Business Statistics

Given LP problem:

Maximize
$Z = 40A + 70B$

Subject to
$3A + 3B \le 36 \quad \Rightarrow \quad A+B \le 12 \quad (1)$ $5A + 2B \le 50 \quad (2)$ $2A + 6B \le 60 \quad \Rightarrow \quad A+3B \le 30 \quad (3)$ $A \ge 0,\ B \ge 0$

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Step 1: Convert constraints into equations and find intercepts

Plotting these lines on a graph and applying (A\ge0,\ B\ge0), the feasible region is bounded by:

• (A)-axis,
• (B)-axis,
• line (5A+2B=50),
• line (A+B=12),
• line (A+3B=30).

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Step 2: Find corner points of the feasible region

The corner points are:

1. (O=(0,0))
2. (P=(10,0)) — intersection of (B=0) and (5A+2B=50)
3. (Q) — intersection of (A+B=12) and (5A+2B=50)

Solving: [ A+B=12 \Rightarrow B=12-A ] [ 5A+2(12-A)=50 ] [ 5A+24-2A=50 ] [ 3A=26 \Rightarrow A=\frac{26}{3} ] [ B=12-\frac{26}{3}=\frac{10}{3} ] So, [ Q=\left(\frac{26}{3},\frac{10}{3}\right) ]

4. (R) — intersection of (A+B=12) and (A+3B=30)

Solving: [ (A+3B)-(A+B)=30-12 ] [ 2B=18 \Rightarrow B=9 ] [ A+9=12 \Rightarrow A=3 ] So, [ R=(3,9) ]

5. (S=(0,10)) — intersection of (A=0) and (A+3B=30)

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Step 3: Evaluate the objective function at each corner point

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Step 4: Conclusion

The maximum value of (Z) occurs at the corner point
[ (A,B)=(3,9) ]

Therefore,

[ \boxed{Z_{\max}=750 \text{ at } A=3,\ B=9} ]

So, the optimal solution is:

[ \boxed{A=3,\quad B=9,\quad Z_{\max}=750} ]

Graphical solution:

Convert the inequalities into equations:

$2x+y=10 \quad \text{and} \quad 3x+2y=12$

Find intercepts:

• For (2x+y=10):
  (x=0 \Rightarrow y=10), and (y=0 \Rightarrow x=5).
  Points: ((0,10)), ((5,0))

• For (3x+2y=12):
  (x=0 \Rightarrow y=6), and (y=0 \Rightarrow x=4).
  Points: ((0,6)), ((4,0))

Test the origin ((0,0)) in both inequalities:

$2(0)+0=0\le 10 \quad \text{and} \quad 3(0)+2(0)=0\le 12$

So the feasible region lies towards the origin.

The feasible region is the triangle bounded by:

$x=0,\quad y=0,\quad 3x+2y=12$

Its corner points are:

$(0,0),\quad (4,0),\quad (0,6)$

Now evaluate the objective function (Z=5x+2y) at each corner point:

The maximum value occurs at ((4,0)).

$\boxed{\text{Maximum } Z=20 \text{ at } x=4,\ y=0}$

(a) Probability that both balls drawn are black

Given:

• Black balls = 4
• Red balls = 4
• Total balls = 4 + 4 = 8
• Number of balls drawn = 2

To find: P(both balls are black)

Formula:

$P(\text{both black}) = \frac{{}^4C_2}{{}^8C_2}$

Substitution:

$P = \frac{\dfrac{4!}{2!\,2!}}{\dfrac{8!}{2!\,6!}} = \frac{6}{28} = \frac{3}{14}$

Alternative method (without replacement):

$P = \frac{4}{8} \times \frac{3}{7} = \frac{12}{56} = \frac{3}{14}$

Result: The probability that both balls drawn are black is $\boxed{\dfrac{3}{14}} \approx 0.2143$.

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(b) Decision by Minimax Regret Criteria

Payoff matrix (rows = courses of action, columns = states of nature):

Step 1: Find the maximum payoff in each state of nature (column maxima)

• Column S₁: max (500, 300, 100) = 500
• Column S₂: max (500, 550, 350) = 550
• Column S₃: max (500, 550, 600) = 600

Step 2: Compute the regret (opportunity loss) for each cell Regret = Column maximum − Payoff of that cell.

Step 3: Apply the minimax regret rule Select the action whose maximum regret is the smallest:

$\text{Minimum of } (100,\ 200,\ 400) = 100$

Decision: The minimum of the maximum regrets is 100, which corresponds to course of action S₁ = 100.

