Statistics Homework Solutions 105

Statistics Homework Solutions 105 - 105 SECTION 4.11 X Y....

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Unformatted text preview: 105 SECTION 4.11 X Y. ¢ The number of particles emitted by both masses together is S ¡ Since X and Y are independent, and both approximately normal, S is approximately normal as well, with mean 100 125 15. µS 100 125 225 and standard deviation σS ¡ ¡ ¡ 1 67. The area to the left of z 1 67 is 0.0475. ¨ ¡ ¨ ¡ ¡ ¢ ¡ £ ©¥ ¤ ¡ 0 0475. ¨ ¡¥ 200 ¤ ¡ PS ¢ 225 15 ¡ The z-score of 200 is 200 ¤ (b) Let X be the number of particles emitted by mass A and let Y be the number of particles emitted by mass B in a two-minute time period. Then X Poisson 40 and Y Poisson 50 . ¥ ¤ ¥ ¤ Now X is approximately normal with mean 40 and variance 40, and Y is approximately normal with mean 50 and variance 50. ¡ Mass B emits more particles than mass A if D Y X. ¡ The difference between the numbers of particles emitted by the two masses is D 0. ¡ Since X and Y are independent, and both approximately normal, D is approximately normal as well, with mean 50 40 9 486833. µD 50 40 10 and standard deviation σD 0 1469 0 8531. ¡ ¨ 1 05 is 1 ¨ ¨ ¡ ¡ ¡ ¤ ¡ 1 05. The area to the right of z ¡ ¢ ¨ ¡ ¡ ¨ £ ¥ ¡ ¡ ¤ ¡ 0 8531. ¨ ¡¥ 0 ¡ ¡ PD 10 9 486833 ¨ The z-score of 0 is 0 ¤ 15. (a) Let X be the number of particles withdrawn in a 5 mL volume. 250, so X Poisson 250 , and X is approximately normal with mean µ X 250 15 8114. ¤ 250 15 8114 ¤ ¨ 250 15 8114 £ ©¥ ¤ ¡ ¨ 0 95. 0 6578. ¨ ¡ 0 1711 ¨ 0 95, and the z-score of 265 is 265 ¡ ¨ ¡ £ ©¥ ¨ 0 95 is 0 8289 ¡ ¡ ¨ ¡ ¤ ¡ ¨ ¡ 0 95 and z ¨ ¡ ¡¥ ¤ ¨ ¡ The area between z ¥ The z-score of 235 is 235 250 ¡ Then the mean of X is 50 5 and standard deviation σX ¡ The probability is 0.6578. (b) Since the withdrawn sample contains 5 mL, the average number of particles per mL will be between 48 and 52 if the total number of particles is between 5 48 240 and 5 52 260. 0 63. ¨ ¡ ¨ ¡ £ ©¥ ¤ ¨ 0 4714. ¨ ¡ 0 2643 15 8114. 250 15 8114 ¡ ¡ ¨ ¡ £ ©¥ ¨ 0 63 is 0 7357 ¡ ¨ ¡ ¤ ¡ ¨ ¡ 0 63 and z ¤ ¤ ¨ ¡ The area between z 0 63, and the z-score of 260 is 260 250 ¡ ¡¥ 250 15 8114 250 and standard deviation σX ¨ ¤ The z-score of 240 is 240 ¡¥ From part (a), X is approximately normal with mean µX ¡ The probability is 0.4714. (c) Let X be the number of particles withdrawn in a 10 mL volume. 500, so X Poisson 500 , and X is approximately normal with mean µ X 500 22 3607. 500 ¡ ¥ ¤ ¨ ¡ ¡¥ ¤ Then the mean of X is 50 10 and standard deviation σX ¤ ¡ The average number of particles per mL will be between 48 and 52 if the total number of particles is between 10 48 480 and 10 52 520. ¡¥ £ ©¥ ¤ ¨ 0 6266. ¨ ¨ ¡ 0 1867 ¡ ¨ ¡ ¡ ¨ ¨ ¨ ¡ £ ©¥ ¤ ¡¥ ¤ ¤ ¨ ¡ ¡ The probability is 0.6266. 0 89 is 0 8133 500 22 3607 0 89. ¨ ¡ 0 89 and z 0 89, and the z-score of 520 is 520 ¡ The area between z 500 22 3607 ¡ The z-score of 480 is 480 ...
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