\(\mu=100, \sigma^{2}=36, \quad \sigma=6, Z=\frac{X-\mu}{\sigma}\)
\(\text {(i) } \mathrm{Z}=\frac{112-100}{6}=2\)

\( \mathrm{P}(\mathrm{X}>112) =\mathrm{P}(\mathrm{Z}>2) \)
\( =0.5-\mathrm{P}(0<\mathrm{Z}<2) \)
= 0.5-0.4772  
= 0.0228
\(\text { (ii) } Z=\frac{106-100}{6}=1\)

\( \mathrm{P}(\mathrm{X}<106) =\mathrm{P}(\mathrm{Z}<1) \)
\(=0.5+\mathrm{P}(0<\mathrm{Z}<1) \)
= 0.5 + 0.3413  
= 0.8413
\(\text {(iii) } \mathrm{Z}=\frac{94-100}{6}=-1\)

\(\mathrm{Z}=\frac{106-100}{6}=1\)
\( \mathrm{P}(94 \leq X \leq 106) =\mathrm{P}(-1 \leq Z \leq 1) \)
\( =2 \mathrm{P}(0 \leq Z \leq 1) \)
= 2(0.3413)  
= 0.6826

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

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

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

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

\( =\mathrm{P}(0 \leq Z \leq 1.12)-\mathrm{P}(0 \leq Z \leq 0.43) \)
 = 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

Let X be a random variable denoting the number of injured persons who will collect their policy.
\(\lambda=\mathrm{np}=4000 \times \frac{10}{100000}=0.4\)
\( \mathrm{P}(\mathrm{X}=\mathrm{X}) =\frac{e^{-\lambda} \lambda^{x}}{x !} \)
\(\mathrm{P}(\mathrm{X}>3) =1-\mathrm{P}(\mathrm{X} \leq 3) \)
\(=1-[\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+ \mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3)] \)
\(=1-\mathrm{e}^{-0.4}\left[1+\frac{0.4}{1 !}+\frac{0.4}{2 !}+\frac{(0.4)^{3}}{3 !}\right]\)

Let X be the random variable denoting the number of accidents in a year
\( \lambda=3 \)
\(\mathrm{P}(\mathrm{X}=\mathrm{x})=\frac{e^{-\lambda} \lambda^{x}}{x !} \)
\(\text {(i) } \mathrm{P}(\mathrm{X}=0)=\frac{e^{-3} 3^{0}}{0 !}=\mathrm{e}^{-3}=0.05\)
Number of drivers with no accident
\(=1000 \times 0.05=50\)
\((ii)\ \mathrm{P}(\mathrm{X}>3)=1-\mathrm{P}(X \leq 3)^{\prime} \)
\(=1-[\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+ \quad \mathrm{P}(\mathrm{X}=2)]+\mathrm{P}(\mathrm{X}=3)] \)
\(=1-\mathrm{e}^{-3}\left[1+3+\frac{3^{2}}{2 !}+\frac{3^{3}}{3 !}\right] \)
\(=1-\mathrm{e}^{-3}[1+3+4.5+4.5] \)
\(=1-\mathrm{e}^{-3}(13)=1-0.65=0.35\)
Number of drivers with more than 3 accidents = 1000 \(\times\) 3.5 = 350

Let X be the random variable denoting the number of fuses.
\( \mathrm{p}=\frac{2}{100} ; \mathrm{n}=200 \)
\(\lambda=\mathrm{np}=4 \)
\( \mathrm{P}(\mathrm{X}=\mathrm{x})=\frac{e^{-\lambda} \lambda^{x}}{n !} \)
\(\mathrm{P}(\mathrm{X} \leq 5)= \mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+ \mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=4) +\mathrm{P}(\mathrm{X}=5) \)
\(= \frac{e^{-4} 4^{0}}{0 !}+\frac{e^{-4} 4^{1}}{1 !}+\frac{e^{-1} 4^{2}}{2 !}+\frac{e^{-4} 4^{3}}{3 !} +\frac{e^{-1} 4^{4}}{4 !}+\frac{e^{-4} 4^{5}}{5 !} \)
\(= e^{-4}\left[1+\frac{4}{1 !}+\frac{4^{2}}{2 !}+\frac{4^{3}}{3 !}+\frac{4^{4}}{4 !}+\frac{4^{5}}{5 !}\right] \)
\(=(0.0183) \times \frac{643}{15}=0.785\)

