Given n = 16, S.D = 2 ⇒ \(\sqrt { npq } \) =2
⇒ npq = 4
∴ 16(pq) = 4 ⇒ pq =\(\frac { 4 }{ 16 } =\frac { 1 }{ 4 } \)
⇒ q = \(\frac { 1 }{ 4p } \)...(1)
Since p + q = 1 ⇒ p+\(\frac { 1 }{ 4p } \)=1
⇒ \(\frac { 4{ p }^{ 2 }+1 }{ 4p } \) = 1
⇒ 4p²+1 = 4p ⇒ 4p² -4p+1 = 0
⇒ (2p-1)² = 0 ⇒ 2p-1 = 0
⇒ 2p = 1 ⇒ p = \(\frac { 1 }{ 2 } \)
∴ q = 1-p = \(1-\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)
Mean = np = 16 x \(\frac { 1 }{ 2 } \) = 8

Let X denote the number of white balls drawn from the urn.
Since there are 4 white balls, X can take values 0,1,2,3,4.
P(X = 0)=p(gettmg no white balls)\(=\frac { { 6C }_{ 4 } }{ 10{ C }_{ 4 } } \)
\(\frac{1}{14}\)
P(X = 1) = P(getting one white ball and 3 red balls) \(\frac { { 4C }_{ 4 }\times { 6C }_{ 3 } }{ { 10 }C_{ 4 } } =\frac { 8 }{ 21 } \)
P(X = 2) = P(getting two white balls and 2 red balls) \(\frac { { 4C }_{ 2 }\times { 6C }_{ 2 } }{ { 10 }C_{ 4 } } =\frac { 8 }{ 21 } \)
P(X = 3) = P(getting 3 white balls and 1 red ball) \(\frac { { 4C }_{ 3 }\times { 6C }l }{ { 10 }C_{ 4 } } =\frac { 4 }{ 35 } \)
P(X = 4) = P(getting 4 white balls) =\(\frac { 4{ C }_{ 4 } }{ 10{ C }_{ 4 } } \)
=\(\frac{1}{210}\)
Thus the probability distribution of X is

(i) Max (50, 10, 5) = 50
∴ A is the best alternative under maximin principle
(ii) Min (100, 80,40) = 40
∴ C is the best alternative under minimax principle

Given y = (A + Bx)e^3x ....(1)
Differentiating w.r.t 'x' we get,
\(\frac { dy }{ dx } \)= (A+Bx)e^3x(3)+e^3x(B)
⇒ \(\frac { dy }{ dx } \) = 3y + Be^3x  [Using (1)]
⇒ Be^3x = \(\frac { dy }{ dx } \)-3y
Differentiating again w.r.t 'x' we get,
\(\frac { d^{ 2 }y }{ { dx }^{ 2 } } =3\left( \frac { dy }{ dx } \right) \)+ Be3x(3)
⇒ \(\frac { d^{ 2 }y }{ { dx }^{ 2 } } =3\left( \frac { dy }{ dx } \right) +3\left[ \frac { dy }{ dx } -3y \right] \) [Using (2)]
⇒ \(\frac { d^{ 2 }y }{ { dx }^{ 2 } } =3\left( \frac { dy }{ dx } \right) +3\left( \frac { dy }{ dx } \right) \)-9y
⇒ \(\frac { d^{ 2 }y }{ { dx }^{ 2 } } =6\left( \frac { dy }{ dx } \right) \)-9y which is the required differential equation.

Since 82 lies at the beginning of the table, we can use Newton's forward interpolation formula
\(\Rightarrow { y }_{ o }+\frac { n }{ n! } \triangle { y }_{ o }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ o }+\frac { n(n+1)(n-2) }{ 3! } { \triangle }^{ 3 }{ (y }_{ o })\)
Also x₀ + nh = 82 ⇒ 75 + n(5) = 82 ⇒ 5n = 82 - 75 = 7
⇒ n = \(\frac75\) = 1.4
The difference table is

\(y=2459+\frac { 1.4 }{ 1! } (-441)+\frac { (1.4)(1.4-1) }{ 2! } (-397)+\frac { (1.4)(1.4-1)(1.4-2) }{ 3! } (457)\)
= 2459 - 617.4 - 111.6 - 25.592
y = 1704. 408 when x = 82.

