Business Statistics

Cost of living index (aggregate expenditure method, base-year quantities)

$\text{CLI}=\frac{\sum p_1q_0}{\sum p_0q_0}\times100=\frac{844}{706}\times100=119.55$

The cost of living in 2024 is 119.55% of that in 2023, an increase of about 19.55%.

Required wage in 2024

To keep the standard of living unchanged, the wage must rise in proportion to the index.

$\text{Wage}_{2024}=37{,}000\times\frac{844}{706}=\text{Rs }44{,}232$

The average wage should be about Rs 44,232 per month (a rise of about Rs 7,232).

(a) Graphical solution of the LPP

Maximize
$Z = 7x + 3y$

Subject to: $x+y \le 5$ $3x+2y \le 12$ $x \ge 0,\ y \ge 0$

Step 1: Convert inequalities into equations

$x+y=5 \quad \Rightarrow \quad (5,0),\ (0,5)$ $3x+2y=12 \quad \Rightarrow \quad (4,0),\ (0,6)$

Step 2: Find intersection of the two lines

Solving $x+y=5$ $3x+2y=12$

From $x+y=5$, $x=5-y$.
Substitute in $3x+2y=12$: $3(5-y)+2y=12$ $15-3y+2y=12$ $y=3$ Then $x=5-3=2$.

So the intersection point is $(2,3)$.

Step 3: Determine the feasible corner points

The feasible region is bounded by: $O(0,0),\ A(4,0),\ B(2,3),\ C(0,5)$

Step 4: Evaluate the objective function at each corner point

The maximum value of $Z$ is $28$, obtained at $(4,0)$.

Answer

$\boxed{\text{Maximum } Z=28 \text{ at } x=4,\ y=0}$

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(b) Probability of selection of Ram and Sita

Let:

• $A$ = event that Ram is selected
• $B$ = event that Sita is selected

Given: $P(A)=\frac14,\qquad P(B)=\frac15$

Assuming the events are independent:

(i) Probability that both will be selected

$P(A\cap B)=P(A)\cdot P(B)$ $=\frac14 \times \frac15=\frac{1}{20}$

(ii) Probability that none will be selected

$P(A'\cap B')=P(A')\cdot P(B')$ $=\left(1-\frac14\right)\left(1-\frac15\right)$ $=\frac34 \times \frac45=\frac{12}{20}=\frac35$

(iii) Probability that at least one will be selected

$P(A\cup B)=1-P(A'\cap B')$ $=1-\frac35=\frac25$

Alternatively: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ $=\frac14+\frac15-\frac{1}{20}$ $=\frac{5+4-1}{20}=\frac{8}{20}=\frac25$

Final Answers

$\boxed{(i)\ \frac{1}{20}\qquad (ii)\ \frac{3}{5}\qquad (iii)\ \frac{2}{5}}$

Given:

• Cost price of a newspaper, (C = \text{Rs } 7)
• Selling price of a newspaper, (S = \text{Rs } 10)
• Profit per sold copy (= 10 - 7 = \text{Rs } 3)
• Unsold copy is worthless, so loss per unsold copy (= \text{Rs } 7)

Let (Q) be the stock level and (D) be the demand.
Payoff for stocking (Q) copies when demand is (D):

[ \text{Payoff} = 10\min(Q,D) - 7Q ]

Expected payoff:

[ E(Q)=\sum P(D)\left[10\min(Q,D)-7Q\right] ]

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Expected payoff table

Sample calculation for (Q=120):

[ E(120)=260(0.17)+360(0.15)+360(0.26)+360(0.27)+360(0.15) ]

[ =44.2+54+93.6+97.2+54=\text{Rs }343 ]

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(a) Optimum stock level

From the table, the maximum expected payoff is at (Q=120).

[ \boxed{\text{Optimum stock level } Q^* = 120 \text{ copies}} ]

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(b) Maximum expected monetary value (EMV)

[ \boxed{\text{Maximum EMV} = \text{Rs }343} ]

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(c) Expected payoff with perfect information (EPPI)

If demand is known in advance, the man should stock exactly the demanded quantity.
Then profit per copy is Rs 3.