Hence, under the minimax regret criterion, the decision maker should choose S₁ = 100.

Given grouped distributions

Use class midpoints:

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(a) Which company is more consistent?

Consistency is judged by the coefficient of variation:

[ CV=\frac{\sigma}{\bar x}\times 100 ]

Company A

[ \bar x_A=\frac{\sum f_Ax}{N_A}=\frac{314}{50}=6.28 ]

[ \sigma_A=\sqrt{\frac{\sum f_Ax^2}{N_A}-\bar x_A^2} ]

[ \sigma_A=\sqrt{\frac{2250}{50}-(6.28)^2} ]

[ \sigma_A=\sqrt{45-39.4384}=\sqrt{5.5616}=2.3583 ]

[ CV_A=\frac{2.3583}{6.28}\times 100=37.55% ]

Company B

[ \bar x_B=\frac{\sum f_Bx}{N_B}=\frac{354}{52}=6.8077 ]

[ \sigma_B=\sqrt{\frac{\sum f_Bx^2}{N_B}-\bar x_B^2} ]

[ \sigma_B=\sqrt{\frac{2844}{52}-(6.8077)^2} ]

[ \sigma_B=\sqrt{54.6923-46.3447}=\sqrt{8.3476}=2.8892 ]

[ CV_B=\frac{2.8892}{6.8077}\times 100=42.44% ]

Since

[ CV_A=37.55% < CV_B=42.44% ]

Company A is more consistent in profit.
Company B has a higher average profit, but its profit is more variable.

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(b) Combined standard deviation

Combined number of observations:

[ N=N_A+N_B=50+52=102 ]

Combined mean:

[ \bar x_c=\frac{\sum f_Ax+\sum f_Bx}{N} ]

[ \bar x_c=\frac{314+354}{102}=\frac{668}{102}=6.5490 ]

Combined variance:

[ \sigma_c^2=\frac{\sum f_Ax^2+\sum f_Bx^2}{N}-\bar x_c^2 ]

[ \sigma_c^2=\frac{2250+2844}{102}-(6.5490)^2 ]

[ \sigma_c^2=\frac{5094}{102}-42.8897 ]

[ \sigma_c^2=49.9412-42.8897=7.0515 ]

Combined standard deviation:

[ \sigma_c=\sqrt{7.0515}=2.6555 ]

[ \boxed{\sigma_c \approx \text{Rs } 2.6555 \text{ million}} ]

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Final answers

(a) Company A is more consistent because its coefficient of variation is lower.

(b) Combined standard deviation is approximately Rs 2.6555 million.

In the table the last row is a fourth commodity (labelled D below). Divide expenditure by price to find quantities.

(a) Laspeyres' index

$P_{01}^{L} = \frac{\sum p_1q_0}{\sum p_0q_0}\times 100 = \frac{202}{91}\times 100 = 221.98$

(b) Fisher's index

(c) Paasche's index

$P_{01}^{P} = \frac{\sum p_1q_1}{\sum p_0q_1}\times 100 = \frac{199}{92}\times 100 = 216.30$

$P_{01}^{F} = \sqrt{P^{L}\times P^{P}} = \sqrt{221.98\times 216.30} = 219.12$

Time reversal test. The test is satisfied when $P_{01}\times P_{10} = 1$.

$P_{01} = \sqrt{\frac{202}{91}\times\frac{199}{92}}, \qquad P_{10} = \sqrt{\frac{\sum p_0q_1}{\sum p_1q_1}\times\frac{\sum p_0q_0}{\sum p_1q_0}} = \sqrt{\frac{92}{199}\times\frac{91}{202}}$

$P_{01}\times P_{10} = \sqrt{\frac{202}{91}\times\frac{199}{92}\times\frac{92}{199}\times\frac{91}{202}} = \sqrt{1} = 1$

So Fisher's index satisfies the time reversal test.

Factor reversal test. The test is satisfied when $P_{01}\times Q_{01} = \dfrac{\sum p_1q_1}{\sum p_0q_0}$.

$Q_{01} = \sqrt{\frac{\sum q_1p_0}{\sum q_0p_0}\times\frac{\sum q_1p_1}{\sum q_0p_1}} = \sqrt{\frac{92}{91}\times\frac{199}{202}} = 0.9980$

$P_{01}\times Q_{01} = 2.1912\times 0.9980 = 2.1868$

$\frac{\sum p_1q_1}{\sum p_0q_0} = \frac{199}{91} = 2.1868$

Both sides are equal, so Fisher's index also satisfies the factor reversal test. Fisher's index is therefore called the "ideal" index number.