Area \(=\int _{ a }^{ b }{ ydx } \)
\(=\int _{ 3 }^{ 9 }{ \frac { 1 }{ x } dx } \)
\(={ [log\quad x] }_{ 3 }^{ 9 }\)
= log9-log3
\(=log\left( \frac { 9 }{ 3 } \right) \)
A = log 3 sq.units.

The non-homogeneous equations are
2x - 3y - 1 = 0, 5x + 2y - 12 = 0
\(\Delta =\left| \begin{matrix} 2 & -3 \\ 5 & 2 \end{matrix} \right| =4+15=19\neq 0\)
Since \(\Delta \neq 0\) Cramer's rule can be applied and the system is consistent with unique solution.
\(\Delta x=\left| \begin{matrix} 1 & -3 \\ 12 & 2 \end{matrix} \right| =2+36=38\)
\(\Delta y=\left| \begin{matrix} 2 & 1 \\ 5 & 12 \end{matrix} \right| =24-5=19\)
\(x=\cfrac { \Delta x }{ \Delta } =\cfrac { 38 }{ 19 } =1\)
\(y=\cfrac { \Delta y }{ \Delta } =\cfrac { 19 }{ 19 } =1\)
\(\therefore \) Solution set is {2, 1}

Let x devotes the set of twins on a day
P(twin birth) = p = 1/80 = 0.0125 and n = 30
The value of mean λ = np = 30 × 0.0125 = 0.375
Hence, X follows poisson distribution with p(x)\(\frac { { e }^{ -\lambda }{ \lambda }^{ x } }{ x! } \)
The probability is 
P(2 or more) = 1 – [p (x = 0) + p (x = )] \(=1-\left[ \frac { { e }^{ -0.375 }{ (0.375) }^{ 0 } }{ 0! } +\frac { { e }^{ -0.375 }{ (0.375) }^{ 1 } }{ 1! } \right] \)
\(=1-{ e }^{ -0.375 }[1+0.375]\)
\(=1-(0.6873\times 1.375)\)
= 0.055

Given supply function p = 7 + x and x = 5
When x₀ = 5, p₀ = 7 + 5 = 12
∴ p₀x₀ = x₁₂ = 60
Producer's Surplus
\(={ p }_{ 0 }{ x }_{ 0 }-\int _{ g }^{ x }{ (x)dx } \)
\(=60-\int _{ 0 }^{ 5 }{ (7+x)dx } \)
\(=60-{ \left[ 7x+\frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 5 }\)
\(=60-\left[ 7(5)+\frac { { 5 }^{ 2 } }{ 2 } \right] \)
\(=60-\left[ 35+\frac { 25 }{ 2 } \right] \)
= 60-(35+12.5)
= 60-47.5
P.S = 12.5 = \(\frac{25}{2}\) units.

Let X is the random variable which counts the number of heads on the upturned faces. The outcomes are stated below

These values are summarized in the following probability table.

(i) p(x_i) \(\ge\)0\(\forall \) i and
(ii) \(\sum _{ i=0 }^{ 3 }{ p({ x }_{ i })=1 } \)
Hence, p(x_i) is a probability mass function.

