IV. Suppose X d N (, , (-) is the cumulative distribution function for the N (0,

Question

IV. Suppose X d N (, , (-) is the cumulative distribution function for the N (0, 1) random variable. Where appropriate, write your answers in terms of (x) with x > 0. (1) P ((X-T)/> 16nt)-? (3) P (X (X- ) (X-n(X-T) (X-2r) > 0) =? (4) Let Give the moment generating function and density function for Z. (5) Suppose Y =-Z with Z being the same Z from(4). Give the moment generating function and density function of Y. (6) what is P (Z =-Z)? Explain why your answer does not contradict the results you obtained in parts (4) and (5)

Explanation / Answer

Given X ~ N(, 2),

Z = {(X - )/} ~ N(0, 1)…………………………………………………….. (1)

Part (1)

(X - )4 > 164

=> (X - ) > 2

=> {(X - )/} > 2.

Thus, P[(X - )4 > 164] = P[{(X - )/} > 2] = P(Z > 2) = 0.0228 ANSWER

[using Excel Function of Standard Normal Distribution]

Part (2)

P[- (2/5) < X < (17/5)]

= P[{- (2/5) – } < (X - ) < {(17/5) - ]

= P[- (7/5) < (X - ) < (12/5)]   

= P[{- (7/5)}/ < {(X - )/} < {(12/5)}/]

= P[- (7/5) < Z < (12/5)]

= P(Z < 2.4) - P(Z < - 1.4)

= 0.9918 – 0.0808 [using Excel Function of Standard Normal Distribution]

= 0.9110 ANSWER

Part (3)

X(X – /2)(X – )(X – 3/2)(X – 2) > 0 => X > 2.

So, P[X(X – /2)(X – )(X – 3/2)(X – 2) > 0] = P(X > 2)

= P[{(X - )/} > 1]

= P(Z > 1)

= 0.1587 ANSWER [using Excel Function of Standard Normal Distribution]

Part (4)

As already stated at the very start, z = {(X - )/} is a Standard Normal Variate and hence its pdf and MGF are pdf and MGF of N(0, 1). ANSWER

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