Given μ = 3500, σ = 500
P(X<3100)
When X = 3100,
Z = \(\frac { X-\mu }{ \sigma } =\frac { 3100-3500 }{ 500 } \)
= -0.8
∴ P(X<3100) = P(Z<-0.8)


= 0.5-P(0 [By symmetry]
= 0.5-0.2881= 0.2119
Hence, the probability that a baby is born with weight less than 3100g is 0.2119.

Given average = λ = \(\frac { 3 }{ 20 } \) = 0.15
P(will not be more than one failure)
= P(X ≤ 1)
= P(X = 0) + P(X = 1)
= \(\frac { { e }^{ -\lambda }.{ \lambda }^{ 0 } }{ 0! } +\frac { { e }^{ -\lambda }.{ \lambda }^{ 1 } }{ 1! } \)
= e^-λ (1+λ) =e^-0.15 (1+0.15)
= (0.8607) (1.15)
= (0.98981 [e^-0.15 =0.8607]
∴ Probability that there will not be more than one failure = 0.98981

Let p be the probability of a patient to recover
Given q = 75% = \(\frac { 75 }{ 100 } \)  = 0.75
∴ P = 1 - q = 1 - 0.75 = 0.25
n = 6
P (4 will recover) = P(X = 4)
= 6C₄ (0.25)⁴ (0.75)²
[∵ P(x) = nC_x p^xq^n-x n = 6, x = 4
= 6C₂ (0.25)⁴ (0.75)²

= 15 (0.25)⁴ (0.75)²
P(X = 4) = 0.03295

Let p be the probability of defective electric bulb.
p = \(\frac { 5 }{ 100 } \) and n = 120

Hence λ follows Poisson distribution with
p(x,λ) = \(\frac { e^{ -\lambda }.{ \lambda }^{ x } }{ x! } \)
∴ P (no defective bulb) = P(X = 0)
= \(\frac { { e }^{ -6 }.6^{ 0 } }{ 0! } \) = e^-6  [λ = 6 and x = 0]
P(X=0) = 0.0025   [∵ e^-6 = 0.0025]

Let X be probability of mortality rate for a certain disease
∴ p = \(\frac { 7 }{ 1000 } \) and n = 400

Hence X follows poisson distribution with p(x, λ)
= \(\frac { e^{ -\lambda }.{ \lambda }^{ x } }{ x! } \)
∴ P (just 2 deaths) = P(X = 2)
= \(\frac { e^{ -2.8 }(2.8)^{ 2 } }{ 2! } \)    [∵ p(x,λ) =\(\frac { e^{ -\lambda }.{ \lambda }^{ x } }{ x! } \), λ = 2.8, x = 2]
= \(\frac { (0.06)(2.8)^{ 2 } }{ 2 } \) = (0.03)(2.8)² = 0.2352
∴ Probability for just 2 deaths on account of this disease = 0.2352.