ST102_Solutions_12.pdf

# ST102_Solutions_12.pdf - ST102 Outline solutions to...

This preview shows pages 1–2. Sign up to view the full content.

ST102 Outline solutions to Exercise 12 1. By the central limit theorem we have: ¯ X N μ, σ 2 n approximately, as n → ∞ . Hence: n ( ¯ X n - μ ) σ = n ( ¯ X n - 5) 3 Z N (0 , 1) . Therefore: P ( | ¯ X n - μ | < 0 . 3) P ( | Z | < 0 . 1 n ) = 2 × Φ(0 . 1 n ) - 1 . However, 2 × Φ(0 . 1 n ) - 1 0 . 95 if and only if: Φ(0 . 1 n ) 1 + 0 . 95 2 = 0 . 975 which is satisfied if: 0 . 1 n 1 . 96 n 384 . 16 . Hence the smallest possible value of n is 385. 2. The following are possible, but not exhaustive, solutions. (a) We could have: 2 X i =1 X i - i i 2 χ 2 2 . (b) We could have: X 1 - 1 s 4 i =2 X i - i i 2 / 3 t 3 . (c) We could have: ( X 1 - 1) 2 4 i =2 X i - i i 2 / 3 F 1 , 3 . 3. (a) Since X 1 + 3 X 2 N (0 , 40), then: P ( X 1 + 3 X 2 > 4) = P X 1 + 3 X 2 40 > 4 40 = P ( Z > 0 . 63) = 0 . 2643 where Z N (0 , 1). 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
(b) X 2 i / 4 χ 2 1 for i = 1 , 2 , 3 , 4, hence ( X 2 1 + X 2 2 + X 2 3 + X 2 4 ) / 4 χ 2 4 , so: P ( X 2 1 + X 2 2 + X 2 3 + X 2 4 < k ) = P ( X < k/ 4) = 0 . 90 where X χ 2 4 . Hence k/ 4 = 7 . 779, so k = 31 . 116. (c) X 1 / 4 N (0 , 1) and ( X 2 2 + X 2 3 ) / 4 χ 2 2 , hence: X 1 / 4 q ( X 2 2 + X 2 3 ) / 4 2 = 2 X 1 p X 2 2 +
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern