(g) F(x)= √√x for 0 < x < 1 and F-¹(x) = x², so X = U² (h) E(et(a+bX)) = E(eªt ebtX) = eªt Mx (bt) (i) [ ² ³x(2² − x)dx = ³ [x² − £x³|²] = (4 − §) = 1
(g) F(x)= √√x for 0 < x < 1 and F-¹(x) = x², so X = U² (h) E(et(a+bX)) = E(eªt ebtX) = eªt Mx (bt) (i) [ ² ³x(2² − x)dx = ³ [x² − £x³|²] = (4 − §) = 1
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
Please answer q's g, h, and i
![For each of the following proof excerpts, explain as fully as possible what the mathematical
statistics context is, and what steps are being taken (i.e. relate it to the content being assessed).
Standard notation for functions and parameters is used throughout. NB: you do not need to
complete the proof.
(a) Σxen (ag(x) + bh(x)) f(x) = a Σxen 9(x)ƒ(x) +bΣxenh(x)ƒ(x)
(b) M"(t) = ²X(X – t)−¹ = &λ(A − t)−² = 2\(\ — t)−³ and M″(0) = 2/1/2
dt2
(c) P(Z² < x) = P( - √x < Z < √√x) = Fz(√x) – Fz( − √x)
∞
d
(d) Σxpq* = pqΣxq-1 = pq
dq
x=0
(e) F(2.5) 1 - e
=
-0.35x2.5
∞
d
-1
Σ9²
Σ² = pq dq (1 — q)−¹
x=0
= 0.583 (3dp)
n
n
tx
n-x
(1) Σeta
* (*) p² (1 − p)¹-² = Σ (*) (pe²)² (1 − p)¹−² = (1 − p + pe²)"
n-x
x=0
x=0
(g) F(x)=√x for 0 < x < 1 and F-¹(x) = x², so X = U²
(h) E(et(a+bX)) = E(eat ebtX) = eªt Mx (bt)
(i) [*²*¾a(2 − x)dx = ³ [x² − }x³[6]
³ [2² - 2²³|²] = ³ (4- ) = 1
(Ax)⁰e-A
(j) P(X > x) = P(0 events in [0, x]) =
=
= e
so F(x) = 1 - e¯λx](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff76b43b9-ccee-44f5-a918-27c1f81bc882%2Feb8e17b1-9572-4774-b17d-55df71a5dcc6%2Fqc6hn1z_processed.png&w=3840&q=75)
Transcribed Image Text:For each of the following proof excerpts, explain as fully as possible what the mathematical
statistics context is, and what steps are being taken (i.e. relate it to the content being assessed).
Standard notation for functions and parameters is used throughout. NB: you do not need to
complete the proof.
(a) Σxen (ag(x) + bh(x)) f(x) = a Σxen 9(x)ƒ(x) +bΣxenh(x)ƒ(x)
(b) M"(t) = ²X(X – t)−¹ = &λ(A − t)−² = 2\(\ — t)−³ and M″(0) = 2/1/2
dt2
(c) P(Z² < x) = P( - √x < Z < √√x) = Fz(√x) – Fz( − √x)
∞
d
(d) Σxpq* = pqΣxq-1 = pq
dq
x=0
(e) F(2.5) 1 - e
=
-0.35x2.5
∞
d
-1
Σ9²
Σ² = pq dq (1 — q)−¹
x=0
= 0.583 (3dp)
n
n
tx
n-x
(1) Σeta
* (*) p² (1 − p)¹-² = Σ (*) (pe²)² (1 − p)¹−² = (1 − p + pe²)"
n-x
x=0
x=0
(g) F(x)=√x for 0 < x < 1 and F-¹(x) = x², so X = U²
(h) E(et(a+bX)) = E(eat ebtX) = eªt Mx (bt)
(i) [*²*¾a(2 − x)dx = ³ [x² − }x³[6]
³ [2² - 2²³|²] = ³ (4- ) = 1
(Ax)⁰e-A
(j) P(X > x) = P(0 events in [0, x]) =
=
= e
so F(x) = 1 - e¯λx
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