Statistics Homework Solutions 80

Statistics Homework Solutions 80 - 80 CHAPTER 4 7. (a) Let...

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Unformatted text preview: 80 CHAPTER 4 7. (a) Let X be the number of messages received in one hour. Poisson 8 . 8 0 0916 ¡ 5! e ¨ ¡¥ 5 ¡ PX 85 ¥¤ Since the mean rate is 8 messages per hour, X ¤ (b) Let X be the number of messages received in 1.5 hours. Since the mean rate is 8 messages per hour, X 0 1048 ¡ 10! ¨ 12 e ¤ ¡¥ 10 Poisson 12 . ¥ ¡ PX 1210 ¤ (c) Let X be the number of messages received in one half hour. 1 PX ¢¥ ¡ ¤ ¢¥ 41 42 2 ¡ ¤ ¡ e4 e4 0! 1! 2! 0 018316 0 073263 0 14653 0 2381 ¢ ¨ ¢ ¢ ¨ e 4 PX ¤ 0 40 ¡ PX ¥ ¢ ¥ 3 PX Poisson 4 . ¥¤ Since the mean rate is 8 messages per hour, X ¤ ¡ ¡ ¨ ¡ ¨ ¡ 9. (ii). Let X Bin n p where µX np 3. Then σ2 np 1 p , which is less than 3 because 1 p 1. Now X let Y have a Poisson distribution with mean 3. The variance of Y is also equal to 3, because the variance of a Poisson random variable is always equal to its mean. Therefore Y has a larger variance than X . 11. Let X represent the number of bacteria observed in 0.5 mL. Let λ represent the true concentration in bacteria per mL. Then X Poisson 0 5λ . The observed value of X is 39. The estimated concentration is λ 39 0 5 78. The uncertainty is σλ 78 0 5 12 49. λ 78 12. ¥ ¡ ¡ ¤ ¡ ¡ ¡ ¡ ¥  ¤ ¡ ¡ ¥ ¡ ¨ £ 0 0516 ¤ ¢ ¡ ¨ ¡ ¡¥ ¡ ¤ (b) Let X be the number that are repairable. Then X Bin 15 0 6 . 15! P X 10 0 6 10 1 0 6 15 10 0 1859 10! 15 10 !  ¤ ¨ ¡ ¥¨ ¡ ¤ ¥¨ ¤ ¥ ¥¨ ¡¥ ¤ ¡ Bin N 0 6 . ¥¨  ¤ ¡ ¤ (c) Given N , X ¨ 15! ¤ ¡ 20 20 ¥ ¨ e £ 15 Poisson 20 . ¨ PN 15 ¡ 13. (a) Let N be the number of defective components produced. Then N ...
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This note was uploaded on 12/20/2011 for the course STA 3163 taught by Professor Mattgilg during the Fall '11 term at UNF.

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