Question

Let X ∼ Normal(μ = 20, σ2 = 4).
(a) Give the mgf MX of X.
(b) Find the 0.10 quantile of X.
(c) Find an interval within which X lies with probability 0.60.
(d) Find the distribution of Y = 3X −10 by finding the mgf MY of Y

Answer 1

X Normal ( = 20, 2 = 4)

a) Answer :

The MGF of X is

MX(t) = exp(t + 0.52t2)     ..........(exp stands for exponential)

         = exp(20t + 0.5 * 4t2)

         = exp(20t + 2t2)

Therefore, MGF of X is exp(20t + 2t2)

b) Answer :

Let c denote the 0.10 quantile of X.

P(X < c) = 0.10

P(Z < (c - 20) / 2) = 0.10

1 - P(Z > (c - 20) / 2) = 0.10

P(Z > (c - 20) / 2) = 0.90

P(Z < -(c - 20) / 2) = 0.90    .............( P(Z > a) = P(Z < -a))

From the standard normal table, P(Z < 1.28) = 0.90

-(c - 20) / 2 = 1.28

c = 20 - (2 * 1.28)

c = 17.44

Therefore, the 0.10 quantile of X is 17.44

c) Answer :

Let P(-m < X < m) = 0.60

P(X < m) - P(X < -m) = 0.60

P(X < m) - P(X > m) = 0.60 ..........( P(Z < -a) = P(Z > a))

P(X < m) - (1 - P(X < m)) = 0.60

2P(X < m) - 1 = 0.60

P(X < m) = 0.80

P(Z < (m - 20) / 2) = 0.80

From the standard normal table, P(Z < 0.84) = 0.80

(m - 20) / 2 = 0.84

m = 20 + 2 * 0.84

m = 21.68

Therefore, X lies between -21.68 and 21.68 with probability 0.60

d) Answer :

Given, Y = 3X - 10

The MGF of Y is

MY(t) = E(eYt)

         = E(e(3X - 10)t)

         = E(e3Xt * e-10t)

         = e-10t * E(e3tX)

         = e-10t * MX(3t)

         = e-10t * exp(20 *3t + 2 * (3t)2)

         = exp(-10t) * exp(60t + 18t2)

         = exp(50t + 18t2)

Therefore, the MGF of Y is exp(50t + 18t2)