A random variable Z = \(\frac { X-\mu }{ \sigma } \) follows the standard normal distribution is called the standard normal variate with mean 0 and standard deviation 1. i.e. Z~ N(0,1). Its,probability density function is given by:
\(\varphi(Z)=\frac{1}{\sqrt{2 \pi}} e^{-\frac{Z^{2}}{2}},-\infty< Z< \infty \)

Poisson distribution is a limiting case of binomial distribution under the following conditions.
(i) n, the number of trials is indefinitely large ie., n⟶∞.
(ii) p, the constant probability of success in each trial is very small ie., p ⟶0.
(iii) np = λ is finite.Thus p = λ/n and q = 1 -(λ/n) where λ is a positive real number.

(i) Number of lightnings per second.
(ii) Number of printing mistakes per page in a textbook.

Given mean of a binomial distribution is 5
np = 5 ....(1)
Also, standard deviation is 2 ⇒ Variance = 2² = 4
∴ npq = 4...(2)
(2) \(\div \) (1) given, \(\frac { npq }{ np } \)=\(\frac { 4 }{ 5 } \)
⇒ q = \(\frac { 4 }{ 5 } \)
p = 1-q = \(1-\frac { 4 }{ 5 } =\frac { 1 }{ 5 } \)
Substituting p = \(\frac { 1 }{ 5 } \) in (1) we get
n\(\left( \frac { 1 }{ 5 } \right) \) = 5 ⇒ n = 25
∴ The binomial distribution is ⁿC_x p^xq^n-x, x = 0,1,2,....n
⇒ ²⁵C_x \(\left( \frac { 1 }{ 5 } \right) ^{ x }\left( \frac { 4 }{ 5 } \right) ^{ 25-x }\), x = 0,1,2,....25.

Let p be the probability of getting doublet in a pair of dice.
∴ p = \(\frac { 6 }{ 36 } \) [∵ favourable events are (1, 1) (2,2) (3,3) (4,4) (5,5) (6,6) and n(S) = 36]
⇒ p = \(\frac { 1 }{ 6 } \) ∴ q=1-p =\(1-\frac { 1 }{ 6 } =\frac { 5 }{ 6 } \)]
∴ P (getting 2 success) = P(X = 2)
=⁴C₂ \(\left( \frac { 1 }{ 6 } \right) ^{ 2 }\left( \frac { 5 }{ 6 } \right) ^{ 2 }\)     [∵ p(x) =ⁿC_x p^xq^n-x, n = 4, x = 2 ]

∴ P(X = 2) =\(\frac { 25 }{ 216 } \).

Let p he the probability of getting a girls
∴ p = \(\frac { 1 }{ 2 } \) [∵ one favourable event and total no of events is 2]
⇒ q = 1-p = 1-\(\frac { 1 }{ 2 } \) = \(\frac { 1 }{ 2 } \) and n = 3
∴ (getting exactly 2 girls)
= P(X = 2)
= ³C₂\(\left( \frac { 1 }{ 2 } \right) ^{ 2 }\left( \frac { 1 }{ 2 } \right) ^{ 1 }\)   [∵ p(x) = ⁿC_x p^xq^n-x, n = 3 and x = 2]
= \(3\left( \frac { 1 }{ 2 } \right) ^{ 3 }=3\times \left( \frac { 1 }{ 2 } \right) ^{ 3 }=3\times \frac { 1 }{ 8 } \) = 0.375
P (getting exactly 2 girls) = 0.375.

Given μ = 225, σ = 55
P(X≤100)
When X = 100, Z = \(\frac { X-\mu }{ \sigma } \)
= \(\frac { 100-225 }{ 55 } \) = -2.272
∴ P(X≤100) = P(Z≤-2.272)

 = 0.5-P(0 [By symmetry]
= 0.5-0.4884 = 0.0116
∴ Probability of one customer to have a bill for Rs.100 or less is 0.0116.
∴ Out of 500 customers, the number of persons to have a bill for Rs. 100 or less is 0.0116 x 500 = 5.8 ≌ 6.

Let X be p probability random variable denoting the number of defective yarns.
Mean = np = 1.2, n = 6
6p = 1.2
\(p=\frac{1.2}{6}=0.2\)
q = 1-p = 1-0.2 = 0.8
P(X <2) = P(X = 0)+ P(X = 1)
P(X = x) = nC_x p^x q^n-x
P(X<2) = 6C₀(0.2)⁰(0.8)⁰+6C₁(0.2)¹ (8)⁵
= (0.32768)2
= 0.6553

Given μ = 30, σ = 4
(a) P(X < 40)
When X = 40, Z = \(\frac { X-\mu }{ \sigma } \)
=\(\frac { 40-30 }{ 4 } =\frac { 10 }{ 4 } \) = 2.5
∴ P(X<40) = P(Z<2.5)

