I(x) = 200 –0.2x C1 = ₹ 3.5 T = 30 Total inventory carrying cost  = 20,685

\(\mathrm{n}=400,\ \bar{x}=171.38, \ \sigma=3.3\)
Null Hypothesis :
\(\mathrm{H}_{0}: \mu=171.17\)
Alternative Hypothesis :
\(\mathrm{H}_{1}: \mu \neq 171.17\)
Test statistic :
\( \mathrm{Z} =\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}} \sim N(0,1) \)
\(=\frac{171.38-171.17}{\frac{3.3}{\sqrt{400}}}=1.273 \)
Since |z| = 1.273 < 1.96 we accept null hypothesis at 5% level of significance. Thus the sample of 400 has come from the population with height of 171.17 cms

\(\text {(a) } \mathrm{P}(0 \leq Z \leq 1.83)=0.4664\)

\(\text {(b) } \mathrm{P}(Z \geq 1.154)\)

\( =0.5-p(0 \leq Z \leq 1.154) \)
\(=0.5-0.4382=0.0618\)
\(\text {(c) } \mathrm{P}(Z \geq-0.86)\)

\( =0.5+\mathrm{P}(-0.86 \leq \mathrm{Z} \leq 0) \)
\(=0.5+0.3051=0.8051\)
\(\text {(d) } \mathrm{P}^{\prime}(0.43 \leq Z \leq 1.12)\)

\( =\mathrm{P}(0 \leq Z \leq 1.12)-\mathrm{P}(0 \leq Z \leq 0.43) \)
 = 0.3686-0.1664  
 = 0.2022
 \(\text {(e) } \mathrm{P}(\mathrm{Z} \leq 0.77)=0.5+\mathrm{P}(0 \leq Z \leq 0.77)\)

= 0.5 + 0.2794
= 0.7794

(i) The Probability distribution of X 

\( \sum P_{i} =1 \)
\(0.1+\mathrm{k}+2 \mathrm{k}+2 \mathrm{k}+\mathrm{k} =1 \)
\(6 \mathrm{k} =1-0.1 \)
\(6 \mathrm{k} =0.9 \Rightarrow \mathrm{k}=0.15 \)
(ii) \(\mathrm{P}(x \geq 2)=\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=4)\)
2k + 2k +k = 5k
= 5(0.15) = 0.75
(iii) P(X = 2) = 2k = 2(0.15) = 0.3
(iv) \(\mathrm{P}(x \leq 2)=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)\)
0.1 + k + 2k = 0.1 + 3k
= 0.1 + 3(0.15) = 0.55

\( x_{0} =60, \mathrm{~h}=20, x_{0}+\mathrm{nh}=90 \\ 20 \mathrm{n} =30 \Rightarrow \mathrm{n}=1.5 \)

\( y=y_{0}+\frac{n}{1 !} \Delta y_{6}+\frac{n(n-1)}{2 !} \Delta^{2} y_{0}+\frac{n(n-1)(n-2)}{3 !} \Delta^{3} y_{0} \) \(+\frac{n(n-1)(n-2)(n-3)}{4 !} \Delta^{4} y_{0} \)
\( y(90)= 250+\frac{(1.5)}{1 !}(250)+\frac{(1.5)(1.5-1)}{2 !}(-20) +\frac{(1.5)(1.5-1)(1.5-2)}{3 !}(-10)+\frac{(1.5)(1.5-1)(1.5-2)(1.5-3)}{4 !}(20)\)
= 250+180.7 .5+0.625+0.46875 
\( =423.59 \sim 424 \)

-cosx-sinx+c

\(\frac{d p}{d x}=\frac{2 p^{3}-x^{3}}{3 x p^{2}}\)
It is a differential equation in x and p of homogeneous type.
\( \therefore \mathrm{p} =\mathrm{v} x \\ \frac{d p}{d x} =v+x \frac{d v}{d x} \)
\(v+x \frac{d v}{d x} =\frac{2 v^{3} x^{3}-x^{3}}{3 x\left(v^{2} x^{2}\right)}=\frac{2 v^{3}-1}{3 v^{2}} \)
\(x \frac{d v}{d x} =\frac{2 v^{3}-1}{3 v^{2}}-v \)
\( =\frac{2 v^{3}-1-3 v^{3}}{3 v^{2}}=-\left(\frac{1+v^{3}}{3 v^{2}}\right) \)
\(\int \frac{3 v^{2}}{1+v^{3}} d v =-\int \frac{d x}{x} \)
\(\log \left(1+v^{3}\right) =-\log x+\log \mathrm{c} \)
\(\log \left(1+\frac{p^{3}}{x^{3}}\right)+\log x =\log \mathrm{c} \)
\(\log \left(\frac{x^{3}+p^{3}}{x^{3}}\right) x =\log \mathrm{c} \)
\(x^{3}+\mathrm{p}^{3} =c^{2} \)
\(\text {If } \mathrm{p} =20, x=10 \)
\( 1000+8000 =c(100) \)
\(\Rightarrow =\frac{9000}{100}=90 \)
Required equation is \(x^{3}+\mathrm{p}^{3}=90 x^{2}\)