Let income be $X$ and expenditure be $Y$.

Here,
$n=7,\quad \sum X=231,\quad \sum Y=185,\quad \sum XY=6275,\quad \sum X^2=7867,\quad \sum Y^2=5019$

a. Coefficient of correlation

$r=\frac{n\sum XY-\sum X\sum Y}{\sqrt{[n\sum X^2-(\sum X)^2][n\sum Y^2-(\sum Y)^2]}}$

$r=\frac{7(6275)-231(185)}{\sqrt{[7(7867)-231^2][7(5019)-185^2]}}$

$r=\frac{43925-42735}{\sqrt{(55069-53361)(35133-34225)}}$

$r=\frac{1190}{\sqrt{1708\times 908}}=\frac{1190}{\sqrt{1550864}}$

$r=\frac{1190}{1245.337}=0.9556$

So, the coefficient of correlation is 0.9556, showing a high degree of positive correlation between income and expenditure.

b. Two regression coefficients

Regression coefficient of $Y$ on $X$:

$b_{yx}=\frac{n\sum XY-\sum X\sum Y}{n\sum X^2-(\sum X)^2}$

$b_{yx}=\frac{1190}{1708}=0.6967$

Regression coefficient of $X$ on $Y$:

$b_{xy}=\frac{n\sum XY-\sum X\sum Y}{n\sum Y^2-(\sum Y)^2}$

$b_{xy}=\frac{1190}{908}=1.3106$

Hence, the two regression coefficients are:

$b_{yx}=0.6967,\qquad b_{xy}=1.3106$

c. Estimate expenditure when income = Rs 45 thousand

Use regression equation of $Y$ on $X$:

$\bar X=\frac{231}{7}=33,\qquad \bar Y=\frac{185}{7}=26.4286$

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

When $X=45$:

$Y=26.4286+0.6967(45-33)$

$Y=26.4286+0.6967\times 12$

$Y=26.4286+8.3604=34.7890$

Therefore, when income is Rs 45 thousand, the estimated expenditure is:

$\boxed{\text{Rs }34.789\text{ thousand} \approx \text{Rs }34,789}$

Let (X = \text{Year} - 2016). Then (n=9), (\sum X=0).

[ \sum Y=736,\quad \sum X^2=60,\quad \sum XY=317 ]

Least square trend line:

[ Y_c=a+bX ]

Normal equations:

[ \sum Y=na+b\sum X ] [ \sum XY=a\sum X+b\sum X^2 ]

Since (\sum X=0),

[ a=\frac{\sum Y}{n}=\frac{736}{9}=81.7778 ]

[ b=\frac{\sum XY}{\sum X^2}=\frac{317}{60}=5.2833 ]

Hence the trend equation is:

[ Y_c=81.7778+5.2833X ]

or,

[ Y_c=81.7778+5.2833(\text{Year}-2016) ]

Trend values

Estimate for 2021

For 2021,

[ X=2021-2016=5 ]

[ Y_c=\frac{736}{9}+\frac{317}{60}(5) ]

[ Y_c=\frac{3895}{36}=108.1944\ldots ]

[ \boxed{Y_{2021}\approx \text{Rs }108.19\text{ million}} ]

So, the estimated profit for 2021 is approximately Rs 108.19 million.

Given system: $x+y+z=1,\qquad x-2y+3z=4,\qquad 4x+y+z=4$

Using Cramer’s rule: $A=\begin{bmatrix}1&1&1\\1&-2&3\\4&1&1\end{bmatrix},\qquad B=\begin{bmatrix}1\\4\\4\end{bmatrix}$

Determinant of coefficient matrix: $D=\begin{vmatrix}1&1&1\\1&-2&3\\4&1&1\end{vmatrix}$ $=1[(-2)(1)-3(1)]-1[(1)(1)-3(4)]+1[(1)(1)-(-2)(4)]$ $=1(-2-3)-1(1-12)+1(1+8)$ $=-5+11+9=15$

Now, $D_x=\begin{vmatrix}1&1&1\\4&-2&3\\4&1&1\end{vmatrix}=15$

$D_y=\begin{vmatrix}1&1&1\\1&4&3\\4&4&1\end{vmatrix}=-9$

$D_z=\begin{vmatrix}1&1&1\\1&-2&4\\4&1&4\end{vmatrix}=9$

Hence, $x=\frac{D_x}{D}=\frac{15}{15}=1$ $y=\frac{D_y}{D}=\frac{-9}{15}=-\frac35$ $z=\frac{D_z}{D}=\frac{9}{15}=\frac35$

