Statistics Homework Solutions 104

Statistics Homework Solutions 104 - 104 Let n be the...

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Unformatted text preview: 104 Let n be the required number of measurements. Let X be the average of the n measurements. 0 95. In any normal population, 95% of the population is within 1.96 n, n 61 47. The smallest value of n is therefore n ¡ ¤£ 1 ¨ ¨ ¢ ¡ ¡ ¨ ¡ 0 25. Since σX ¡ ¡¥ ¨ ¤ ¡ ¡ Therefore 1 96σX n. ¨ Now P µX 0 25 X µX 0 25 standard deviations of the mean. 1 ¡ Then the true value is µX , and the standard deviation is σX ¤£ 9. CHAPTER 4 62. ¨ 11. (a) Let X represent the number of defective parts in a shipment. ¤ ¡ ¥ ¡ ¥ ¨ ¦¥ ¨ ¤ ¤ ¨ ¤ ¡ ¥ 1 31. £ ©¥ 0 0951. ¡ 0 9049 ¨ ¨ ¡ ¨ ¨ ¡ ¡ ¤ ¤ 1 31 is 1 ¡ 80 8 ¨ The area to the right of z ¡ 0 0951. ¨ ¡¥ 90 80 and standard deviation 90 , use the continuity correction and find the z-score of 90.5. The z-score of 90.5 is 90 5 ¡ PX ¨ To find P X 400 0 20 ¡¥ Then X Bin 400 0 20 , so X is approximately normal with mean µX σX 400 0 2 0 8 8. ¤ (b) Let Y represent the number of shipments out of 500 that are returned. ¤ ¨ ¡¥ ¨ ¤ ¡ ¨ ¡¥ ¨ ¤ ¦¥ ¨ ¥ ¡ 0 0344. ¡ £ ¥ 0 9656 ¨ ¨ ¨ ¡ ¨ ¨ 1 82 is 1 1 82. ¨ 47 55 6 5596 ¡ ¤ § ¡ ¤ ¨ ¤ ¡ The area to the right of z 0 0344. ¨ ¡¥ 60 47 55 and standard deviation 60 , use the continuity correction and find the z-score of 59.5. The z-score of 59.5 is 59 5 § PX  To find P Y 500 0 0951 ¨ It follows that Y is approximately normal with mean µY σY 500 0 0951 0 9049 6 5596. Bin 500 0 0951 . ¥ From part (a) the probability that a shipment is returned is 0.0951, so Y ¤ (c) Let p be the required proportion of defective parts, and let X represent the number of defective parts in a shipment. 400 p and standard deviation ¡ Then X Bin 400 p , so X is approximately normal with mean µX σX 400 p 1 p . ¥  ¤ ¥ 0 01. ¡¥ 90 ¨ ¡ ¤ ¡ The probability that a shipment is returned is P X ¡ ¤ Using the continuity correction, 90.5 is the 1st percentile of the distribution of X . ©£¥ 8190 25 ¡ ¨ ¢ ¨ ¤ ¡ ¡ ¨ 74 571 56 p 400 p 1 ¤ 400 p p. ¥ 90 5 ¡ ¨  ¡ ¡ ¨  0. 0 181. (0.278 is a spurious root.) ¨ ¡ Solving for p yields p 2 33 ¡ This equation can be rewritten as 162 171 56 p2 ¡ The z-score can be expressed in terms of p by 2 33. ¨ The z-score of the 1st percentile is approximately z 13. (a) 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 five-minute time period. Then X Poisson 100 and Y Poisson 125 . ¥ ¤ ¥ ¤ Now X is approximately normal with mean 100 and variance 100, and Y is approximately normal with mean 125 and variance 125. ...
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