Statistics Homework Solutions 106

Statistics Homework Solutions 106 - 106 CHAPTER 4 (d) Let v...

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Unformatted text preview: 106 CHAPTER 4 (d) Let v be the required volume. Let X be the number of particles withdrawn in a volume of v mL. 50v and standard deviation σX ¡ ¡ Poisson 50v , so X is approximately normal with mean µX ¤ ¥ ¤ Then X 50v. The average number of particles per mL will be between 48 and 52 if the total number of particles X is between 48v and 52v. ¤ £ ¤ ©¥ ¤ ¡ ¡ ¡ ¤ ¨ ¨ ¡¥ ¨ ¡ ¡ ¤ ¡ ¥ ¨  50 ¤ ¨ ¡ §¥ ¨ ¤¥ ¨ ¤ § ¥ 3 55. ¨ 0 9998 0 0002. ¨ ¡ £ ©¥ ¨ ¡ ¨ ¨ ¡ ¤ ¤ 3 55 is 1 ¡ 50 6 89202 ¨ ¡ ¡ The area to the right of z 0 0002. ¨ ¡¥ 75 1000 0 05 75 , use the continuity correction and find the z-score of 74.5. The z-score of 74.5 is 74 5 § PX 1 96 50 2. 48 02 ml. 17. (a) If the claim is true, then X Bin 1000 0 05 , so X is approximately normal with mean µ X and σX 1000 0 05 0 95 6 89202. To find P X v ¡ 50v, or equivalently, ¨ ¡¥ Solving for v yields v 52v 50v ¤ ¨ The z-score of 1.96 therefore satisfies the equation 1 96 1 96 and the z-score of 52v is 1 96. £ 0 95 if the z-score of 48v is ¡ 52v ¡ X ¨ 50v, P 48v ¨ Since µX ¤ (b) Yes. Only about 2 in 10,000 samples of size 1000 will have 75 or more nonconforming tiles if the goal has been reached. (c) No, because 75 nonconforming tiles in a sample of 1000 is an unusually large number if the goal has been reached. ¤ ¡ ¥ ¨  ¤ ¨ ¡ §¥ ¨ ¤¥ ¨ § ¥ ¡ ¨ 0 6406 0 3594. ¨ ¡ ¨ ¡ £ ©¥ 0 36 is 1 0 36. ¨ 50 6 89202 ¡ ¨ ¤ ¡ ¤ ¨ ¡ ¤ The area to the right of z 0 3594. ¨ ¡¥ 53 50 53 , use the continuity correction and find the z-score of 52.5. The z-score of 52.5 is 52 5 § PX ¨ To find P X 1000 0 05 ¡¥ (d) If the claim is true, then X Bin 1000 0 05 , so X is approximately normal with mean µ X and σX 1000 0 05 0 95 6 89202. ¤ (e) No. More than 1/3 of the samples of size 1000 will have 53 or more nonconforming tiles if the goal has been reached. (f) Yes, because 53 nonconforming tiles in a sample of 1000 is not an unusually large number if the goal has been reached. Let X be the number of rivets from vendor A that meet the specification, and let Y be the number of rivets from vendor B that meet the specification. Bin 500 0 8 . ¤ Bin 500 0 7 , and Y  ¥¨  ¤ Then X ¥¨ 19. ...
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