Therefore, the required solution is: $\boxed{x=1,\quad y=-\frac35,\quad z=\frac35}$

Let
(x) = number of article A produced daily
(y) = number of article B produced daily

Linear programming model

Maximize profit:

[ Z = 600x + 1000y ]

Subject to the machine-hour constraints:

[ 4x + 8y \le 24 \quad \text{(Machine M)} ]

[ 8x + 4y \le 24 \quad \text{(Machine N)} ]

[ x \ge 0,\quad y \ge 0 ]

Dividing the constraints by 4:

[ x + 2y \le 6 ]

[ 2x + y \le 6 ]

[ x \ge 0,\quad y \ge 0 ]

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Finding the optimal solution

The feasible region is bounded by:

[ x + 2y = 6 ]

[ 2x + y = 6 ]

[ x = 0,\quad y = 0 ]

The corner points are:

1. ((0,0))
   [ Z = 600(0)+1000(0)=0 ]

2. ((3,0))
   [ Z = 600(3)+1000(0)=1800 ]

3. ((0,3))
   [ Z = 600(0)+1000(3)=3000 ]

4. Intersection of
   [ x+2y=6 ] and
   [ 2x+y=6 ]

Solving:

[ x+2y=6 ]

[ 2x+y=6 ]

From the first equation:

[ x=6-2y ]

Substitute in the second:

[ 2(6-2y)+y=6 ]

[ 12-4y+y=6 ]

[ -3y=-6 ]

[ y=2 ]

Then,

[ x=6-2(2)=2 ]

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

Profit at ((2,2)):

[ Z = 600(2)+1000(2) ]

[ Z = 1200+2000 = 3200 ]

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Conclusion

The maximum profit is obtained at:

[ x=2,\quad y=2 ]

Therefore, the factory should produce daily:

[ \boxed{2 \text{ articles of A and } 2 \text{ articles of B}} ]

Maximum daily profit:

[ \boxed{Rs.\ 3200} ]

Let the class midpoints be (x).

For Model A

[ \bar x_A=\frac{\sum f_Ax}{n_A}=\frac{61000}{60}=1016.67 ]

[ \sigma_A^2=\frac{\sum f_Ax^2}{n_A}-\bar x_A^2 ]

[ =\frac{64920000}{60}-(1016.67)^2 ]

[ =1082000-1033611.11=48388.89 ]

[ \sigma_A=\sqrt{48388.89}=219.97 ]

Coefficient of variation:

[ CV_A=\frac{\sigma_A}{\bar x_A}\times 100 ]

[ =\frac{219.97}{1016.67}\times 100=21.64% ]

For Model B

[ \bar x_B=\frac{\sum f_Bx}{n_B}=\frac{60000}{60}=1000 ]

[ \sigma_B^2=\frac{\sum f_Bx^2}{n_B}-\bar x_B^2 ]

[ =\frac{62400000}{60}-(1000)^2 ]

[ =1040000-1000000=40000 ]

[ \sigma_B=\sqrt{40000}=200 ]

[ CV_B=\frac{200}{1000}\times 100=20% ]

a) Which model has greater uniformity?

A model has greater uniformity if its coefficient of variation is smaller.

[ CV_B=20% < CV_A=21.64% ]

Therefore, Model B has greater uniformity of life.

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b) Combined standard deviation

Combined mean:

[ \bar x_c=\frac{61000+60000}{60+60}=\frac{121000}{120}=1008.33 ]

Let

[ d_A=\bar x_A-\bar x_c=1016.67-1008.33=8.34 ]

[ d_B=\bar x_B-\bar x_c=1000-1008.33=-8.33 ]

Combined variance:

[ \sigma_c^2= \frac{n_A(\sigma_A^2+d_A^2)+n_B(\sigma_B^2+d_B^2)}{n_A+n_B} ]

[ =\frac{60(48388.89+8.34^2)+60(40000+(-8.33)^2)}{120} ]

[ =\frac{60(48388.89+69.56)+60(40000+69.39)}{120} ]

[ =\frac{60(48458.45)+60(40069.39)}{120} ]

[ =\frac{2907507+2404163.4}{120} ]

[ =44263.89 ]

[ \sigma_c=\sqrt{44263.89}=210.39 ]

[ \boxed{\text{Combined standard deviation}=210.39\text{ hours}} ]

Answer:
a) Model B has greater uniformity.
b) Combined standard deviation = 210.39 hours.