Let x be the random variable denoting the number of successes.
\( \mathrm{P}(\mathrm{X}=\mathrm{x})=n C_{x} p^{x} q^{n-x} \)
\(\mathrm{n}=5, \mathrm{p}=0.3, \mathrm{q}=0.7 \)
\(=5 C_{3}(0.3)^{3}(0.7)^{2}+5 C_{4}(0.3)^{4}(0.7) +5 \mathrm{C}_{5}(0.3)^{5} \)
\(=0.1631 \)
\(\text {(ii) } \mathrm{P}(\mathrm{X} \leq 3)=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)^{\circ}+\mathrm{P}(\mathrm{X}=2) +\mathrm{P}(\mathrm{X}=3) \)
\(=5 C_{0}(0.3)^{0} \cdot(0.7)^{5}+ 5 \mathrm{C}_{1}(0.3)^{1}(0.7)^{4}+5 \mathrm{C}_{2}(0.3)^{2}(0.7)^{3} \cdots+5 \mathrm{C}_{3}(0.3)^{3}(0.7)^{2} \)
= 0.9692

Let X be the random variable denoting the number of successes.
p probability of getting an odd number
\( =\frac{3}{6}=\frac{1}{2} \)
\( \mathrm{q}=1-\mathrm{p}=1-\frac{1}{2}=\frac{1}{2} \)
\( \mathrm{n}=6 \)
\( \mathrm{P}(\mathrm{X}=\mathrm{x})=n C_{x} p^{x} q^{n-x} \)
\( (i)\ \mathrm{P}(\mathrm{X}=5)=6 C_{5}\left(\frac{1}{2}\right)^{6}=\frac{6}{64}=\frac{3}{32} \)
\((ii)\ \mathrm{P}(X \geq 5)=\mathrm{P}(\mathrm{X}=5)+\mathrm{P}(\mathrm{X}=6) \)
\( =6 C_{5}\left(\frac{1}{2}\right)^{6}+6 C_{6}\left(\frac{1}{2}\right)^{6} \)
\(=\frac{1}{64}\left[6 C_{1}+1\right]=\frac{7}{64}\)
\(\text {(iii) } \mathrm{P}(X \leq 5) =1-\mathrm{P}(\mathrm{X}>5) \)
\(=1-\mathrm{P}(\mathrm{X}=6) \)
\(=1-6 \mathrm{C}_{6}\left(\frac{1}{2}\right)^{6}=\frac{63}{64} \)
\((iv) \ \mathrm{P}(X \geq 1)=1-\mathrm{P}(\mathrm{X}<1) \)
\( =1-P(X=0) \)
\(=1-6 C_{0}\left(\frac{1}{2}\right)^{6}=1-\frac{1}{64}=\frac{63}{64} \)
\((v)\ \mathrm{P}(\mathrm{X}=0)=6 C_{0} \frac{1}{2^{6}}=\frac{1}{64}\)

Let X be the random variable denoting the number of successes 
p = probability of getting a total of 7 in a single throw of a pair of dice
= {(1, 6), (6, 1), (2, 5) (5, 2), (3, 4) (4, 3)}
\( =\frac{6}{36}=\frac{1}{6} \)
\(q =\frac{5}{6} \)
\( \mathrm{P}(\mathrm{X}=\mathrm{x})=n C_{x} p^{x} q^{n-x} \)
\((i)\ \mathrm{P}(\mathrm{X}=0)=7 C_{0}\left(\frac{1}{6}\right)^{0}\left(\frac{5}{6}\right)^{7}=\left(\frac{5}{6}\right)^{7} \)
\((ii)\ \mathrm{P}(\mathrm{X}=6)=7 C_{6}\left(\frac{1}{6}\right)^{6}\left(\frac{5}{6}\right)^{1}=35\left(\frac{1}{6}\right)^{7} \)
\( \text {(iii) } \mathrm{P}(X \geq 6)=\mathrm{P}(\mathrm{X}=6)+\mathrm{P}(\mathrm{X}=7) \)
\( =7 C_{6}\left(\frac{1}{6}\right)^{6}\left(\frac{5}{6}\right)^{1}+7 C_{-}\left(\frac{1}{6}\right)^{7} \)
\( =\frac{1}{6^{7}}\left[7 C_{1}(5)+1\right]=\frac{36}{6^{7}}=\frac{1}{6^{5}} \)
\( \text {(iv) } \mathrm{P}(X \leq 6)=1-\mathrm{P}(\mathrm{X}>6) \)
\( =1-\mathrm{P}(\mathrm{X}=7) \)
\( =1-7 C_{-}\left(\frac{1}{6}\right)^{-}=1-\frac{1}{6^{7}} \)