Expected demand:

[ E(D)=110(0.17)+120(0.15)+130(0.26)+140(0.27)+150(0.15) ]

[ =18.7+18+33.8+37.8+22.5=130.8 ]

Therefore,

[ \text{EPPI}=3E(D)=3(130.8)=\text{Rs }392.4 ]

[ \boxed{\text{EPPI} = \text{Rs }392.4} ]

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(d) Expected value of perfect information (EVPI)

[ \text{EVPI}=\text{EPPI}-\text{Maximum EMV} ]

[ =392.4-343=\text{Rs }49.4 ]

[ \boxed{\text{EVPI} = \text{Rs }49.4} ]

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Final answers

[ \boxed{ \begin{aligned} \text{(a) Optimum stock level} &= 120 \text{ copies}\ \text{(b) Maximum EMV} &= \text{Rs }343\ \text{(c) EPPI} &= \text{Rs }392.4\ \text{(d) EVPI} &= \text{Rs }49.4 \end{aligned} } ]

Cumulative frequencies: 5, 20, 40, 71, 95, 110, 120. So $N=120$.

(a) Coefficient of quartile deviation

$Q_1$: position $N/4=30$, lying in the class 20 – 30 (cf before = 20, $f=20$).

$Q_1=20+\frac{30-20}{20}\times10=25$

$Q_3$: position $3N/4=90$, lying in the class 40 – 50 (cf before = 71, $f=24$).

$Q_3=40+\frac{90-71}{24}\times10=47.92$

$\text{Coefficient of Q.D.}=\frac{Q_3-Q_1}{Q_3+Q_1}=\frac{47.92-25}{47.92+25}=\frac{22.92}{72.92}=\mathbf{0.314}$

(b) Coefficient of kurtosis (percentile coefficient)

$K=\frac{Q.D.}{P_{90}-P_{10}},\qquad Q.D.=\frac{Q_3-Q_1}{2}=\frac{22.92}{2}=11.46$

$P_{10}$: position $0.1N=12$, in the class 10 – 20 (cf before = 5, $f=15$).

$P_{10}=10+\frac{12-5}{15}\times10=14.67$

$P_{90}$: position $0.9N=108$, in the class 50 – 60 (cf before = 95, $f=15$).

$P_{90}=50+\frac{108-95}{15}\times10=58.67$

$K=\frac{11.46}{58.67-14.67}=\frac{11.46}{44}=\mathbf{0.260}$

(c) Interpretation

For a normal (mesokurtic) curve, $K=0.263$. Here $K=0.260<0.263$, so the distribution is platykurtic, but only slightly. It is flatter than the normal curve and very close to mesokurtic.

Let $X$ = sales and $Y$ = advertisement cost (both Rs '000), $n=10$.

$\bar X=\frac{1490}{10}=149$, $\bar Y=\frac{545}{10}=54.5$

Using deviations from the means: $\sum dx^2=10{,}240$, $\sum dy^2=572.5$, $\sum dx\,dy=2{,}245$.

(a) Regression lines

Regression coefficients

$b_{XY}=\frac{\sum dx\,dy}{\sum dy^2}=\frac{2245}{572.5}=3.921,\qquad b_{YX}=\frac{\sum dx\,dy}{\sum dx^2}=\frac{2245}{10240}=0.219$

Regression of sales on advertisement cost

$X-149=3.921(Y-54.5)\Rightarrow X=3.921Y-64.72$

Regression of advertisement cost on sales

$Y-54.5=0.219(X-149)\Rightarrow Y=0.219X+21.83$

(b) Estimated sales when $Y=63$

$X=3.921(63)-64.72=182.33$

Estimated sales are about Rs 182.3 thousand (Rs 1,82,330).

(c) Karl Pearson's correlation coefficient

$r=\frac{\sum dx\,dy}{\sqrt{\sum dx^2\cdot\sum dy^2}}=\frac{2245}{\sqrt{10240\times572.5}}=\frac{2245}{2421.2}=\mathbf{0.927}$

(Check: $r=\sqrt{b_{XY}\,b_{YX}}=\sqrt{3.921\times0.219}=0.927$.)

Interpretation: $r=+0.927$ shows a very high positive correlation. Sales rise with advertisement cost. $r^2=0.86$, so about 86% of the variation in sales is associated with variation in advertisement cost.