Let $X$ be family income and $Y$ food expenditure (both in Rs '000). Use class midpoints and the step deviations $u = \dfrac{X - 650}{100}$ and $v = \dfrac{Y - 37.5}{5}$.

Midpoints of $X$: 450, 550, 650, 750, 850 ($u = -2, -1, 0, 1, 2$). Midpoints of $Y$: 32.5, 37.5, 42.5, 47.5 ($v = -1, 0, 1, 2$). Total frequency $N = 100$.

Row totals for $Y$: 10, 20, 30, 40. Column totals for $X$: 10, 20, 40, 20, 10.

Correlation coefficient

$r = \frac{N\sum fuv - \sum fu\sum fv}{\sqrt{N\sum fu^2 - (\sum fu)^2}\,\sqrt{N\sum fv^2 - (\sum fv)^2}} = \frac{100(-48) - 0}{\sqrt{100(120)}\,\sqrt{100(200) - 100^2}}$

$r = \frac{-4800}{109.54\times 100} = -0.438$

There is a moderate negative correlation between family income and expenditure on food in this locality.

Estimating food expenditure at income Rs 750,000

Means: $\bar X = 650 + 100(0) = 650$ and $\bar Y = 37.5 + 5(1) = 42.5$.

Standard deviations: $\sigma_x = 100\sqrt{120/100} = 109.54$ and $\sigma_y = 5\sqrt{200/100 - 1} = 5$.

The regression coefficient of $Y$ on $X$ is

$b_{yx} = r\frac{\sigma_y}{\sigma_x} = -0.438\times\frac{5}{109.54} = -0.02$

The regression line of $Y$ on $X$ is $Y - 42.5 = -0.02(X - 650)$.

For $X = 750$:

$Y = 42.5 - 0.02(750 - 650) = 42.5 - 2 = 40.5$

The estimated expenditure on food is Rs 40.5 thousand, that is Rs 40,500.

Use class midpoints $m$. Total frequency $N = \sum f = 170$.

$\bar X = \frac{\sum fm}{N} = \frac{11{,}770}{170} = 69.24$

Central moments, from the deviations $m - \bar X$:

$\mu_2 = \frac{\sum f(m-\bar X)^2}{N} = \frac{77{,}400.6}{170} = 455.30$

$\mu_3 = \frac{\sum f(m-\bar X)^3}{N} = \frac{342{,}392.1}{170} = 2{,}014.07$

$\mu_4 = \frac{\sum f(m-\bar X)^4}{N} = \frac{77{,}833{,}187}{170} = 457{,}842.3$

The standard deviation is $\sigma = \sqrt{455.30} = 21.34$.

Coefficient of skewness

$\beta_1 = \frac{\mu_3^2}{\mu_2^3} = \frac{(2014.07)^2}{(455.30)^3} = 0.043, \qquad \gamma_1 = \sqrt{\beta_1} = +0.207$

Coefficient of kurtosis

$\beta_2 = \frac{\mu_4}{\mu_2^2} = \frac{457{,}842.3}{(455.30)^2} = 2.21, \qquad \gamma_2 = \beta_2 - 3 = -0.79$

Comment. Because $\mu_3$ is positive, the distribution is slightly positively skewed. The value 0.207 is small, so it is almost symmetrical, with a slightly longer tail on the higher salary side. Since $\beta_2 = 2.21 < 3$, the distribution is platykurtic, meaning flatter than the normal curve, with salaries spread fairly evenly over the classes rather than peaked at the centre.

Let the middle year 2019 be taken as the origin, so
[ X=\text{Year}-2019 ]

The straight-line trend equation is
[ Y_c=a+bX ]

where
[ a=\frac{\sum Y}{n},\qquad b=\frac{\sum XY}{\sum X^2} ]

Calculation table

Now, [ a=\frac{212}{9}=23.5556 ]

[ b=\frac{19}{60}=0.3167 ]

Hence the trend equation is
[ Y_c=23.5556+0.3167X ] where (X=\text{Year}-2019).

Trend values

Estimate for 2024

For 2024,
[ X=2024-2019=5 ]

[ Y_c=23.5556+0.3167(5) ]

[ Y_c=23.5556+1.5835=25.1391 ]

[ \boxed{Y_c \approx \text{Rs }25.14\text{ million}} ]

Therefore, the estimated profit for the year 2024 is approximately Rs 25.14 million.

