← Back
ExamFirst midtermExam paper only

27 11 14

Study material for Statistica per L'ingegneria, shared by the Studwiz community and reviewed by moderators.

Statistica per L'ingegneriaFirst midterm

Document information

What's included in this study material

Study material for Statistica per L'ingegneria, shared by the Studwiz community and reviewed by moderators.

Import quality: text was extracted directly from the original document.

Extracted content from the document

Representative passages recognised in different parts of the material. The full extracted text remains available to search, while this compact preview makes the page easier to read.

Page 1

d v v=h·d h h km/sM p c h [0 10) [10 20) [20 30) [30 40) [40 50) [50 60) [60 70) [70 80) [80 90) [90 100) [100 110) [0,10) [10 20) [20,30) [30,40) [40,50) [50 60) [60 70) [70 80) [80 90) [90 100) [100 110) Istogramma di H − 11 classi H [km/s * Mpc] Density 020406080100120 0.0000.0050.0100.0150.020 n= 36 pn 1 + log2n [0,20) [20 40) [40,60) [60,80) [80,10) [100 120) Istogramma di H − 6 classi H [km/s * Mpc] Density 020406080100120 0.0000.0050.0100.015 h36'(2·5+0·15 + 2·25 + 2·35 + 6·45 + 7·55 + 5·65 + 6·75 + 3·85 + 2·95 + 1·105)/36 = 58.61 Q32[70 80) 0.75'0.6667 + (Q3 70)·0,0167 =)Q3'75. X fX(x)= ( ↵(x+ 1)3, 1x1, 0, . ↵ fX FX X X X 8 >< >: ↵ 0, 1= Z1 1 ↵(x+ 1)3dx=↵(x+ 1)4 4 1 1 =)↵=1/4. 1x1 FX(x)= Zx 1 1 4(t+ 1)3dt=(x+ 1)4 16 FX(x)= 8 >< >: 0,x < 1, 1 16(x+ 1)4, 1x1, 1,x > 1. −2−1012 0 2 4 6 8 Densità x f(x) mediamediana −2−1012 0.00.20.40.60.81.0 Ripartizione x F(x) mediamediana P(X>0) = 1 P(X0) = 1 FX(0) = 1 1 16=0.9375 m>0 P(X0)<0.5 m<E[X] fX FX(m)=0.5=)1 16(m+ 1)4=1 2=)m+1=4p 8=)m=0.6817. E[X]= Z1 1 x·1 4(x+1)3dx= 1 4 x5 5+3 4 x4 4+3 4 x3 3+1 4 x2 2 1 1 =1 4 1 5+1 5 +1 4[1 + 1] =6 10=0.6. n X X⇠B(363,0.92) 363·0.08 = 29.04>5 363·0.92 = 333.96>5 X⇠B(363,0.92)'N(333.96,26.7168). P(X= 363) = 0.92363=7.2·10 14 P(X325) =P(X325.5)' ⇣325.5 333.96p 26.7168 ⌘ = ( 1.64) = 1 (1.64) = 0.0508 = 5.08% P(X325) = 0.0550 = 5.5% Y n Y⇠B(n,0.92) n P(Y325)<0.025 n>363 Y⇠B(n,0.92)'N(0.92n,0.0736n). 0.025>P(Y325) = ✓325.5 0.92np 0.0736n ◆ 325.5 0.92np 0.0736n< z0.025= 1.96 n>364.84 n= 365 P(Y>330)'1 ⇣330.5 365·0.92p 365·0.92·0.08 ⌘ =0.8467 = 84.67%. P(Y>330) = 0.8467 = 84.67%.

Preview

First page of the document.

First page: 27 11 14