y = x² is a parabola open upwards
\( \mathrm{y} =|x| \Rightarrow \mathrm{y}=x \text { if } x \geq 0 \)
\(=-x \text { if } x<0 \)
Since the curve is symmetrical about y-axis

\( \text {Area } =\int_{a}^{b} y d x \\ =2 \int_{0}^{1}\left(x-x^{2}\right) d x \)
\( =2\left[\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{1}=2\left[\frac{1}{2}-\frac{1}{3}\right] \)
\( =2\left(\frac{3-2}{6}\right)=\frac{1}{3} \text { sq. units } \)

\( \int\left[\frac{2+x+x^{2}}{x^{2}(2+\mathrm{x})}+\frac{2 x-1}{(\mathrm{x}+1)^{2}}\right] d x \)
\( =\int\left[\frac{2+x}{x^{2}(2+\mathrm{x})}+\frac{x^{2}}{\mathrm{x}^{2}(2+x)}+\frac{2(\mathrm{x}+1)-3}{(\mathrm{x}+1)^{2}}\right] d x \)
\( =\int\left[\frac{1}{x^{2}}+\frac{1}{2+x}+\frac{2}{\mathrm{x}+1}-\frac{3}{(\mathrm{x}+1)^{2}}\right] d x \)
\(=\int\left[x^{-2}+\frac{1}{2+x}+\frac{2}{x+1}-3(\mathrm{x}+1)^{-2}\right] d x \)
\(=\frac{x^{-1}}{-1}+\log |2+x|+2 \log |x+1|-\frac{3(x+1)^{-1}}{-1}+c \)
\(=\frac{-1}{x}+\log |2+x|+2 \log |\mathrm{x}+1|+\frac{3}{(\mathrm{x}+1)}+c\)

In matrix form
\(\left(\begin{array}{ccc} 1 & 2 & -3 \\ 3 & -1 & -2 \\ 2 & 3 & -5 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} -2 \\ 1 \\ k \end{array}\right)\)
AX = B
Augmented matrix
\((A, B)=\left(\begin{array}{cccc} 1 & 2 & -3 & -2 \\ 3 & -1 & -2 & 1 \\ 2 & 3 & -5 & k \end{array}\right)\)
\(\sim\left(\begin{array}{cccc} 1 & 2 & -3 & -2 \\ 0 & -7 & 7 & 7 \\ 0 & -1 & 1 & K+4 \end{array}\right) \begin{aligned} &R_{2} \rightarrow R_{2}-3 R_{1} \\ &R_{3} \rightarrow R_{3}-2 R_{1} \end{aligned}\)
\(\sim\left(\begin{array}{cccc} 1 & 2 & -3 & -2 \\ 0 & -1 & 1 & 1 \\ 0 & -1 & 1 & k+4 \end{array}\right) \quad R_{2} \rightarrow R_{2} / 7\)
\(\sim\left(\begin{array}{cccc} 1 & 2 & -3 & -2 \\ 0 & -1 & 1 & 1 \\ 0 & 0 & 0 & k+3 \end{array}\right) \quad R_{3} \rightarrow R_{3}-R_{2}\)
Since the system is consistent
\(\rho(A, B)=\rho(A)=2\)
k + 3 = 0
k = -3

Fisher's Ideal Index = \(\sqrt\frac{{\Sigma p_1q_0}\times{\Sigma p_1q_1}}{{\Sigma p_0q_0}\times{\Sigma p_0q_1}}\times100\)
\(= \sqrt\frac{1380\times1500}{510\times571}\times100\)
= 266.61
Time reversaltest:
\(P_{01}\times P_{10}=\sqrt\frac{{\Sigma p_1q_0}\times{\Sigma p_1q_1}\times{\Sigma p_0q_1}\times{\Sigma p_0q_0}}{{\Sigma p_0q_0}\times{\Sigma p_0q_1}\times{\Sigma p_1q_1}\times{\Sigma p_1q_0}}\)