Graphical solution of the LPP

Minimize
[ Z=3x+5y ] Subject to
[ x+2y\ge 40 ] [ x+y\ge 30 ] [ x\ge 0,\quad y\ge 0 ]

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Step 1: Convert inequalities into equations

[ x+2y=40 ] [ x+y=30 ]

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Step 2: Find intercepts

For (x+2y=40):

So the line passes through ((0,20)) and ((40,0)).

For (x+y=30):

So the line passes through ((0,30)) and ((30,0)).

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Step 3: Find intersection point

Solve: [ x+2y=40 ] [ x+y=30 ]

Subtracting: [ y=10 ]

Then: [ x+10=30 \Rightarrow x=20 ]

So the intersection point is: [ (20,10) ]

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Step 4: Identify feasible corner points

The feasible region is unbounded, but its corner points are:

[ A(0,30),\quad B(20,10),\quad C(40,0) ]

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Step 5: Evaluate the objective function

[ Z=3x+5y ]

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Conclusion

The minimum value of (Z) is:

[ \boxed{Z_{\min}=110} ]

and it occurs at:

[ \boxed{x=20,\quad y=10} ]

So the optimal solution is:

[ \boxed{(x,y)=(20,10)} ]

with minimum value:

[ \boxed{Z=110} ]

Given:

• Purchase cost per magazine, (C = \text{Rs }10)
• Selling price per magazine, (S = \text{Rs }15)
• Unsold magazines have no salvage value, so loss per unsold copy (= \text{Rs }10)
• Profit per sold copy (= 15 - 10 = \text{Rs }5)

Demand distribution:

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(a) Optimum stock level

For a stock level (Q), expected profit is:

[ E(Q)=15E[\min(Q,D)]-10Q ]

Calculate expected profit for each possible stock level.

Stock (Q = 200)

All demands are at least 200, so sales (=200).

[ E(200)=15(200)-10(200)=3000-2000=\text{Rs }1000 ]

Stock (Q = 210)

[ E(210)=900(0.19)+1050(0.81) ]

[ =171+850.50=\text{Rs }1021.50 ]

Stock (Q = 220)

[ E(220)=800(0.19)+950(0.16)+1100(0.65) ]

[ =152+152+715=\text{Rs }1019 ]

Stock (Q = 230)

[ E(230)=700(0.19)+850(0.16)+1000(0.25)+1150(0.40) ]

[ =133+136+250+460=\text{Rs }979 ]

Stock (Q = 240)

[ E(240)=600(0.19)+750(0.16)+900(0.25)+1050(0.27)+1200(0.13) ]

[ =114+120+225+283.50+156=\text{Rs }898.50 ]

Summary:

The highest expected profit occurs at (Q = 210).

[ \boxed{\text{Optimum stock level} = 210 \text{ copies}} ]

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(b) Maximum expected monetary value (EMV)

The maximum EMV is the expected profit at the optimum stock level (Q=210).

[ \boxed{\text{Maximum EMV} = \text{Rs }1021.50} ]

So the vendor should stock 210 magazines, giving a maximum expected profit of Rs 1,021.50.

Working table (class width $h = 10$, mid-values $m$, $N = 120$)

$\bar{X} = \frac{6{,}480}{120} = 54$

Moments about the mean:

$\mu_2 = \frac{29{,}880}{120} = 249,\quad \mu_3 = \frac{-42{,}240}{120} = -352,\quad \mu_4 = \frac{17{,}051{,}640}{120} = 142{,}097$

$\sigma = \sqrt{249} = 15.78$

(a) Coefficient of skewness

Modal class is 50–60 (highest frequency 30):

$Z = 50 + \frac{30 - 21}{2(30) - 21 - 24} \times 10 = 50 + \frac{9}{15} \times 10 = 56$

Karl Pearson's coefficient:

$Sk_P = \frac{\bar{X} - Z}{\sigma} = \frac{54 - 56}{15.78} = -0.127$

(Using moments, $\beta_1 = \mu_3^2/\mu_2^3 = 0.008$ and $\gamma_1 = \sqrt{\beta_1} = -0.09$, the sign again being negative.)

Interpretation: the coefficient is negative but very close to zero, so the distribution is slightly negatively skewed, almost symmetrical. The longer tail is on the lower-wage side, with the mean a little below the mode.

