sol12 - Chapter 12 Exercises 12.1 1 4 4 1. E( X ) = 0 + 1 =...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
381 Chapter 12 Exercises 12.1 1. E ( X ) = 0 1 5 ⎟ + 1 4 5 ⎟ = 4 5 V ( X ) = 0 4 5 2 1 5 ⎟ + 1 4 5 2 4 5 = 20 125 = .16 Standard deviation .16 .4 == 2. E ( X ) = 1 4 9 + 2 4 9 + 3 1 9 = 15 9 = 5 3 V ( X ) = 1 5 3 2 4 9 + 2 5 3 2 4 9 + 3 5 3 2 1 9 = 36 81 = 4 9 Standard deviation 36 6 2 81 9 3 = 3. a. E ( X ) = 4(.5) + 6(.5) = 5 V ( X ) = (4 5) 2 (.5) + (6 5) 2 (.5) = 1 b. E ( X ) = 3(.5) + 7(.5) = 5 V ( X ) = (3 2 (.5) + (7 5) 2 (.5) = 4 c. E ( X ) = 1(.5) + 9(.5) = 5 V ( X ) = (1 2 (.5) + (9 5) 2 (.5) = 16 As the difference between maximum and minimum values increases, so does the variance. 4. a. E ( X ) = 2(.1) + 4(.4) + 6(.4) + 8(.1) = 5 V ( X ) = (2 5) 2 (.1) + 2 (.4) + 5) 2 (.4) + (8 5) 2 (.1) = 2.6 b. E ( X ) = 2(.3) + 4(.2) + 6(.2) + 8(.3) = 5 V ( X ) = 5) 2 (.3) + 5) 2 (.2) + 5) 2 (.2) + 5) 2 (.3) = 5.8 Values of (b) farther from E ( X ) have higher probabilities. Thus the variance is larger.
Background image of page 1

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

View Full DocumentRight Arrow Icon
Chapter 12: Probability and Calculus ISM: Calculus & Its Applications, 11e 382 5. a. Outcome 0 1 2 3 Probability 11 52 26 52 13 52 2 52 b. E ( X ) = 0 11 52 ⎟ + 1 26 52 ⎟ + 2 13 52 ⎟ + 3 2 52 = 58 52 = 29 26 1.12 c. E ( X ) is the average number of accidents per week in the given year. 6. a. Outcome 0 1 2 Probability 30 60 20 60 10 60 b. E ( X ) = 0 30 60 + 1 20 60 + 2 10 60 = 2 3 c. E ( X ) is the average number of calls coming into the switchboard each minute. 7. a. () 2 11 22 2 area within unit of center Total area (1) π = π 125 = 1 4 = .25 Thus 25% of the points in the circle are within 1 2 unit of the center. b. 100 × π c 2 π 2 = 100 c 2 % 8. a. 1 2 b. 1 4 c. 1 100 d. 0 9. Let X be the profit that the grower makes if he does not protect the fruit. Then E ( X ) = 100,000(.75) + 60,000(.25) = 90,000 < 95,000. Therefore he should spend the $5000 to protect the fruit. 10. Let X be the random variable defined in Ex. 9. In this case, E ( X ) = 100,000(.80) + 85,000(.1) + 75,000(.1) = 96,000 > 95,000, so the grower should not spend the money to protect the fruit. Exercises 12.2 1. I. 1 18 x 0 for all 0 x 6 II. 1 18 xdx 0 6 = 1 18 x 2 2 0 6 = 1 0 = 1 2. I. 2( x – 1) = 2 x – 2 0 for all 1 x 2 II. (2 x 2) dx 1 2 = x 2 2 x 1 2 = 0 + 1 = 1 3. I. 1 4 0 II. 1 4 dx 1 5 = 1 4 x 1 5 = 5 4 1 4 = 4. I. 8 9 x 0 for all 0 x 3 2 . II. 8 9 0 3/2 = 4 9 x 2 0 = 1 0 = 5. I. 5 x 4 0 for all x . II. 5 x 4 dx 0 1 = x 5 0 1 = 1 0 = 6. I. 3 2 x 3 4 x 2 = 3 2 x 1 1 2 x 0 for all 0 x 2. II. 3 2 x 3 4 x 2 dx 0 2 = 3 4 x 2 1 4 x 3 0 2 = 1 – 0 = 1 7. kxdx 1 3 = k 2 x 2 1 3 = 9 k 2 k 2 = 4 k = 1; so k = 1 4 8. kx 2 dx 0 2 = k 3 x 3 0 2 = 8 3 k = 1; so k = 3 8
Background image of page 2
ISM: Calculus & Its Applications, 11e Chapter 12: Probability and Calculus 383 9.
Background image of page 3

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

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/24/2010 for the course MATH 9374 taught by Professor Smith during the Spring '10 term at University of California, Berkeley.

Page1 / 19

sol12 - Chapter 12 Exercises 12.1 1 4 4 1. E( X ) = 0 + 1 =...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online