= \(\sqrt{1}=1\)
Hence, time reversal test is satisfied.
Factor reversaltest:
\(P_{01}\times Q_{01}=\sqrt\frac{{\Sigma p_1q_0}\times{\Sigma p_1q_1}\times{\Sigma q_1p_0}\times{\Sigma q_1p_1}}{{\Sigma p_0q_0}\times{\Sigma p_0q_1}\times{\Sigma q_0p_0}\times{\Sigma q_0p_1}}\)

\(P_{01}\times Q_{01}=\frac{{\Sigma p_1q_1}}{{\Sigma p_0q_0}}\)
Hence, Fisher's ideal index satisfies factor reversal test also.

Since the number of rows is less then the number of columns, given assignment problem is unbalanced one.
To balance it, introduce a dummy row with all the entries zero.
The revised assignment problem is

Here only 3 tasks can be assigned to 3 men.
Step 1 :
Select the smallest element in each row and subtract it will all the elements in its row.

Here each row and column has atleast one zero.
Step 2:
Examine the row with only one zero, mark that zero by \(\Box\) and draw a vertical line.
After examining all the rows, examine the column with single zero, mark that zero by \(\Box\) and draw a horizontal line.
Step 3:
Only two assignment have been made.
The elements not lying on the line are
\(\begin{matrix} 6 & 10 & 14 \\ 5 & 9 & 11 \\ 5 & 5 & 12 \end{matrix}\)
and minimum is 5.
Subtract 5 from all these numbers. Other numbers remains the same.
A new cost matrix will be found .and repeat step 2.
∴ The new cost matrix is

Thus, 3 assignments have been made.
The optical assignment schedule and total cost is

Let the diameter be x and area be y.
To find y when x = 82, use Newton's forward interpolation form
∴ x0 + nh = 82 ⇒ 80 + n(5) ⇒ 82 - 80 = 2
⇒ n = \(\frac{2}{5}\) = 0.4
The difference table is

∴ y(82) = y0 + \(\frac { n }{ 1! } { \triangle y }_{ 0 }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n+1)(n+2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }\)+ \(\frac { n(n+1)(n+2)(n-3) }{ 4! } ​​{ \triangle }^{ 4 }{ y }_{ 0 }​​​​​​​​​​​​​​\)

+ \(\frac{(0.4)(0.4-1)(0.4-2)}{6}\)(-2)

= 5026 + 259.2 + (0.4) (-0.6) (20) + \(\frac{(0.4)(-0.6)(-1.6)(-1)}{3}\) + \(\frac{(0.4)(-0.6)(-1.6)(-2.6)}{3}\)
= 5026 + 259.2 - 4.8 - 0.128 - 0.1664
= 5280.10
∴ When the diameter is 82, area of circle is 5280.1 (≅ 5281)
To find y when x = 91, use Newton's backward interpolation formula.
∴ x_n + nh = 91 ⇒ 100 + n(5) = 91
⇒ 5n = 91 - 100
⇒ 5n = -9 ⇒ n = \(\frac{-9}{5}\) = 1.8
Newton's backward interpolation formula is
y_(x=xn+nh) = yn + \(\frac { n }{ 1! } \triangledown { y }_{ n }+\frac { n(n+1) }{ 2! } { \triangledown }^{ 2 }{ y }_{ n }+\frac { n(n+1)(n+2) }{ 3! } { \triangledown }^{ 3 }{ y }_{ n }​​​​​​​​​​​​​​\)
y(x = 91) = 7854 - 1.8 (766) +

\(+\frac{(-1.8)(-1.8+1)(-1.8+2)(-1.8+3)}{3!}\)(4)
⇒ y_{(x = 91)} = 7854 - 1378.8 + (-1.8) (-0.8) (20) + (-1.8) (-0.8) (0.2) + \(\frac{(-1.8)(-0.8)(0.2)(1.2)}{6}\) (4)
⇒ y_{(x = 91)} = 7854 - 1378.8 + 28.8 + 0.288 + 0.2304
⇒ y_{(x = 91)} = 6504.5
Hence when the diameter is 91, area is 6504.5 ≅ 6504

