Chap007 - CHAPTER 7 CONTINUOUS PROBABILITY DISTRIBUTIONS 1....

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
C HAPTER 7 C ONTINUOUS P ROBABILITY D ISTRIBUTIONS 1. a. a = 6 b = 10 b. 8, found by (6 + 10)/2 c. 1.1547 found by 2 (10 6) 12 - d. [1/(10 – 6)](10 – 6) = 1 e. 0.75, found by [1/(10 – 6)](10 – 7) f. 0.5, found by [1/(10 – 6)](9 – 7) 2. a. a = 2 b = 5 b. 3.5, found by (2 + 5)/2 c. 0.8660 found by 2 (5 2) 12 - d. [1/(5 – 2)](5 – 2) = 1 e. 0.8, found by [1/(5 – 2)](5 – 2.6) f. 0.2667, found by [1/(5 – 2)](3.7 – 2.9) 3. a. b. Mean is 65 found by (60 + 70) / 2; Variance is 8.3333 found by [(70 – 60) ^ 2] / 12 c. 0.8 found by [1 / (70 – 60)] (68 – 60) d. 0.6 found by [1 / (70 – 60)] (70 – 64) 4. a. Mean is 2100 found by (400 + 3800) / 2 b. 981.50 found by 2 (3800 400) 12 - c. 0.4706 found by [1 / (3800 - 400)](2000 – 400) d. 0.2353 found by [1 / (3800 - 400)](3800 – 3000) Chapter 7 61
Background image of page 1

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

View Full DocumentRight Arrow Icon
5. a. a = 0.5, b = 3.0 b. Mean is 1.75, found by (0.5 + 3.0)/2 Standard deviation is 0.72, found by 2 (3.0 0.5) 12 - c. 0.2, found by [1/(3.0 – 0.5](1.0 – 0.5) d. 0.0, found by [1/(3.0 – 0.5](1.0 – 1.0) e. 0.6, found by [1/(3.0 – 0.5](3.0 – 1.5) 6. a. a = 0.5, b = 10.0 (using minutes as the units) b. Mean is 5.25, found by (0.5 + 10)/2 Standard deviation is 2.74, found by 2 (10 0.5) 12 - Chapter 7 62
Background image of page 2
c. 0.5263, found by [1/(10 – 0.5]*(10 – 5) d. 2.875, found from [1/(10 – 0.5]*( x – 5) = 0.25 and 7.625 found from [1/(10 – 0.5]*(10 – x ) = 0.25 7. The actual shape of a normal distribution depends on its mean and standard deviation. Thus, there is a normal distribution, and an accompanying normal curve, for a mean of 7 and a standard deviation of 2. There is a another normal curve for a mean of $25,000 and a standard deviation of $1742, and so on. 8. It is bell shaped and symmetrical about its mean. It is asymptotic. There is a family of normal curves. The mean, median, and the mode are equal. 9. a. 490 and 510, found by 500 ± 1(10) b. 480 and 520, found by 500 ± 2(10) c. 470 and 530, found by 500 ± 3(10) 10. a. about 68 percent b. about 95 percent c. over 99 percent 11. 50,000 60,000 2.00 5000 Rob Z - = = - 50,000 35,000 1.875 8000 Rachel Z - = = Adjusting for their industries, Rob is well below average and Rachel well above. 12. 1 75 90 0.68 22 Z - = = - 2 100 90 0.45 22 Z - = = The first is slightly less expensive than average and the second is slightly more. 13. a. 1.25 found by 25 20 1.25 4.0 z - = = b. 0.3944, found in Appendix D c. 0.3085, found by 18 20 0.5 2.5 z - = = - Find 0.1915 in Appendix D for z = – 0.5 then 0.5000 – 0.1915 = 0.3085 Chapter 7 63
Background image of page 3

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

View Full DocumentRight Arrow Icon
14. a. z = 0.84, found by 10 12.2 0.84 2.5 - = = z b. 0.2995, found in Appendix D c. 0.1894, found by 10 12.5 0.88 2.5 z - = = - Find 0.3106 in Appendix D for z = – 0.88, then 0.5000 – 0.3106 = 0.1894 15. a. 0.3413, found by $24 $20.50 1.00 $3.50 z - = = Then find 0.3413 in Appendix D for a z = 1 b. 0.1587, found by 0.5000 – 0.3413 = 0.1587 c. 0.3386, found by $19.00 $20.50 0.43 $3.50 z - = = - Find 0.1664 in Appendix D, for a
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 8

Chap007 - CHAPTER 7 CONTINUOUS PROBABILITY DISTRIBUTIONS 1....

This preview shows document pages 1 - 5. Sign up to view the full document.

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