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(d) Σ apq* = pq Σκg- x=0 x=1 (e) F(2.5) = 1 – e-0.35x2.5 d Σq = pa4 (1 – q)-1 x=0 tx = pa da = 0.583 (3dp) n n (Σe (?)p*(1-1)^- - Σ (?) (pe)*(1 - p)n-* = (1 - p+ pen = x=0 x=0 (0) Fla) - Γfor Πολ F-1/3) m2 any - 12

(d) Σ apq* = pq Σκg- x=0 x=1 (e) F(2.5) = 1 – e-0.35x2.5 d Σq = pa4 (1 – q)-1 x=0 tx = pa da = 0.583 (3dp) n n (Σe (?)p*(1-1)^- - Σ (?) (pe)*(1 - p)n-* = (1 - p+ pen = x=0 x=0 (0) Fla) - Γfor Πολ F-1/3) m2 any - 12

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.5: Graphs Of Functions
Problem 14E
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Please answer q's d, e, and f

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
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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