Since the pressure required is at the end of the table, we apply Backward interpolation formula. Let temperature be x and the pressure be y.
\({ y }_{ \left( x={ x }_{ 0 }+nh \right) }={ y }_{ 0 }+\frac { n }{ n! } \Delta { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \Delta }^{ 2 }{ y }_{ 0 }+\frac { n(n-)(n-2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 }+...\)
To find y at x = 175
\(\therefore\) x_n + nh = 175 , x_n = 180, h = 10 \(\Rightarrow\) n = −0.5

\({ y }_{ (x=1750 }=10.225+\left( -0.5 \right) (2.149)+\frac { (-0.5)(0-5) }{ 2! } (0.375)+\frac { (-0.5)(0-5)(1.5) }{ 3! } (0.049)+\frac { (-0-5)(0.5)(1.5)(2.5) }{ 4! } (0.002)\)

= 10.225−1.0745−0.046875−0.0030625 − 0.000078125
= 9.10048438
= 9.1

Here, the number of rows and columns are equal.
∴ The given assignment problem is balanced.
Step 1 : Select a smallest element in each row and subtract this from all the elements in its row.
∴ The cost matrix: of the given assignment problem is

Column 1 contains no zero. Go to step 2.
Step 2 : Select the smallest element (1) in column 1 and subtract this from all the elements in its column.

Since each row and column contains atleast one zero, assignments can be made.
Step 3 : Examine the rows with only one zero.
Row P, Q and S contains exactly one zero, mark them by 0 and mark the other zeros in the column byX.
Thus, all the four assignments have been made.
∴ The optimal assignment schedule and total cost is

Here the number of rows and columns are equal.
\(\therefore\)The given assignment problem is balanced.
Now let us find the solution.
Step 1: Select a smallest element in each row and subtract this from all the elements in its row.

Look for atleast one zero in each row and each column.Otherwise go to step 2.
Step 2 : Select the smallest element in each column and subtract this from all the elements in its column.

Since each row and column contains atleast one zero, assignments can be made.
Step 3 (Assignment):
Examine the rows with exactly one zero. First three rows contain more than one zero. Go to row D.
There is exactly one zero. Mark that zero by \(\square\) (i.e) job D is assigned to machine
I . Mark other zeros in its column by\(\text { X }\).

Step 4: Now examine the columns with exactly one zero. Already there is an assignment in column I. Go to the column II. There is exactly one zero. Mark that zero by \(\square\) . Mark other zeros in its row by\(\text { X }\).

Column III contains more than one zero. Therefore proceed to Column IV, there is exactly one zero. Mark that zero by \(\square\) . Mark other zeros in its row by \(\text { X }\).

Step 5: Again examine the rows. Row B contains exactly one zero. Mark that zero by \(\square\).

Thus all the four assignments have been made. The optimal assignment schedule and total cost is
\(\begin{array}{|c|c|c|} \hline \text { Job } & \text { Machine } & \text { cost } \\ \hline \text { A } & \text { II } & 12 \\ \hline \text { B } & \text { III } & 7 \\ \hline \text { C } & \text { IV } & 11 \\ \hline \text { D } & \text { I } & 8 \\ \hline {\text { Total cost }} \ && 38 \\ \hline \end{array}\)
The optimal assignment (minimum) cost
= Rs. 38

Auxiliary equation is m² - 3m + 2 = 0
∴ (m - 1)(m - 2) = 0
⇒ m = 1,2
∴ Complementary function CF is A^ex + Be^2x
PI = \(\frac { 1 }{ \phi (D) } \)(x)
= \(\frac { 1 }{ { D }^{ 2 }-3D+2 } \).e3x
= \(\frac { 1 }{ (D-1)(D-2) } \).e3x
= \(\frac { e^{ 3x } }{ (3-1)(3-2) } \).e3x
= \(\frac { { e }^{ 3x } }{ 2 } \)
General solution is y = CF + PI
y = A^ex+Be^2x+\(\frac { 1 }{ 2 } \)e^3x ...(1)
Given when x = 0, y = 0
∴ 0 = Ae⁰+Be₀+\(\frac { 1 }{ 2 } \)e₀ ⇒ 0 = A+B-\(\frac { 1 }{ 2 } \) ..(2)
Also, when x = log2, y = 0
⇒ 0 = Ae^log2+Be^2log2+e^3log3
⇒ 0 = A(2)+\(Be^{ log2^{ 2 }+ }+\frac { 1 }{ 2 } elog2^{ 3 }\)
⇒ 0 = 2A+4B+\(\frac{8}{2}\) ⇒ 2A+4B=-4
⇒ A+2B = -2
(2)-(3) ⇒ A+B = -\(\frac { 1 }{ 2 } \)