Mean+ Variance = 24
np+ npq = 24
\( \mathrm{np}(1+\mathrm{q}) =24 \Rightarrow \mathrm{np}=\frac{24}{1+q} \)
\(\text {Mean }(\text {variance }) =128 \)
\(\mathrm{np}(\mathrm{npq}) =128 \)
\(\mathrm{n}^{2} \mathrm{p}^{2} =\frac{128}{q} \)
\(\left(\frac{24}{1+q}\right)^{2} =\frac{128}{q} \)
\(9 \mathrm{q} =2\left(1+2 \mathrm{q}+\mathrm{q}^{2}\right)\)
\( 2 q^{2}-5 q+2 =0 \)
\((2 q-1)(q-2) =0 \Rightarrow q=\frac{1}{2} ; p=\frac{1}{2} \)
\(n p(1+q) =24 \)
\(n\left(\frac{1}{2}\right)\left(1+\frac{1}{2}\right) =24 \)
\(\mathrm{n} =\frac{24 \times 2 \times 2}{3}=32\)
The Binomial distribution n
\( \mathrm{P}(\mathrm{X}=\mathrm{x}) =n C_{x} p^{x} q^{n-x} \)
\(= 32 C_{x}\left(\frac{1}{2}\right)^{x}\left(\frac{1}{x}\right)^{32-x}, \mathrm{x}=0,1,2 \ldots 32 \)

Mean = np = 4
Variance = npq = 2
\( \Rightarrow \ \mathrm{q} =2 \Rightarrow \mathrm{q}=\frac{2}{4}=\frac{1}{2} \)
\(\mathrm{p}= \frac{1}{2} \)
\(n\left(\frac{1}{2}\right)= 4 \Rightarrow \mathrm{n}=8 \)
\(\mathrm{P}(\mathrm{X}= \mathrm{x})=n C_{x} p^{x} q^{n-x} \)
\(\mathrm{P}(|\mathrm{X}-4| \leq 2)= \mathrm{P}^{\mathrm{P}}(-2 \leq \mathrm{X}-4 \leq 2)=\mathrm{P}(2 \leq X \leq 6) \)
\(= \mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3) +\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5)+\mathrm{P}(\mathrm{X}=6) \)
\(= 8 C_{2}\left(\frac{1}{2}\right)^{8}+8 C_{3}\left(\frac{1}{2}\right)^{8}+8 C_{4}\left(\frac{1}{2}\right)^{8}+ 8 C_{5}\left(\frac{1}{2}\right)^{8}+8 C_{6}\left(\frac{1}{2}\right)^{8} \)
\(= \frac{1}{2^{8}}\left[8 C_{2}+8 C_{3}+8 C_{4}+8 C_{5}+8 C_{6}\right] \)
\(= \frac{119}{128} \)

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.

Let X denote the height of the student
Given μ = 64.5 inches and σ = 3.3 inches
Given that P(-∞ < Z < C) = 0.99
⇒ P( -∞ < Z < 0) + P (0 < Z < C) = 0.99
⇒ 0.5 + P (0 < Z < C) = 0.99
⇒ P(0 < Z < C) = 0.99 - 0.5 = 0.49...(1)
From the standard normal distribution table
P(0 < Z < 2.33) = 0.49...(2)
From (1) & (2), C = 2.33
we know that Z = \(\frac { X-\mu }{ \sigma } \)
⇒ 2.33 = \(\frac { X-64.5 }{ 3.3 } \)
⇒ X = (2.33) (3.3) + 64.5
⇒ X = 72.19 inches
Hence, the height below which 99% of the student lie is 72.19 inches.

Let X denotes the weight of the male students
Given μ = 151, σ = 15 and N = 500
(i) between 120 and 155 pounds
P(120 < X < 155)
When X = 120, Z = \(\frac { X-\mu }{ \sigma } \)
= \(\frac { 120-151 }{ 15 } \)
= \(\frac { -31 }{ 15 } \) = -2.067
When x = 155, Z = \(\frac { 155-151 }{ 15 } \)
= \(\frac { 4 }{ 15 } \) = 0.2667