Given sample size n = 400
Sample mean \(\bar { x } \) = 171.38
Population mean μ = 171.17
Population Standard deviation σ = 3.3
Null hypotheses: H₀ : μ = 171.17
Alternative hypotheses: H₁ : μ ≠ 171.17
The test statistic, z = \(\frac { \bar { x } -\mu }{ \frac { \sigma }{ \sqrt { n } } } =\frac { 171.38-171.17 }{ \frac { 3.3 }{ \sqrt { 400 } } } \)
=\(\frac { 0.21 }{ 0.165 } \) = 1.273
As the level of significance is α = 0.005, \(Z_{ \frac { \alpha }{ 2 } }\) = 1.96
Here z < \(Z_{ \frac { \alpha }{ 2 } }\) as 1.273 < 1.96
Inference: since z < \(Z_{ \frac { \alpha }{ 2 } }\), we accept the null hypotheses at 5% level of significance.
Hence, we can conclude that the sample of 400 has taken from the population with mean height of 171.17 cm.

Mean = E(X)=\(\int _{ -\infty }^{ \infty }{ x.f(x)dx=\int _{ -12 }^{ 12 }{ x.\left( \frac { 1 }{ 24 } \right) } dx } \)
\(=\frac { 1 }{ 24 } \int _{ -12 }^{ 12 }{ x.dx } \)
\(=0[\because \int _{ -a }^{ a }{ f(x)dx=0 } when\ f(x)\ is\ an\ odd\ function]\)
\(\therefore E(X)=0\)

Min (45, 70, 53) = 45
∴ Choosing Bonds is the best decision under minimax principle.

x dy = -y dx
Separating the variables we get
\(\frac { dy }{ y } =-\frac { dx }{ x } \)
Integrating, \(\int { \frac { dy }{ y } } =-\int { \frac { dx }{ x } } \)
⇒ log y = -log x + log C
⇒ log y = log\(\left( \frac { C }{ x } \right) \Rightarrow y=\frac { C }{ x } \) ⇒ xy = C.

We know Δ (f(x)) = f(x + h) -f(x)
Since h = 1, ∆ (f)) = f(x + 1) - f(x)
[∵ (x + 1)³ = x³ + 3x² + 3x + 1]
⇒ Δ (x³) = (x + 1)³ - x³

⇒ Δ (x³) = 3x² + 3x + 1

Area under the curve is
\(A=\int _{ c }^{ d }{ xdy } =\int _{ 2 }^{ 4 }{ \sqrt { 4y } dy } \)
\(=2\int _{ 2 }^{ 4 }{ { y }^{ \frac { 1 }{ 2 } }dy } =2{ \left( \frac { { y }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } \right) }_{ 2 }^{ 4 }\)
\(=\frac { 4 }{ 3 } \left( { y }^{ \frac { 3 }{ 2 } } \right) \)
\(=\frac { 4 }{ 3 } \left( { 4 }^{ \frac { 3 }{ 2 } }-{ 2 }^{ \frac { 3 }{ 2 } } \right) \)
\(=\frac { 4 }{ 3 } \left( 4\sqrt { 4 } -2\sqrt { 2 } \right) \)
\(=\frac { 4 }{ 3 } (8-2\sqrt { 2 } )\)sq.units

\(\Delta =\left| \begin{matrix} 2 & 3 \\ 4 & 6 \end{matrix} \right| =12-12=0\)
\(\Delta x=\left| \begin{matrix} 4 & 3 \\ 8 & 6 \end{matrix} \right| =24-24=0\)
\(\Delta x=\left| \begin{matrix} 4 & 3 \\ 8 & 6 \end{matrix} \right| =24-24=0\)
\(\therefore \Delta =\Delta x=\Delta y=0\)
\(\therefore \) The system is consistent with infinite number of solutions
let y = k, \(k\epsilon R\)
\(\therefore 2x+3k=4\Rightarrow 2x=4-3k\)
\(\Rightarrow x=\cfrac { 1 }{ 2 } \left( 4-3k \right) ,k\epsilon R\)
\(\therefore \) Solution set is \(\left\{ \cfrac { 4-3k }{ 2 } ,k \right\} ,k\epsilon R\)

Give n = 100, \(\bar x\) =162 cm, s = 8 cm is known in this problem
since σ is unknown , so we consider \(\hat{\sigma}\) = s and \(\varphi\) = 160 cm
\(S.E = \frac { \hat{\sigma} }{ \sqrt { n } } =\frac { s }{ \sqrt { n } } =\frac { 8 }{ \sqrt { 100 } } =0.8\)
Therefore the standard error for the average height of large group of students of 160 cm is 0.8.