Substituting B = -\(\frac { 3 }{ 2 } \) in  (2) we get
\(A-\frac { 3 }{ 2 } =-\frac { 1 }{ 2 } \Rightarrow A=-\frac { 1 }{ 2 } +\frac { 3 }{ 2 } \Rightarrow \frac { 2 }{ 2 } \)=1
∴ y = \(1.e^{ x }-\frac { 3 }{ 2 } e^{ 2x }+\frac { e^{ 3x } }{ 2 } \).

\(\overline {\overline {X}} = \frac{15+17+15+18+17+14+18+15+17+15+17+16}{10}\) = \(\frac {162}{10}\) = 16.2
\(\overline {R}\) =  \(\frac{7 + 7 + 4+ 9+ 8+ 12 + 7 + 4+ 11+ 15}{10}\)  = \(\frac {74}{10}\) = 7.4
Control limits for mean chart
UCL = \(\overline {\overline{X}} + A_{2} \overline {R}\)
= 16.2  + 0.58(7.4)
= 16.2  + 4.292 = 20.492
[∴ A₂  = 0.58]
CL = \(\overline {\overline{X}}\) = 16.2
LCL =  \(\overline {\overline{X}} - A_{2} \overline {R}\)
= 16.2 - 0.58(7.4)
= 16.2 - 4.292 = 11.91

Control limits for R-chart
UCL = D₄  \(\overline{R}\) 
= 2.115 (7.4) = 15.65
[∴ D₄  = 2.115]
CL =  \(\overline{R}\) = 7.4
LCL = D₃ \(\overline{R}\)
= 0(\(\overline{R}\)) = 0
[∴ D₃  = 0]

Conclusion : The above diagram shows all the control lines with the data points plotted. Since all the points lie within the control limits, we can say that the process is in control.

Smce \(\sum\)X = 0, a =\(\frac {\sum Y}{n}\) = \(\frac {244}{5}\) = 48.8
b = \(\frac {\sum XY}{\sum X^2}\) = \(\frac {20}{10}\) = 2
∴ The required equation of the straight line trend is given by
Y = a + bX  \(\Rightarrow \) Y = 48.8 + 2X
\(\Rightarrow \) Y = 48.8 +2 (X - 1994) .... (1)
∴ Number of salesmen in 1997 is put X = 1997 in (1)
∴ Y = 48.8 + 2 (1997 - 1994) 
= 48.8 + 2 (3) 
= 48.8 + 6 = 54.8
∴  Number of salesmen in 1997 is 54.8

Given that \(MR=\frac { dy }{ dq } =\frac { { q }^{ 2 }+{ 3 }y^{ 2 } }{ 2qy } \)  (1)
Put y = vq and \(\frac { dy }{ dq } =v+q\frac { dv }{ dq } \) in (1)
Now (1) becomes
\(v+q\frac { dy }{ dq } =\frac { { q }^{ 2 }+{ 3 }v^{ 2 }{ q }^{ 2 } }{ 2yvq } \)
\(=\frac { 1+3{ v }^{ 2 } }{ 2v } \)
\(q\frac { dv }{ dq } =\frac { 1+3{ v }^{ 2 } }{ 2v } -v\)
\(=\frac { 1+3{ v }^{ 2 }-2{ v }^{ 2 } }{ 2v } \)
\(=\frac { { 1+v }^{ 2 } }{ 2v } \)
\(\frac { 2v }{ { 1+v }^{ 2 } } dv=\frac { dq }{ q } \)
On Integration
\(ഽ\frac { 2v }{ { 1+v }^{ 2 } } dv=ഽ\frac { dq }{ q } \)
log (1+ v²) = log q + log c
1+ v² = cq
Replace \(v=\frac { y }{ q } \)
\(1+\frac { { y }^{ 2 } }{ { q }^{ 2 } } \) = cq
q² + y² = c q³ (2)
Given output is 1 unit and revenue is Rs. 5
∴ (2) ⇒ 1 + 25 = c ⇒ c = 26
∴ The total revenue function is q² + y² = 26q³

Sample size n = 100
Sample mean \(\bar { X } \)= 72
Population mean μ = 76
Population standard deviation σ = 8
Null Hypotheses H₀:
μ = 76(i.e., There is no Significant difference between the state scores and the national scores)
\(Z=\frac { \bar { X } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \)
\(\Rightarrow Z=\frac { 72-76 }{ \frac { 8 }{ \sqrt { 100 } } } =\frac { -4 }{ \frac { 8 }{ 10 } } =\frac { -4 }{ 8 } =-5\)
\(\Rightarrow |Z|=5\)
\({ Z }_{ \frac { \alpha }{ 2 } }=1.96\)
\(Z>{ Z }_{ \frac { \alpha }{ 2 } }i.e.,\ 5>1.96\)
Inference : Since \(Z>{ Z }_{ \frac { \alpha }{ 2 } }\)  at 5% level of significance, the null hypothesis H₀ is rejected. 
Hence, we conclude that there is significant difference between the state scores and the national scores.

The first thing is to number the population (8585 children). The numbering has already been provided by the frequency table. There are 2 children with height of 105 cm, therefore we assign number 1 and 2 to the children those in the group 105 cm, number 3 to 6 is assigned to those in the group 107 cm and similarly all other children are assigned the numbers. In the last group 145 cms there are two children with assigned number 8584 and 8585.

Now we take 10 samples from the tables, since the population size is in 4 digits we can use the given random number table. Select the10 random numbers from 1 to 8585 in the table, Here, we consider column wise selection of random numbers, starting from first column.

The children with assigned number 2952 is selected and then see the cumulative frequency table where 2952 is present, now select the corresponding row height which is 123 cm, similarly all the selected random numbers are considered for the selection of the child with their corresponding height. The following table shows all the selected 10 children with their heights.

(a) P(X<40,000)
When X = 40,000.
Z = \(\frac { X-\mu }{ \sigma } \)
= \(\frac { 40,000-50,000 }{ 20,000 } \) = -0.5
∴ P(X<40,000) = P(Z<-0.5)
= P(-∞ =P(-∞ < Z <0) - P(-0.5 = 0.5-P(0 [By symmetry]
= 0.5-0.1915 = 0.3085
∴ 30.85 % pf people earn less than $40,000

(b) P(45,000 < X < 65,000)
When X = 45,000
Z₁ = \(\frac { 45,000-50,000 }{ 20,000 } \) = -0.25
When X=65,000
Z=\(\frac { 65,000-50,000 }{ 20,000 } \)=0.75
∴ P(45000 < X < 65,000) = P(-0.25 < Z < 0.75)

= P (-0.25 < Z < 0) + P(0 < Z < 0.75)
= P(0 < Z < 0.25) + P (0 < Z < 0.75)
= 0.0987 + 0.2734 = 0.3721
Hence, 37.21% of people earn between Dallor 45,000 and Dallor 65,000.
(c) P(X>7000)
When X = 7000
Z = \(\frac { 75000-50,000 }{ 20,000 } \) = 1.25
∴ P(X > 7000) = P(Z > 1.25)
= 0.5 - 0.3944
= 0.1056

 Hence, 10.56% of people earn more than Dallor 75000.

Let X be the normal variate showing the score of the candidate with mean 45 and standard deviation 10.
(i) less than 35 marks
When X = 35
\(Z=\frac { X-\mu }{ \sigma } =\frac { 35-45 }{ 10 } =-1\)
P(X < 35) = P(Z < –1)
P(Z > 1) = 0.5 – P(0 < Z < 1)
= 0.5 – 0.3413
= 0.1587
Expected number of students scoring less than 35 marks are 0.1587 × 1300
= 206

(ii) more than 65 marks
When X = 65
\(Z=\frac { X-\mu }{ \sigma } =\frac { 65-45 }{ 10 } =2.0\)
P(X > 65) = P(Z > 2.0)
0.5 – P(0 < Z < 2.0)
0.5 – 0.4772
= 0.0228
Expected number of students scoring more than 65 marks are 0.0228 x 1300 = 30

ഽ\(\left[ \frac { 1 }{ \log x } -\frac { 1 }{ { \left( \log x \right) }^{ 2 } } \right] dx\) = ഽ\(\left[ \frac { 1 }{ z } -\frac { 1 }{ { z }^{ 2 } } \right] { e }^{ z }dx\)
= ഽ e^x [ f(z) +f'(z)] dx
= e^z f(z) + c
= e^z \(\left[ \frac { 1 }{ z } \right] \) + c
= \(\frac { x }{ \log\ x } \) + c

