson Applications & Axiomatic Probability [5]

Poisson used for: rare/discrete events over a fixed interval — customer arrivals/hour, defects/batch, accidents/day, calls/minute.

Axiomatic (Kolmogorov) approach: for any event $A$ in sample space $S$:

1. $0 \le P(A) \le 1$
2. $P(S) = 1$
3. For mutually exclusive $A_1, A_2, \ldots$: $P(A_1\cup A_2\cup\cdots) = \Sigma P(A_i)$

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