\(\int _{ 0 }^{ \infty }{ { e }^{ -\frac { x }{ 2 } } } dx=-2{ \left[ { e }^{ -\frac { x }{ 2 } } \right] }_{ 0 }^{ \infty }\)
= − 2[0 −1] = 2

Expected number of success is
E(X)\(E(X)=\sum _{ x }^{ }{ x } { P }_{ X }(x)\)
\(=\left( 0\times \frac { 6 }{ 11 } \right) +\left( 1\times \frac { 9 }{ 22 } \right) +\left( 2\times \frac { 1 }{ 22 } \right) \)
\(=\frac { 11 }{ 22 } \)
= 0.5
Therefore, the expected number of success is 0.5. Approximately one success.

Given sample size n = 200
Sample mean \(\bar { x } \) = 0.824
Sample S.D. s = 0.042
Standard error = \(\frac { s }{ \sqrt { n } } =\frac { 0.042 }{ \sqrt { 200 } } \)
= \(\frac { 0.042 }{ 14.14 } \) = 0.00270
(a) As the level of significance is α = 0.05, \(Z_{ \frac { \alpha }{ 2 } }\) =  1.96
∴ 95% confidence limits for μ are given by \(\bar { x } -Z_{ \frac { \alpha }{ 2 } }(S.E)\le \mu \le \bar { x } +Z_{ \frac { \alpha }{ 2 } }(S.E)\)
⇒ 0.824 - (1.96) (0.00270) ≤ μ ≤ 0.824 + (1.96) (0.00270)
⇒ 0.824 - 0.00582 ≤ μ ≤ 0.824 + 0.00582
⇒ 0.818 ≤ μ ≤ 0.832
Hence, the 95% confidence limits for μ is (0.818,0.832)
(b) As the level of significance is α =0.001, \(Z_{ \frac { \alpha }{ 2 } }\) = 2.58
∴ 99% confidence limits for μ are given by \(\bar { x } -Z_{ \frac { \alpha }{ 2 } }(S.E)\le \mu \le \bar { x } +Z_{ \frac { \alpha }{ 2 } }(S.E)\)
⇒ 0.824 - (2.58) (0.00270) ≤ μ ≤ 0.824 + (2.58) (0.00270)
⇒ 0.824 - 0.00582 ≤ μ ≤ 0.824 + 0.005825
⇒ 0.816 ≤ μ ≤ 0.832
Hence, the 99% confidence limits for μ is (0.816, 0.832)

Let X denote the height of the student
Given P (X < 40) = 10% = \(\frac { 10 }{ 100 } \) =0.1
P(X> 90) = 10% = \(\frac { 10 }{ 100 } \) =0.1

∴ P(40 < X < 90) = P(-∞ < X < ∞) - [P(X < 40) + P(X < 90)]
= 1 - (0.1 + 0.1)
= 1 - 0.2 = 0.8
∴ out of 800 students, number of students scored between 40 and 90 = 800 x 0.8
= 640 students.

