Statistics Homework Solutions 111

Statistics Homework Solutions 111 - 111 SUPPLEMENTARY...

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Unformatted text preview: 111 SUPPLEMENTARY EXERCISES FOR CHAPTER 4 0 0228 ¨ 0 50 is 0 6915 0 6687. ¡ ¨ ¡ ¨ ¡ 2 00 and z ¨ ¨ ¡ ¡ The area between z The probability that the clearance is between 180 and 205 µm is 0.6687. (c) Let X be the number of valves whose clearances are greater than 215 µm. From part (a), the probability that a valve has a clearance greater than 215 µm is 0.0668, so X Bin 6 0 0668 . 6! 0 0668 2 1 0 0668 6 2 0 0508. PX 2 2! 6 2 ! ¥ ¨ ¡ ¥ ¨ ¤¥ ¡ ¨ ¤ ¨ ¥  ¡¥ ¤ ¡ ¤ ¡ ¤ 11. (a) Let X be the number of assemblies in a sample of 300 that are oversize. ¤ ¡ ¥ ¨ ¨ ¡¥ ¤ ¦¥  ¨ ¤ ¨ ¥ 1 19. ¡ 15 3 7749 ¨ ¨ £ ©¥ ¨ ¡ ¤ ¡ ¨ ¡ 0 8830. ¨ Bin 10 0 05 .  ¤ ¡¥ ¤ ¡ (b) Let Y be the number of assemblies in a sample of 10 that are oversize. Then X 10! PX 1 1 PX 0 1 0 05 0 1 0 05 10 0 0 4013. 0! 10 0 ! ¨ ¡ ¤ 1 19 is 0.8830. ¥ ¤ The area to the left of z 20 15 and standard deviation 20 , use the continuity correction and find the z-score of 19.5. The z-score of 19.5 is 19 5 PX ¨ To find P X 300 0 05 ¡¥ Then X Bin 300 0 05 , so X is approximately normal with mean µX 300 0 05 0 95 3 7749. σX ¨ ¡ ¥ ¨ ¤¥ ¡ ¨ ¤ ¡¥ ¡ ¥ ¤ ¡ ¡ ¤ § ¡¥ ¡ ¤ (c) Let p be the required probability, and let X represent the number of assemblies in a sample of 300 that are oversize.  ¤ ¥ ¤ ¡ 0 01. ¨ ¡¥ 20 ¥ ¡ § PX 300 p and standard deviation ¡ Then X Bin 300 p , so X is approximately normal with mean µX σX 300 p 1 p . ¤ Using the continuity correction, 19.5 is the 1st percentile of the distribution of X . ©£¥ 380 25 ¡ ¨ ¨ ¡ ¤ ¡ ¨ ¡ 13 328 67 p 300 p ¢ ¨  ¡ ¨  0 0390. (0.1065 is a spurious root.) ¨ ¡ ¡ ¡ ¢ ¡ £ £ ¡ ¢ ¡ ¨ ¨ ¡ ¢ ¡ 28 for λ and solving for v yields v ¡ ¡ ¡ Substituting λ λv (b) Let v be the required volume, in mL. Then σλ 28 2 1. 28 mL. £ λ 2. Substituting λ for λ, σλ ¡ 3 7. 28. 3 7. ¨ 28 0 ¡ λ £ The uncertainty is σλ 56 2 ¡ 13. (a) Let λ be the true concentration. Then λ 300 p 1 ¤ 19 5 0. ¡ 2 33 p. ¥ Solving for p yields p ¡ This equation can be rewritten as 91 628 67 p2 ¡ The z-score can be expressed in terms of p by 2 33. ¨ The z-score of the 1st percentile is approximately z ...
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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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