Chapter 7 For Students

# Chapter 7 For Students - Chapter 7 Continuous Probability...

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Unformatted text preview: Chapter 7 Continuous Probability Distributions 1. a. a = 6 b = 10 b. 8, found by (6 + 10)/2 c. 1.1547 found by 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 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 c. 0.4706 found by [1 / (3800 - 400)] * (2000 – 400) d. 0.2353 found by [1 / (3800 - 400)] * (3800 – 3000) 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 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 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. Adjusting for their industries, Rob is well below average and Rachel well above. 12. The first is slightly less expensive than average and the second is slightly more. 13. a. 1.25 found by b. 0.3944, found in Appendix B.1 c. 0.3085, found by Find 0.1915 in Appendix B.1 for z = – 0.5 then 0.5000 – 0.1915 = 0.3085 14. a. z = 0.84, found by b. 0.2995, found in Appendix B.1 c. 0.1894, found by Find 0.3106 in Appendix B.1 for z = – 0.88, then 0.5000 – 0.3106 = 0.1894 15. a. 0.3413, found by Then find 0.3413 in Appendix B.1 for a z = 1 b. 0.1587, found by 0.5000 – 0.3413 = 0.1587 c. 0.3336, found by Find 0.1664 in Appendix B.1, for a z = – 0.43, then 0.5000 – 0.1664 = 0.3336 16. a. About 0.4332 from Appendix B.1, where z = 1.50 b. About 0.1915, where z = – 0.50 c. About 0.3085, found by 0.5000 – 0.1915 17. a. 0.8276, first find z = – 1.5, found by ((44 – 50)/4) and z = 1.25 = (55 – 50)/4). The area between –1.5 and 0 is 0.4332 and the area between 0 and 1.25 is 0.3944, both from Appendix B.1. Then adding the two area we find that 0.4332 + 0.3944 = 0.8276.Then adding the two area we find that 0....
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## This note was uploaded on 12/30/2010 for the course BUSS 202 taught by Professor Ritageutchian during the Three '10 term at Notre Dame AU.

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Chapter 7 For Students - Chapter 7 Continuous Probability...

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