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

i) E(X) = Σx_ip_i
= 0(0.1)+1(0.25)+2(0.3)+3(0.2)+4.(0.15)
= 2.05
E(X²) = ∑x_i²p_i
= 0(0.1)+1(0.25)+4(0.3)+9(0.2)+16(0.15)
= 5.65
Now, V(X) = E(X²)-[E(X)]²
= 5.65 - (2.05)² = 1.4475
ii) \(V\left( \frac { X }{ 2 } \right) =\frac { 1 }{ 4 } v(X)\quad [\because V(aX)={ a }^{ 2 }V(X)]\)
\(=\frac { 1 }{ 4 } (1.4475)\)
\(V\left( \frac { X }{ 2 } \right) =0.361875\)

Let I =  ഽ x³ sin (x⁴) dx
Put t = x⁴
⇒ dt = 4x³ dfx
\(\Rightarrow \frac { dt }{ 4 } ={ x }^{ 3 }dx\)
\(\therefore I=\int { sin } t.\frac { dt }{ 4 } =\frac { 1 }{ 4 } sint\quad dt\)
= \(-\frac { 1 }{ 4 } cos\quad t+c\)
= \(-\frac { 1 }{ 4 } cos\left( { x }^{ 4 } \right) +c\)      \(\left[ \because t={ x }^{ 4 } \right] \)

Given \(\frac { dp }{ dx } =\frac { 2{ p }^{ 3 }-{ x }^{ 3 } }{ 3x{ p }^{ 2 } } \)
The numerator and denominator are homogeneous functions of 3,
∴ Put p=vx and \(\frac { dp }{ dx } =v+x\frac { dv }{ dx } \)

= \(\frac { 2{ v }^{ 3 }-1 }{ 3{ v }^{ 2 } } \)
⇒ \(\frac { dv }{ dx } =\frac { 2{ v }^{ 3 }-1 }{ 3{ v }^{ 2 } } -v=\frac { 2{ v }^{ 3 }-1-3{ v }^{ 3 } }{ 3{ v }^{ 2 } } \)
= \(\frac { -1-{ v }^{ 3 } }{ 3{ v }^{ 2 } } \)
⇒ \(\left( \frac { 3{ v }^{ 2 } }{ 1+{ v }^{ 3 } } \right) dv=-\frac { dx }{ x } \)
Integrating, \(\int { \frac { 3{ v }^{ 2 } }{ 1+{ v }^{ 3 } } } dv=-\int { \frac { dx }{ x } } \)
⇒ log(1+v³) = -log x + log c
⇒ 1+v³ = \(\frac { c }{ x } \)
Replacing v by \(\frac { p }{ x } \) we get
\(1+\frac { { p }^{ 3 } }{ { x }^{ 3 } } =\frac { c }{ x } \Rightarrow \frac { { x }^{ 3 }+{ p }^{ 3 } }{ { x }^{ 3 } } =\frac { c }{ x } \)
⇒ \(\frac { { x }^{ 3 }+{ p }^{ 3 } }{ { x }^{ 2 } } \)= c ⇒ x³+p³ = cx²...(1)
When x = 10, p = 20
⇒ 10³ + 20³ = c(10)² ⇒ 1000 + 8000 = 100 c
⇒ 9000 = 100 c
⇒ c = 90
∴ (1) becomes,
x³+p³ = 90x²
⇒ p³ = 90x²-x³
⇒ p³ = x²(90-x) which is the required relationship.

Since e^1.75 lies at the beginning of the table, we can use Newton's forward interpolation formula
∴ x_o + nh = x ⇒ 1.7 + n(0.1) = 1.75
⇒ n(0.1) = 1.75 - 1.7 = 0.05
⇒ n = \(\frac{0.05}{0.1}\) = 0.5
\({ y }_{ x }={ y }_{ o }+\frac { n }{ n! } \triangle { y }_{ o }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ o }+\frac { n(n+1)(n-2) }{ 3! } { \triangle }^{ 3 }{ (y }_{ o })+.......\)
The difference table is

∴ \(y\left( { e }^{ 1.75 } \right) =5.74+\frac { 0.5 }{ 1! } (0.576)+\frac { (0.5)(0.5-1) }{ 2! } (0.06)+\frac { (0.5)(0.5-1)(0.5-2) }{ 3! } (0.007)\)
y(e^1.75) = 5.474 + 0.288 - 0.0075 + 0.0004375
= 5.7549375