= P(-∞) = 0.5+.4938
P(X<40) = 0.9938
(b) P(X>21)
When X = 21, Z = \(\frac { 21-30 }{ 4 } \quad \)
= \(\frac { -9 }{ 4 } \) = -2.25
∴ P(X>21) = P(Z>-2.25)

= 0.4878 + 0.5
P(X > 21) = 0.9878
(c) P(30 When X = 30, Z₁ =\(\frac { 30-30 }{ 4 } \) = 0
When X = 35, Z₂ = \(\frac { 35-30 }{ 4 } \) = 1.25
P(30 = 0.3944

Given μ = 20, σ = 2
a) P(X<19.5)
When X = 19.5, Z = \(\frac { X-\mu }{ \sigma } \)
= \(\frac { 19.5-20 }{ 20 } \) = -0.25
∴ P(Z<-0.25) = P(-∞ =P(-∞

= 0.5-P(0 = 0.5-0.0987
P(X<19.5) = 0.4013
b) P(between 20 and 22 hours)
= P(20 When X = 20, Z₁ = \(\frac { 20-20 }{ 2 } \) = 0
When X = 22, Z₂ = \(\frac { 22-20 }{ 2 } \) = 1

∴ P (20 < X < 22) - P(0 < Z < 1)
∴ P (20 < X < 22) = 0.3413

Let p be the probability getting his piston rejected
Given p = 12% = \(\frac { 12 }{ 100 } \) = 0.12
∴ q = 1-p = 1-0.12 = 0.88
n = 10
(a) P (not more than 2 rejects)
= P(X≤2) = P(X = 0) + P(X = 1) + P(X = 2)
= 10C₀(0.12)⁰ (0.88)¹⁰ + 10C₁ (0.12)¹ (0.88)⁹ + 10C₂ (0.12)² (0.88)⁸
[∵ P(x) = nC_x p^xq^n-x]

= (0.88)⁸ [(0.88)² + 10(0.12) (0.88) + 45 (0.12)²]
= (0.88)⁸ [0.7744 + 1.056 + 0.648]
= (0.3596) (2.4784) = 0.8913
∴ Probability of not more than 2 rejects = 0.8913
b) P(at least 2 rejects)
= P(X ≥ 2) = 1 - P (X < 2)
= 1 - [P(X = 0) + P (X = 1)]
= 1-[10C₀(0.12)⁰ (0.88)¹⁰ + 10C₁ (0.12)¹ (0.88)⁹ ]
= 1 - [(0.88)¹⁰ + 10 (0.12) (0.88)⁹]
= 1 - (0.88)⁹ [0.88 + 1.2] = 1 - (0.31647) (2.08)
= 1 - 0.6583 = 0.34173
P (atleast 2 rejects) = 0.34173

Given μ = 12 and σ = 4
(1) P( ≤ 20)
When X = 20, Z = \(\frac { X-\mu }{ \sigma } \)
= \(\frac { 20-12 }{ 4 } =\frac { 8 }{ 4 } \) = 2
∴ P(X≤20) = P(Z≤2)
= P(-∞ = 0.5+0.4772
P(X≤20) = 0.9772

(ii) P(0≤X≤12)
When X=0, Z=\(\frac { X-\mu }{ \sigma } \)
\(\frac { 0-12 }{ 4 } =\frac { -12 }{ 4 } \)=-3
When X=12, Z=\(\frac { X-\mu }{ \sigma } \)
=\(\frac { 12-12 }{ 4 } =\frac { 0 }{ 4 } \)=0
∴ P(0≤X≤12) = P(-3≤Z≤0)
= P(0≤Z≤3) (By symmetry)
P(0≤X≤12) = 0.4987.

Plot the variable X = 50 on the left side and
X = 86 on the right side of the curve.
Given P( -∞< Z₁ < -Z₁) = 0.3
⇒ P(-Z₁ < Z < Z₁) = 0.2       [∵ 0.5 - 0.3 = 0.2]
⇒ P(0 < Z < Z₁) = 0.2 [By symmetry]
⇒ Z₁ = -0.52 [From the normal distribution table 'and it lies on the negative side]
⇒ -0.52 = \(\frac { X-\mu }{ \sigma } \) ⇒ -0.52 σ = 50-μ
⇒ 50-μ = -0.52 σ ....(1)
Also given P(Z₂ ∴ P(Z₂2) = 0.4    [∵ 0.5-0.1 = 0.4]
⇒ Z₂ = 1.28 (from the table)
⇒ 1.28 =\(\frac { 86-\mu }{ \sigma } \)
⇒ 86-μ = 1.28 σ ...(2)

Substituting σ = 20 in (2) we get,
86-μ = (1.28)(20)
86-μ = 25.6
μ = 86-25.6
μ = 60.4

Hence, the mean is 60.4 and standard deviation is 20.