∴ P(120< X < 155) = P(-2.067 < X < 0.2667)
= P (-2.067 < Z < 0) + P (0 < Z < 0.2667)
= P (0 < Z < 2.067) + P (0 < Z < 0.2667)
(By symmetry)
= 0.4803 + 0.1026 = 0.5829
Probability for a student weigh between 120 and 155 pounds is 0.5829
∴ Out of 500 students, number of students weighing between 120 and 155 pounds
= 500 \(\times\) 0.5829 = 291 students
(ii) more than 185 pounds
P(X > 185)
When X = 185, Z = \(\frac { 185-151 }{ 15 } \)
=\(\frac { 34 }{ 15 } \) = 2.2667

∴ P(X >185) = P (Z > 2.2667)
= P (2.2667 < Z < ∞)
= P(0 < Z < ∞) - (0 < Z < 2.2667)
= 0.5 - 0.4881
= 0.0119
i.e Probability for a student weighing above 185 is 0.0119
∴ out of 500 male students, number of students weighing more than 185 pounds.
= 500 \(\times\) 0.0119 = 6 students.

Given n = 10, p = \(\frac { 20 }{ 100 } =\frac { 1 }{ 5 } \)
∴ q = 1-p = \(1-\frac { 1 }{ 5 } =\frac { 4 }{ 5 } \)
Let
X denote the number of defective bolts chosen
∴ X = 2
(i) Using binomial distribution
P(X = 2) = \(10{ C }_{ 2 }\left( \frac { 1 }{ 5 } \right) ^{ 2 }\left( \frac { 4 }{ 5 } \right) ^{ 8 }\)
= \(\frac { 10\times 9 }{ 2\times 1 } \left( \frac { { 4 }^{ 8 } }{ { 5 }^{ 10 } } \right) =45\left( \frac { 4^{ 8 } }{ { 5 }^{ 10 } } \right) \)
(ii) Using Poisson distribution
λ = np = 10 \(\times\) \(\frac { 1 }{ 5 } \) = 2
P(X = x) \(\times\)\(\frac { { e }^{ -\lambda }.{ \lambda }^{ 0 } }{ x! } \)
x = 0,1,2,......n
∴ P(X = 2) = \(\frac { e^{ -2 }(2^{ 2 }) }{ 2 } =e^{ -2 }\left( \frac { 4 }{ 2 } \right) \)
= 2e^-2
= 2(0.1353) = 0.2706
∴ P(X = 2) = 0.2706

Given n = 4
p = probability of getting a head when a coin is tossed
= \(\frac { 1 }{ 2 } \)
∴ q=1-p =\(1-\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)
X is a ran dom verticical having
p(X = x) = nC_x p^xq^n-x, x = 0,1,2,3,4
a) atleast 2 heads
P(X ≥ 2) = 1-P(X<2)
= 1- [P(X = 0) + P(X = 1)]
=1-\(\left[ 4C_{ 0 }\left( \frac { 1 }{ 2 } \right) ^{ 0 }\left( \frac { 1 }{ 2 } \right) ^{ 4 }+4C_{ 1 }\left( \frac { 1 }{ 2 } \right) ^{ 1 }\left( \frac { 1 }{ 2 } \right) ^{ 3 } \right] \)
=1-\(\left[ 1\left( \frac { 1 }{ 16 } \right) +4\left( \frac { 1 }{ 8 } \right) \left( \frac { 1 }{ 2 } \right) \right] \)
=1-\(1-\left( \frac { 4 }{ 16 } +\frac { 1 }{ 16 } \right) =1-\frac { 5 }{ 16 } =\frac { 11 }{ 16 } \)
P(X ≥ 2) = \(\frac { 11 }{ 16 } \)
b) atmost 2 heads
P(X≤2) = 1-P(X>2)
= 1 - [P(X = 3) + P (X = 4)]
= 1-\(\left[ { 4C }_{ 3 }\left( \frac { 1 }{ 2 } \right) ^{ 3 }\left( \frac { 1 }{ 2 } \right) ^{ 1 }+{ 4C }_{ 4 }\left( \frac { 1 }{ 2 } \right) ^{ 4 }\left( \frac { 1 }{ 2 } \right) ^{ 0 } \right] \)
= 1-\(\left[ 4\left( \frac { 1 }{ 8 } \right) \left( \frac { 1 }{ 2 } \right) +1\left( \frac { 1 }{ 16 } \right) \right] \)
= \(1-\left( \frac { 4 }{ 16 } +\frac { 1 }{ 16 } \right) =1-\frac { 5 }{ 16 } =\frac { 11 }{ 16 } \)
P(X≤2) =\(\frac { 11 }{ 16 } \).