Statistics Homework Solutions 107

Statistics Homework Solutions 107 - 107 SECTION 4.12 It...

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Unformatted text preview: 107 SECTION 4.12 It follows that X is approximately normal with mean µX 500 0 7 350 and variance σ2 500 0 7 0 3 X 105, and Y is approximately normal with mean µY 500 0 8 400 and variance σ2 500 0 8 0 2 80. X ¡¥¨ ¡ ¡¥¨ ¤ ¡ ¤ ¡ 350 ¢ ¡ ¢ ¡ 13 601471. ¡ ¢ ¡ ¤ ¡ ¥ 0 0307. ¨ ¡ ¨ £ ©¥ 0 9693 ¡ ¨ ¡ ¡ ¨ 1 87 is 1 1 87. ¨ 750 13 601471 ¡ ¢ ¤ ¨ ¡ ¤ 0 0307. ¨ ¡¥ ¡ ¤ Section 4.12 Bin 100 0 05 ¥ ¨  ¤ Bin 100 0 03 , Y ¥ ¨  ¤ (b) Answers will vary. 0 72 (c) ¨ (d) 0 18 ¨ (e) The distribution deviates somewhat from the normal. ¡ 0 16 ¨ (c) The distribution is approximately normal. 0 25 ¨ 0 61 ¨ (c) 0 25 (b) ¨ 5. (a) 7. (a–c) Answers will vary. (b) 6 exactly (simulation results will be approximate), σ2 A 0 25. ¨ 3. (a) µA 750, and standard deviation σT 775 , use the continuity correction and find the z-score of 775.5. The area to the right of z 1. (a) X 400 ¨ 80 µY ¡ 105 µX ¡ ¢ To find P T 775 ¡ 2 σY The z-score of 775.5 is 775 5 PT ¤ Y be the total number of rivets that meet the specification. Then T is approximately normal with mean µT σ2 X ¡ ¥ ¨ ¦¥ ¨ ¤ ¤¥ ¨ ¤ X ¡ §¥ ¨ Let T ...
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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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