Let A represents the percent of commuters who use the transit system and B represents the percent of commuters who use their own car. Transition probability matrix

Given 50% of commuters use the transit system and 50% of the commuters use their own car this year.
(i) Percentage of commuters after one year
\(\left( \cdot 5\cdot 5 \right) \left( \begin{matrix} \cdot 9 & \cdot 1 \\ \cdot 2 & \cdot 8 \end{matrix} \right) \)
= (-5\(\times\)·9+·5\(\times\).2 ·5\(\times\)·1+·5\(\times\)·8)
= (-45 + ·10 ·05 +.40)
= (-55 - 45)
A = 55% and B = 45%
(ii) Equilibrium will be reached in the long run at equilibrium, we must have
(A B)T = (A B) wher A+B = 1
\(\Rightarrow \left( \begin{matrix} A & B \end{matrix} \right) \left( \begin{matrix} \cdot 9 & \cdot 1 \\ \cdot 2 & \cdot 8 \end{matrix} \right) =\left( \begin{matrix} A & B \end{matrix} \right) \)
\(\left( \begin{matrix} \cdot 9A+\cdot 2B & \cdot 1A+8B \end{matrix} \right) =\left( \begin{matrix} A & B \end{matrix} \right) \)
Equating the corresponding entries on both sides we get,
\(\cdot 9A+\cdot 2B=A\Rightarrow \cdot 9A+\cdot 2(1-A)=A\)
[Since A + B = 1, B = 1 -A]
\(\Rightarrow \cdot 9A+\cdot 2-\cdot 2A=A\)
\(\Rightarrow \cdot 2=A-\cdot 9A+\cdot 2A\)
\(\Rightarrow \cdot 2=A(1-\cdot 9+\cdot 2)\)
\(\Rightarrow \cdot 2=A=(\cdot 3)\)
\(\therefore\) 67% of the commuters will be using the transit system in the long run.

The given equation can be reduced to
\(\frac { dy }{ dx } +\frac { 2x }{ { x }^{ 2 }+1 } y=\frac { 4{ x }^{ 2 } }{ { x }^{ 2 }+1 } \)
It is of the form \(\frac { dy }{ dx } \) + Py = Q
Here P \(=\frac { 2x }{ { x }^{ 2 }+1 } ,Q=\frac { 4{ x }^{ 2 } }{ { x }^{ 2 }+1 } \)
ഽPdx = ഽ\(\frac { 2x }{ { x }^{ 2 }+1 } \)dx = log(x² +1)
I.F = e^ഽpdx = e^log(x2+1) = x² + 1
The required solution is y(IF) = ഽQ(I.F)dx + c
y(x² +1) = ഽ\(\frac { 4{ x }^{ 2 } }{ { x }^{ 2 }+1 } \)(x² + 1)dx + c
y(x² +1) = \(\frac { 4{ x }^{ 3 } }{ 3 } \) + c

