Let X denote the life of army shoes.
Given μ = 8, σ = 2 and N = 5000
When μ = 12, Z = \(\frac { X-\mu }{ \sigma } =\frac { 12-8 }{ 2 } \) = 2
∴ P(X≤12) = P(Z≤2)
= P(-∞ = P(-∞ =0.5+0.4772
P(X≤12) = 0.9772

∴ The probability for a shoe need to be replaces is 0.9772.
∴ Out of 5000 pairs of shoes, number of pairs need to be replaced = 5000 \(\times\) 0.9772
 = 4886.

Let X be a random vertical. which is normally distributed
Given that μ = 4.35 and σ = 0.59
When X = 5.20, Z + \(\frac { X-\mu }{ \sigma } =\frac { 5.2-4.35 }{ 0.59 } \)
= \(\frac { 0.85 }{ 0.59 } \) = 1.44

∴ P(X > 5.20) = P(Z > 1.44)
= P(1.44 = P(0 = 0.5 - 0.4251
P(X > 5.20) = 0.749

Average rate of particles emitted in
20 minutes = 5
∴ λ = 5
Let X be the number of particles emitted in 20 minutes
∴ P(X = x) = \(\frac { { e }^{ -\lambda }.{ \lambda }^{ 0 } }{ x! } \), x = 0,1,2,....n
∴ P(X≥2) = 1-P(X<2)
= 1-[P(X = 0) + P(X = 1)]
= 1-\(\left[ \frac { e^{ -5 }.5^{ 0 } }{ 0! } +\frac { e^{ -5 }.5^{ 1 } }{ 1! } \right] \)
= 1-e^-5 (1+5)
= 1-e^-5(6)
= 1-(0.0067)(6)
= 1-0.0402
P(X ≥ 2) = 0.9598

Given n = 500
Letp = Probability of number of ships arrive safely = \(\frac { 9 }{ 10 } \)
∴ q = 1 - p = \(1-\frac { 9 }{ 10 } =\frac { 1 }{ 10 } \)
Mean = np = 500 \(\times\) \(\frac { 9 }{ 10 } \)=450
Variance = npq = 450 \(\times\) \(\frac { 1 }{ 10 } \)= 45
∴ Standard deviation =\(\sqrt { npq } =\sqrt { 45 } =3\sqrt { 5 } \).

Given n = 20
P(getting 1 or 5) = \(\frac { 2 }{ 6 } \Rightarrow p=\frac { 1 }{ 3 } \)
∴ q = 1 - p = \(1-\frac { 1 }{ 3 } =\frac { 2 }{ 3 } \)
Using Binomial distribution,
Mean = np = 120 \(\times\) \(\frac { 1 }{ 3 } \) = 40
Variance = npq = 40 \(\times\) \(\frac { 2 }{ 3 } =\frac { 80 }{ 3 } \)