# HW 20 Answers (6.1) - is given by P X ≤ 1 25 = F(1 25 = 5...

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Math331, Spring 2008 Instructor: David Anderson Section 6.1 Homework Answers Homework: pgs. 228 - 239, #’s 2, 3, 4. 2. (a) To fnd f , we simply need to diFerentiate F . This gives f ( x ) = b 32 x - 3 x 4 0 x < 4 . (c) The probability that the soap opera takes at most 50 hours is P ( X 5) = F (5) = 1 - 16 / 25 = 9 / 25 . At least 60: P ( X > 6) = 1 - F (6) = 16 / 36 = 4 / 9 . Between 50 and 70: P (5 < X < 7) = F (7) - F (5) = - 16 / 49 + 16 / 25 = 384 / 1225 = . 3135 . Between 10 and 35: P (1 < X < 3 . 5) = F (3 . 5) - F (1) = 0 - 0 = 0 . 3. (a) We are told that the density ±unction is given by f ( x ) = b c ( x - 1)(2 - x ) i± 1 < x < 2 0 else Integrating f and using that 1 = i -∞ f ( x ) dx = i 2 1 f ( x ) dx = c/ 6, we see that c = 6. (b) We know that ±or any t R , F ( t ) = i t -∞ f ( x ) dx . Using that ±or any t (1 , 2) we have I t 1 f ( x ) dx = - 2 t 3 + 9 t 2 - 12 t + 5 , we see that F ( t ) is given by F ( t ) = 0 t < 1 - 2 t 3 + 9 t 2 - 12 t + 5 i± 1 t < 2 1 i± 2 t . Note that F (1) = 0 and F (2) = 1, so F is continuous as expected. (c) The probability that a random student will fnish in less than 75 minutes (= 1.25 hours)

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Unformatted text preview: is given by P ( X ≤ 1 . 25) = F (1 . 25) = 5 32 = 0 . 156250 . 1 The probability that a student fnishes between 1.5 and 2 hours is P (1 . 5 < X < 2) = F (2)-F (1 . 5) = 1-. 5 = 1 2 . 4. We need the distribution Function. Using that For t ∈ (1 , 2) we have i t 1 f ( x ) dx = 2-2 t , we have F ( t ) = t < 1 2-2 /t 1 ≤ t < 2 1 2 ≤ t . To answer part (a) we want P ( X < 1 . 5) = F (1 . 5) = 2-2 * 2 3 = 2 / 3 . ±or part (b) we want P (1 < X < 1 . 25 | X < 1 . 5) = P (1 < X < 1 . 25 , X < 1 . 5) P ( X < 1 . 5) = P (1 < X < 1 . 25) P ( X < 1 . 5) = F (1 . 25)-F (1) F (1 . 5) = 2-2 / 1 . 25 (2 / 3) = 3 / 5 . 2...
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HW 20 Answers (6.1) - is given by P X ≤ 1 25 = F(1 25 = 5...

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