Given probability distribution function
\(F(x)\begin{cases} 0,\quad ​​​​\text{for}\quad x<0 \\ \frac { 1 }{ 2 } ,\quad ​​​​\text{for}\quad 0\le x<1 \\ 0,\quad ​​​​\text{for}\quad 1\le x<2\quad \\ \frac { 1 }{ 4 } ,\quad ​​​​\text{for}\quad 2\le x<4 \\ 0,\quad ​​​​\text{for}\quad x\ge 4 \end{cases}\)
a) The distribution function F(x) is continuous since it is a step function
We know f'(x) = f(x)
∴ Probability density function
\(F(x)\begin{cases} 0,\quad \text{for}\quad x\le 0 \\ \frac { 1 }{ 2 } ,\quad \text{for}\quad 0\le x<1 \\ 0,\quad \text{for}\quad 1\le x<2\quad \\ \frac { 1 }{ 4 } ,\quad \text{for}\quad 2\le x<4 \\ 0,\quad \text{for}\quad x\ge 4 \end{cases}...(1)\)
(b) (i) Probability that a person will have to wait more than 3 minutes is P(X > 3)
∴ P(X>3) = P(3≤X<4)+P(X≥4)
\(\frac{1}{4}+0=\frac{1}{4}\) [from (1)]
ii) Probability that a person will have to wait less than 3 minutes is P(X < 3)
∴ P(X<3) = P(X≤0)+P(0≤x≤1)+P(1≤x≤2)+P(2≤x≤3)
0 + \(\frac{1}{2}+0+\frac{1}{4}\)[from (1)]
∴ P(X<3) = \(\frac{3}{4}\)
= P(1≤X<2)+P(2≤X<3)
= 0 +\(\frac{1}{4}=\frac{1}{4}\) [from (1)]

Given marginal revenue function.
\(MR=\frac { DR }{ dx } =20{ e }^{ \frac { -x }{ 10 } }\left( 1-\frac { x }{ 10 } \right) \)
\(dR={ 20e }^{ \frac { -x }{ 10 } }\left( 1-\frac { x }{ 10 } \right) dx\)
\(R=\int { { 20e }^{ \frac { -x }{ 10 } } } \left( 1-\frac { x }{ 10 } \right) dx\)
We know that \(\int { { e }^{ ax }[af(x)+f'(x)]dx={ e }^{ ax }f(x) } +c\)
Here \(\\ a=\frac { -1 }{ 10 } ,f(x)=x,\quad f'(x)=1\)
\(R=20\int { { e }^{ \frac { -x }{ 10 } } } \left[ -\frac { 1 }{ 10 } x+1 \right] dx=20{ e }^{ \frac { -x }{ 10 } }x+k\)
\(\Rightarrow R=20{ e }^{ \frac { -x }{ 10 } }x+k\quad ...(1)\)
When x = 0, R = 0
0 = 0 + k ⇒ k = 0
(1) becomes 
\(R=20x{ e }^{ \frac { -x }{ 10 } }\)
Demand function P
\(=\frac { R }{ x } =\frac { 20x{ e }^{ \frac { -x }{ 10 } } }{ x } \)
\(\Rightarrow P=20{ e }^{ \frac { -x }{ 10 } }\)

(a) We know that
\(\int _{ -\infty }^{ \infty }{ f(x) } dx=1\)
\(\int _{ 0 }^{ 10 }{ Axdx } +\int _{ 10 }^{ 20 }{ A(20-x)dx=1 } \)
\(A\left\{ { \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 10 }+{ \left[ 20x-\frac { { x }^{ 2 } }{ 2 } \right] }_{ 10 }^{ 20 } \right\} =1\)
A[(50-0)+(400-200)-(200-50)] = 1
\(A=\frac{1}{100}\)
(b) (i) The probability that the number of pounds of bread that will be sold tomorrow is more than 10 pounds is given by
\(P(10\le X\le 20)=\int _{ 10 }^{ 20 }{ \frac { 1 }{ 100 } (20-x) } dx\)
\(=\frac { 1 }{ 100 } { \left[ 20x-\frac { { x }^{ 2 } }{ 2 } \right] }_{ 10 }^{ 20 }\)
\(=\frac { 1 }{ 100 } [(400-200)-(200-50)]\)
= 0.5
(ii) The probability that the number of pounds of bread that will be sold tomorrow is less than 10 pounds, is given by
\(P(0\le X\le 20)=\int _{ 0 }^{ 10 }{ \frac { 1 }{ 100 } } xdx\)
\(=\frac { 1 }{ 100 } { \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 10 }\)
\(=\frac { 1 }{ 100 } (50-0)\)
= 0.5
(ii) The probability that the number of pounds of bread that will be sold tomorrow is between 5 and 15 pounds is
\(P(5\le X \le15)=\int _{ 5 }^{ 10 }{ \frac { 1 }{ 100 } xdx } +\int _{ 10 }^{ 15 }{ \frac { 1 }{ 100 } (20-x)dx } \)
\(=\frac { 1 }{ 100 } { \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 5 }^{ 10 }+\frac { 1 }{ 100 } { \left[ 20x-\frac { { x }^{ 2 } }{ 2 } \right] }_{ 10 }^{ 15 }\)
= 0.75

