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Statistics Homework Solutions 27

# Statistics Homework Solutions 27 - 27 SECTION 2.4 Â Â£ Â¨...

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Unformatted text preview: 27 SECTION 2.4 Â¡ Â£ Â¨ Â¡Â¥ Â¤ âˆž 0 2t e Â¨ Â¡ Â¡ dt âˆž 0 2t Â¡ Â¡ Â¡ Â¡ Â¡ Â¡ 5 5e 52 426. 0 0 0 0 5, so xm dt âˆž 0 2t Â¢ te 0 2t 50 Â¡ 0 2te 0 10xm 1800 Â¨ âˆž x2 2 m Â¡ 15. (a) Âµ 0 5. Therefore Â¨ (e) The median xm solves F xm 0 Â¡ Â¡ 0 2t 0 2te Â¨ Â¢ Â¡ Â¢ Â¡ Â¡ Â¥Â¤ Â¡ dt 25 dt 25 25 5 Â¡ 25 0 Â¡ Â¡ Â¤ f t dt . x x 0 2t dt 1 e 0 2e Â¨ Â¢ 0 0 dt 1 Â¡ Â¨ Â¡ Â¡Â¥Â¤ âˆž 0. e Â¡ Â¡Â¥Â¤ 0, F x 0 dt 0 2x âˆž 0 Â¡Â¥Â¤ 0, F x Â¡ âˆž Â¥Â¤ . 0 8647 0 2x90 Â¡ e 0 9, so x90 Â¨ Â¡ Â¡ Â¨ 0 9. Therefore 1 3 466. Â¡ 0 5, so xm 11 5129. Â¨ 0 2xm Â¡ Â¨ Â¡Â¥ Â¡Â¥ Â¡Â¥ Â¨ Â¤ Â¡Â¥ Â¡ Â¤ Â¤ (f) The 90th percentile x90 solves F x90 e Â¨ 0 5. Therefore 1 Â¡ 2 Â¨ Â¤ (e) The median xm solves F xm Â¡Â¥ F 10 Â¡ 10 PT Â¡ 10 Â¨ (d) P T Â¡ 10 5 0 x (c) F x If x âˆž 10 0 2t 0 25 ÏƒX If x 2te Â¡ Â¡ 0 âˆž 0 2t Âµ2 Â¨ te dt âˆž 2 0 0 2t 0 Â¡ 0 2t 2 e âˆž Â¡ (b) Ïƒ2 Â¤ ...
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