Given average number of phone calls between 10 am and 2.30,pm per minute is 2.5.
∴ λ = 2.5
Hence X follows a Poisson distribution with
p(x, λ) = \(\frac { { e }^{ -\lambda }.{ \lambda }^{ x } }{ x! } \)
(i) P (no phone at all) = P(X = 0)
P(X = 0) = \(\frac { { e }^{ -2.5 }.(2.5)^{ 0 } }{ 0! } \) = e^-2.5
= 0.08208   [∵ e^-2.5 = 0.08208]
(ii) P (exactly 3 calls)
P(X = 3) = \(\frac { { e }^{ -2.5 }.(2.5)^{ 3 } }{ 3! } \)  [∵ λ = 2.5 and x=3]
= \(\frac { (0.08208)(2.5)^{ 3 } }{ 3\times 2 } \) = 0.21375
∴ P(X = 3) = 0.2138
(iii) P (atleast 5 calls) = P(X ≥ 5)
P(X≥5) = 1-P(X<5)
=1 - [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)]
= 1-\(\left[ \frac { { e }^{ -\lambda }.{ \lambda }^{ 0 } }{ 0! } +\frac { { e }^{ -\lambda }.{ \lambda }^{ 1 } }{ 1! } +\frac { { e }^{ -\lambda }.{ \lambda }^{ 2 } }{ 2! } +\frac { { e }^{ -\lambda }.{ \lambda }^{ 3 } }{ 3! } +\frac { { e }^{ -\lambda }.{ \lambda }^{ 4 } }{ 4! } \right] \)
= 1-e^-λ \(\left( 1+\lambda +\frac { { \lambda }^{ 2 } }{ 2 } +\frac { { \lambda }^{ 3 } }{ 6 } +\frac { { \lambda }^{ 4 } }{ 24 } \right) \)
= 1-e^-2.5 (1+2.5+\(\frac { ({ 2.5) }^{ 2 } }{ 2 } +\frac { (2.5)^{ 3 } }{ 6 } +\frac { (2.5)^{ 4 } }{ 24 } \))
= 1-e^-2.5 (1 + 2.5 + 3.125 + 2.604) + 1.6276)
= 1-e^-2.5 (10.8566)
= 1 - 0.08208 (10.8566) = 1 - 0.89111
P(X ≥ 5) = 0.1089.

Mean E(X) =\(\overset { \infty }{ \underset { x=0 }{ \Sigma } } x.p(x,\lambda )\)
= \(\overset { \infty }{ \underset { x=0 }{ \Sigma } } x.\frac { e^{ -\lambda }.\lambda ^{ x } }{ x! } \)
Taking out λ from the numerator and x from the denominator

= \(\lambda .{ e }^{ -\lambda }\overset { \infty }{ \underset { x=1 }{ \Sigma } } \frac { { \lambda }^{ x-1 } }{ (x-1)! } \)
= \(\lambda .e^{ -\lambda }(1+\lambda +\frac { { \lambda }^{ 2 } }{ 2! } +...)\)
= \(\lambda .e^{ -\lambda }.e^{ \lambda }\)   [∵ e^x =1+x+\(\frac { x^{ 2 } }{ 2! } \)+..]
= λ.e⁰ =λ(1) =λ
∴ E(X) = λ
E(X²) = \(\overset { \infty }{ \underset { x=0 }{ \Sigma } } p(x,\lambda )\)
= \(\overset { \infty }{ \underset { x=0 }{ \Sigma } } { x }^{ 2 }.\frac { e^{ -\lambda }.{ \lambda }^{ x } }{ x! } \)
=\(\overset { \infty }{ \underset { x=0 }{ \Sigma } } (x(x-1)+x)\frac { e^{ -\lambda }{ \lambda }^{ x } }{ x! } \)
[∵ x(x-1)+x = x²-x+x = x²]
=\(\overset { \infty }{ \underset { x=0 }{ \Sigma } } (x(x-1).\frac { e^{ -\lambda }{ \lambda }^{ x } }{ x! } +\overset { \infty }{ \underset { x=0 }{ \Sigma } } \frac { x.e^{ -\lambda }{ \lambda }^{ x } }{ x! } \)
Taking λ² from the numerator and x(x-1) from. the denominator we get,
E(X²)=\(\overset { \infty }{ \underset { x=0 }{ \Sigma } } \frac { { \lambda }^{ 2 } }{ x(x-1) } .(x)(x-1).\frac { { e }^{ -\lambda }.{ \lambda }^{ x-2 } }{ (x-2)! } +E(X)\)
= \({ \lambda }^{ 2 }.e^{ -\lambda }\overset { \infty }{ \underset { x=0 }{ \Sigma } } \frac { { \lambda }^{ x-2 } }{ (x-2)! } +\lambda \)   [∵ E(X)=λ]
= λ².e^-λ.e^λ+λ   [∵ e^λ =1+λ+\(\frac { { \lambda }^{ 2 } }{ 2! } \)+....]
= λ²+λ
V(X) = E(X²)-[E(X)]² = λ²+ λ - λ² = λ
∴ Mean = Variance = λ