Let the weight assigned to the three varieties be Rs. x, Rs. y and Rs. z respectively By the given data,
x + 2y + 3z = 11
2x + 4y + 5z = 21
3x + 5y + 6z = 27
\(\Delta =\left| \begin{matrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \end{matrix} \right| =1\left| \begin{matrix} 4 & 5 \\ 5 & 6 \end{matrix} \right| -2\left| \begin{matrix} 2 & 5 \\ 3 & 6 \end{matrix} \right| +3\left| \begin{matrix} 2 & 4 \\ 3 & 5 \end{matrix} \right| \)
= 1(24 - 25) -2(12 - 15) + 3(10 - 12)
= 1(-1) -2 (-3) + 3(-2)
= -1+6 - 6 = -1\(\neq \) 0.
Since \(\Delta \neq 0\) the system is consistent with unique solution and Cramer's rule can be applied.
\(\Delta x=\left| \begin{matrix} 11 & 2 & 3 \\ 21 & 4 & 5 \\ 27 & 5 & 6 \end{matrix} \right| \)
\(=11\left| \begin{matrix} 4 & 5 \\ 5 & 6 \end{matrix} \right| -2\left| \begin{matrix} 21 & 5 \\ 27 & 6 \end{matrix} \right| +3\left| \begin{matrix} 21 & 4 \\ 27 & 5 \end{matrix} \right| \)
= 11(24 - 25) - 2(126 - 135) + 3(105 - 108)
= 11(-1) - 2(-9) + 3 (-3)
= 11+18-9
= -2
\(\Delta y=\left| \begin{matrix} 1 & 11 & 3 \\ 2 & 21 & 5 \\ 3 & 27 & 6 \end{matrix} \right| \)
= \(\left| \begin{matrix} 21 & 5 \\ 27 & 6 \end{matrix} \right| -11\left| \begin{matrix} 2 & 5 \\ 3 & 6 \end{matrix} \right| +3\left| \begin{matrix} 2 & 21 \\ 3 & 27 \end{matrix} \right| \)
= 1(126 - 135) - 11(12 -15) + 3(54 - 63)
= - 9 - 11(-3) + 3(-9)
= - 9 + 33 - 27
= 3
\(\Delta z=\left| \begin{matrix} 1 & 2 & 11 \\ 2 & 4 & 21 \\ 3 & 5 & 27 \end{matrix} \right| =1\left| \begin{matrix} 4 & 21 \\ 5 & 27 \end{matrix} \right| -2\left| \begin{matrix} 2 & 21 \\ 3 & 27 \end{matrix} \right| +11\left| \begin{matrix} 2 & 4 \\ 3 & 5 \end{matrix} \right| \)
= 1(108 - 105) - 2(54 - 63) + 11(10 - 12)
= 1(3) - 2(-9) + 11(-2)
= 3 + 18 - 22
= -1
\(x=\cfrac { \Delta x }{ \Delta } =\cfrac { -2 }{ -1 } =2\)
\(y=\cfrac { \Delta y }{ \Delta } =\cfrac { -3 }{ 1 } =3\)
and \(z=\cfrac { \Delta z }{ \Delta } =\cfrac { -1 }{ -1 } =1\)
Hence, the weights assigned to the three varieties are 2, 3 and 1 respectively

The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 0 & 1 & k \end{matrix} \right) \left( \begin{matrix} X \\ Y \\ Z \end{matrix} \right) =\left( \begin{matrix} 7 \\ 18 \\ 6 \end{matrix} \right) \)
 AX = B

For the equations to be inconsistent
\(\rho ([A,B])\neq \rho (A)\)
It is possible if k − 2 = 0.
\(\therefore \)  k = 2