Fisher’s price index number
\({ P }_{ 01 }^{ F }=\left( \sqrt { \frac { \sum { { p }_{ 1 }{ q }_{ 10 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } } } \right) \times 100=\left( \sqrt { \frac { 623\times 517 }{ 584\times 484 } } \right) \times 100=106.74\)
Time Reversal Test: P₀₁ × P₁₀ = 1
\({ P }_{ 01 }\times { P }_{ 10 }=\sqrt { \left( \frac { \sum { { p }_{ 1 }{ q }_{ 0 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } \times \sum { { p }_{ 0 }{ q }_{ 0 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } \times \sum { { p }_{ 1 }{ q }_{ 0 } } } \right) } \)
\({ P }_{ 01 }\times { P }_{ 10 }=\sqrt { \left( \frac { 623\times 517\times 484\times 584 }{ 584\times 487\times 517\times 623 } \right) } \)
P₀₁ x P₁₀ = 1
Factor Reversal Test
\({ P }_{ 01 }\times { Q }_{ 01 }=\frac { \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } } \)
\({ P }_{ 01 }\times { P }_{ 01 }=\sqrt { \left( \frac { \sum { { p }_{ 1 }{ q }_{ 0 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } \times \sum { { p }_{ 1 }{ q }_{ 0 } } \times \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } \times \sum { { p }_{ 0 }{ q }_{ 0 } } \times \sum { { p }_{ 0 }{ q }_{ 1 } } } \right) } \)
\({ P }_{ 01 }\times { P }_{ 10 }=\sqrt { \left( \frac { 623\times 517\times 484\times 517 }{ 584\times 484\times 584\times 623 } \right) } \)
\({ P }_{ 01 }\times { P }_{ 01 }=\sqrt { \left( \frac { 517\times 517 }{ 585\times 584 } \right) } =\frac { 517 }{ 584 } \)
\({ P }_{ 01 }\times { P }_{ 01 }=\frac { \sum { { p }_{ 1 }{ q }_{ 1 } } }{ \sum { { p }_{ 0 }{ q }_{ 0 } } } \)

Sample size n =100, Sample mean \(\bar x\) = 390 pages, Population mean \(\mu\) = 400 pages
Population SD \(\sigma\) = 20 pages
The sample is a large sample and so we apply Z -test
Null Hypothesis:
There is no significant difference between the sample mean and the population mean of writing life of pen he manufactures, i.e., H₀ : \(\mu\) = 400
Alternative Hypothesis:
There is significant difference between the sample mean and the population mean of writing life of pen he manufactures, i.e., H₁:\(\mu\neq\) 400 (two tailed test)
The level of significance \(\alpha\) = 1% = 0.01
Applying the test statistic
\(Z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}} \sim N(0,1) ; \)
\(Z=\frac{390-400}{\frac{20}{\sqrt{100}}}=\frac{-10}{2}=-5, \therefore|Z|=5\)
Thus the calculated value |Z| = 5 and the significant value or table value \({ Z }_{ \frac { \sigma }{ 2 } }=2.58\)
Comparing the calculated and table values, we found Z > \({ Z }_{ \frac { \sigma }{ 2 } }\) i.e., 5 > 2.58
Inference: Since the calculated value is greater than table value i.e., \(Z>{ Z }_{ \frac { \sigma }{ 2 } }\) at 1% level of significance, the null hypothesis is rejected and Therefore we concluded that \(\mu \neq400\) and the manufacturer’s claim is rejected at 1% level of significance.

Given mean = λ = 1.5
X follows poisson distribution with
p(x,λ) = \(\frac { e^{ -\lambda }.{ \lambda }^{ x } }{ x! } \)
(i) ∴ P(neither car is used)
= P(X = 0) = \(\frac { e^{ -1.5 }.(1.5)^{ 0 } }{ 0! } \) =e^-1.5
= 0.2231    [∵ e^-1.5 =0.2231]
(ii) P (Some demand is refused)
The demand may be either 0 car, 1 car or 2 cars
∴ P (Some demand is refused) = 1 - P(X ≤ 2)
= 1 - [P(X = 0) + P (X = 1) + P(X = 2)]
= 1-\(\left[ \frac { e^{ -\lambda }.{ \lambda }^{ 0 } }{ 0! } +\frac { e^{ -\lambda }.{ \lambda }^{ 1 } }{ 1! } +\frac { e^{ -\lambda }.{ \lambda }^{ 2 } }{ 2! } \right] \)
= 1-e^-λ (1+λ+\(\frac { { \lambda }^{ 2 } }{ 2 } \))
= 1-e^-1.5 (1+1.5+\(\frac { (1.5)^{ 2 } }{ 2 } \))
= 1 - 0.2231 (3.625) = 1 - 0.8087 = 0.1912
∴ Probability of some demand is refused = 0.1912.