Statistics Homework Solutions 88

Statistics Homework Solutions 88 - 88 CHAPTER 4 (b) Let σ...

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Unformatted text preview: 88 CHAPTER 4 (b) Let σ be the required standard deviation. The value of σ must be chosen so that the 1st percentile of the distribution is 65. The z-score of the 1st percentile is approximately z 2 33. Therefore 2 33 65 75 σ. Solving for σ yields σ 4 292 N/m2 . ¡ ¨ ¨ ¡ ¡ ¡ ¨ ¡ £ ¥ ¤ ¡ (c) Let µ be the required value of the mean. This value must be chosen so that the 1st percentile of the distribution is 65. The z-score of the 1st percentile is approximately z 2 33. Therefore 2 33 65 µ 5. Solving for µ yields µ 76 65 N/m2 . £ ©¥ ¤ ¡ ¡ ¨ ¨ ¡ ¡ ¡ ¨ ¡ Let X be the lifetime of bulb A and let Y be the lifetime of bulb B. ¡ ¥ £¤ ¡ ¢ ¢ ¡ £  ¤ ¥ ¡ £ ¡  ¤ ¥  ¤ 100. 1002 ¤ 2 σY 1502 ¢ ¡ σ2 X 180 278. ¡ ¢ ¡ ¡ ¡ ¡ ¡ ¡ Since X and Y are independent, σD ¨ 800 ¡ 900 0. ¡ ¡ N 900 1502 . X . The event that bulb B lasts longer than bulb A is the event that D µX 1 σ 2 σ2 ¥ µY £ µD Y 0, and a 2 σ2 ¡ (a) Let D N 800 1002 and Y µσ ¡ Then X 1 σµ £¤ 21. b ¥ Let a 1 σ and let b µ σ. Then Z aX b. Now aµ Equation (4.25) now shows that Z N 0 1 . ¡ 19. ¡ Since D is a linear combination of independent normal random variables, D is normally distributed. ¡ ¡ ¨ ¡ ¡ ¨ £ ¥ ¤ ¡ ¡ ¤ 100. ¤ 2 σY 1002 ¢ ¡ σ2 X 1502 ¡ ¢ ¡ ¡ ¡ ¡ ¡ ¡ ¡ Since X and Y are independent, σD 200. 180 278. ¨ 800 ¡ 900 ¡ ¡¥ X . The event that bulb B lasts longer than bulb A is the event that D µX 0 2912 0 7088. ¨ µY 0 55 is 1 ¨ µD Y 0 55. The area to the right of z ¡ (b) Let D 100 180 278 ¨ 0 ¡ Therefore P D Since D is a linear combination of independent normal random variables, D is normally distributed. ¡ 0 7088 0 2912. ¨ ¨ ¡ ¤ ¨ ¡ 0 2912. ¨ ¤ ¤ 1700. ¡ 2 σY ¤ ¢ σ2 X 1002 ¢ ¡¡ ¡ ¢ ¢ ¡ ¡ Since X and Y are independent, σS 1502 ¡ 900 2000 . 180 278. ¨ ¢ 800 ¡ µY ¡ ¡ Y be the total lifetime of the two bulbs. It is necessary to find P S ¥ ¡¥ µX ¡ µS X £ ©¥ (c) Let S 200 ¨ Therefore P D 0 55 is 1 0 55. ¡ The area to the right of z 100 180 278 ¨ The z-score of 200 is 200 Since S is a linear combination of independent normal random variables, S is normally distributed. 0 7088. ¨ The z-score of 0 is 0 1. ...
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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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