(b) Coefficient of kurtosis

$\beta_2 = \frac{\mu_4}{\mu_2^2} = \frac{142{,}097}{(249)^2} = \frac{142{,}097}{62{,}001} = 2.29$

$\gamma_2 = \beta_2 - 3 = -0.71$

Interpretation: $\beta_2 = 2.29 < 3$ (or $\gamma_2 < 0$), so the distribution is platykurtic, flatter-topped than the normal curve, with wages spread widely rather than concentrated around the centre.

Let income be (X) and expenditure be (Y), both in Rs '000.

[ n=10,\quad \bar X=\frac{690}{10}=69,\quad \bar Y=\frac{460}{10}=46 ]

[ \sum X=690,\quad \sum Y=460,\quad \sum X^2=52650,\quad \sum Y^2=23750,\quad \sum XY=34800 ]

a. Karl Pearson's correlation coefficient

[ r=\frac{n\sum XY-\sum X\sum Y} {\sqrt{\left[n\sum X^2-(\sum X)^2\right]\left[n\sum Y^2-(\sum Y)^2\right]}} ]

[ r=\frac{10(34800)-690(460)} {\sqrt{\left[10(52650)-690^2\right]\left[10(23750)-460^2\right]}} ]

[ r=\frac{348000-317400}{\sqrt{(526500-476100)(237500-211600)}} ]

[ r=\frac{30600}{\sqrt{50400\times 25900}} =\frac{30600}{36129.77} ]

[ \boxed{r=0.8469\approx 0.847} ]

There is a high degree of positive correlation between income and expenditure.

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b. Two regression coefficients

Regression coefficient of (Y) on (X):

[ b_{yx}=\frac{n\sum XY-\sum X\sum Y}{n\sum X^2-(\sum X)^2} ]

[ b_{yx}=\frac{30600}{50400}=0.6071 ]

Regression coefficient of (X) on (Y):

[ b_{xy}=\frac{n\sum XY-\sum X\sum Y}{n\sum Y^2-(\sum Y)^2} ]

[ b_{xy}=\frac{30600}{25900}=1.1815 ]

[ \boxed{b_{yx}=0.6071,\qquad b_{xy}=1.1815} ]

Check:

[ r^2=b_{yx}\times b_{xy}=0.6071\times 1.1815=0.7173 ]

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c. Estimate expenditure when income is Rs 85 thousand

Regression equation of (Y) on (X):

[ Y-\bar Y=b_{yx}(X-\bar X) ]

[ Y-46=0.607142857(X-69) ]

For (X=85):

[ Y=46+0.607142857(85-69) ]

[ Y=46+0.607142857(16) ]

[ Y=46+9.7142857=55.7142857 ]

Since the values are in Rs '000,

[ \boxed{Y=\text{Rs }55.714\text{ thousand}\approx \text{Rs }55,714} ]

So, when income is Rs 85 thousand, the estimated expenditure is approximately Rs 55,714.

Given data

Here,
(n_1=n_2=10).

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(a) Which section has more uniform marks?

For uniformity, compare the coefficient of variation (CV):

[ CV=\frac{\sigma}{\bar X}\times 100 ]

Section A

Let assumed mean (A=55).

[ \sum d=-2,\qquad \sum d^2=222 ]

[ \bar X_A=55+\frac{-2}{10}=54.8 ]

[ \sigma_A=\sqrt{\frac{222}{10}-\left(\frac{-2}{10}\right)^2} ]

[ \sigma_A=\sqrt{22.16}=4.707 ]

[ CV_A=\frac{4.707}{54.8}\times 100=8.59% ]

Section B

Let assumed mean (A=57).

[ \sum d=4,\qquad \sum d^2=72 ]

[ \bar X_B=57+\frac{4}{10}=57.4 ]

[ \sigma_B=\sqrt{\frac{72}{10}-\left(\frac{4}{10}\right)^2} ]

[ \sigma_B=\sqrt{7.04}=2.653 ]

[ CV_B=\frac{2.653}{57.4}\times 100=4.62% ]

Since,

[ CV_B < CV_A ]

Section B has more uniform marks.

