Question

In: Statistics and Probability

1. Assume that X and Y are two independent discrete random variables and that X~N(0,1) and...

1. Assume that X and Y are two independent discrete random variables and that X~N(0,1) and Y~N(µ,σ2).

                a. Derive E(X3) and deduce that E[((Y-µ)/σ)3] = 0

                b. Derive P(X > 1.65). With µ = 0.5 and σ2 = 4.0, find z such that P(((Y-µ)/σ) ≤ z) = 0.95. Does z depend on µ and/or σ? Why

Solutions

Expert Solution

Solution

Back-up Theory

If a random variable X ~ N(µ, σ2), i.e., X has Normal Distribution with mean µ and variance σ2,

then, Z = (X - µ)/σ ~ N(0, 1), i.e., Standard Normal Distribution .................................................................................... (1)

and hence

P(X ≤ or ≥ t) = P[{(X - µ)/σ} ≤ or ≥ {(t - µ)/σ}] = P[Z ≤ or ≥ {(t - µ)/σ}] .……….................................................………...…(1a)

Probability values for the Standard Normal Variable, Z, can be directly read off from Standard Normal Tables............ (2a)

or can be found using Excel Function: Statistical, NORMSDIST(z) which gives P(Z ≤ z) ..........................................…(2b)

Percentage points of N(0, 1) can be found using Excel Function: Statistical, NORMSINV,

which gives values of t for which P(Z ≤ t) = given probability………………. …........................................................……(2c)

Skewness, α = [{E(X – µ)3}/σ3] .......................................................................................................................................(3)

Skewness for Normal distribution and for any symmetric distribution is zero ................................................................ (4)

Now to work out the solution,

Part (a)

Sub-part (i)

Given, X ~ N(0, 1), vide (3) and (4),

α = [{E(X – 0)3}/13] = 0

i.e., E(X3) = 0 Answer 1

Sub-part (ii)

Given, Y ~N(µ,σ2), vide (1),

(Y-µ)/σ) ~ N(0, 1)

The above along-with Answer 1 =>

E[((Y-µ)/σ)3]

= 0 Answer 2

Part (b)

Sub-part (i)

Given, X ~ N(0, 1), vide (2b),

P(X > 1.66)

= 0.0485 Answer 3

Sub-part (ii)

Given, Y~N(µ,σ2), vide (1),

(Y-µ)/σ) ~ N(0, 1)

Hence, vide (2c),

P[{(Y-µ)/σ)} ≤ z)] = 0.95

=> z = 1.645

Answer 4

Sub-part (iii)

Given, Y~N(µ,σ2), vide (1),

(Y-µ)/σ) ~ N(0, 1)

Hence, whatever be the value of µ and σ, the value of z does not change. Answer 5

DONE


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