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(b) Combined standard deviation

Combined mean:

[ \bar X_c=\frac{n_1\bar X_1+n_2\bar X_2}{n_1+n_2} ]

[ \bar X_c=\frac{10(54.8)+10(57.4)}{20} ]

[ \bar X_c=\frac{548+574}{20}=56.1 ]

Now,

[ d_1=\bar X_1-\bar X_c=54.8-56.1=-1.3 ]

[ d_2=\bar X_2-\bar X_c=57.4-56.1=1.3 ]

Combined variance:

[ \sigma_c^2=\frac{n_1\sigma_1^2+n_2\sigma_2^2+n_1d_1^2+n_2d_2^2}{n_1+n_2} ]

[ \sigma_c^2=\frac{10(22.16)+10(7.04)+10(1.69)+10(1.69)}{20} ]

[ \sigma_c^2=\frac{221.6+70.4+16.9+16.9}{20} ]

[ \sigma_c^2=\frac{325.8}{20}=16.29 ]

[ \sigma_c=\sqrt{16.29}=4.036 ]

[ \boxed{\sigma_c \approx 4.04 \text{ marks}} ]

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Final Answer

(a) Section B has more uniform marks.
(b) Combined standard deviation (= \boxed{4.04 \text{ marks}}).

Let the straight line trend be
[ Y_c=a+bX ] where (X=\text{Year}-2021) (taking 2021 as origin).

[ n=7,\quad \sum Y=177,\quad \sum X=0 ]

[ a=\frac{\sum Y}{n}=\frac{177}{7}=25.2857 ]

[ \sum XY=62,\quad \sum X^2=28 ]

[ b=\frac{\sum XY}{\sum X^2}=\frac{62}{28}=2.2143 ]

Hence the trend equation is
[ \boxed{Y_c=25.2857+2.2143X} ] where (X=\text{Year}-2021).

Equivalently, taking 2018 as origin: [ \boxed{Y_c=18.6429+2.2143(\text{Year}-2018)} ]

Trend values and short-term fluctuations

Short-term fluctuation (=Y-Y_c)

Thus the trend shows an average annual increase of about Rs 2.2143 million in profit. Positive fluctuations mean actual profit is above the trend, and negative fluctuations mean actual profit is below the trend.

Let 2023 be the base year, so (p_0,q_0) = 2023 price and quantity, and (p_1,q_1) = 2024 price and quantity.

(a) Laspeyre's formula

[ P_L=\frac{\sum p_1q_0}{\sum p_0q_0}\times 100 ]

[ P_L=\frac{429000}{247500}\times 100=173.33 ]

[ \boxed{P_L=173.33} ]

(b) Paasche's formula

[ P_P=\frac{\sum p_1q_1}{\sum p_0q_1}\times 100 ]

[ P_P=\frac{436300}{255100}\times 100=171.03 ]

[ \boxed{P_P=171.03} ]

(c) Fisher's formula

[ P_F=\sqrt{P_L \times P_P} ]

[ P_F=\sqrt{173.33 \times 171.03} ]

[ P_F=\sqrt{29645.37}=172.18 ]

[ \boxed{P_F=172.18} ]

Hence, the price index numbers are:

• Laspeyre's index = 173.33
• Paasche's index = 171.03
• Fisher's index = 172.18

Pay-off table

(i) Maximax approach (optimistic): take the maximum pay-off of each strategy and choose the largest.

Maximum of S1 = 90, S2 = 100, S3 = 120. The largest is 120, so the decision is strategy S3.

(ii) Maximin approach (pessimistic): take the minimum pay-off of each strategy and choose the largest.

Minimum of S1 = 40, S2 = 50, S3 = 50. The largest is 50, which both S2 and S3 give, so the decision is S2 or S3 (a tie).

(iii) Minimax regret approach: regret = best pay-off in the state of nature − actual pay-off.

Best pay-offs: N1 = 120, N2 = 100, N3 = 60.

The smallest of the maximum regrets is 40, so the decision is strategy S3.

Working table (class width $h = 5$, $N = 135$)

Mean

$\bar{X} = \frac{\sum fm}{N} = \frac{3{,}867.5}{135} = 28.65 \text{ thousand} \approx \text{Rs } 28,648$

Median

$N/2 = 67.5$, which lies in the class 25–30 (cumulative frequency 80).

$Md = L + \frac{N/2 - cf}{f} \times h = 25 + \frac{67.5 - 45}{35} \times 5 = 25 + 3.21 = 28.21 \text{ thousand} \approx \text{Rs } 28,214$

Mode

The modal class is 25–30 (highest frequency 35), with $f_0 = 25$ and $f_2 = 27$.

$Z = L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 25 + \frac{35 - 25}{70 - 25 - 27} \times 5 = 25 + 2.78 = 27.78 \text{ thousand} \approx \text{Rs } 27,778$

The mean, median and mode are close (28.65, 28.21, 27.78), so the distribution is nearly symmetrical, with a slight positive skew (mean > median > mode).