Statistics Homework Solutions 83

# Statistics Homework Solutions 83 - 83 SECTION 4.4 0 0314 Â...

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Unformatted text preview: 83 SECTION 4.4 0 0314 Â¡ 01 09 Â¨ Â¥Â¨ Â¤ Â¦Â¥ Â¨ Â¤ 12 Geom 0 1 . Â¡Â¥ PX 11 Â¤ Let X be the number of the day on which the computer crashes. Then X Â¥Â¨ 3. Â¡ Â¤ Â¤ Â¥Â¨ Â¤ Â¡ Â¡ Â¡ Â¡Â¥ Â¡ Â¤ Â¥Â¨  1 73 04 0 1244 75 Â¨ 11 25 Â¡Â¥ 0 42 Â¨ 04 Â¨ Â¡ Â¤ Â¨ Â¤ Â¥ Â¨ Â£ Â£ Â¡ 31 3 04 Â¡ Â¤ Â¡ 2 (c) ÏƒY 1 1 Â¥Â¨ 3 04 7 3 (b) ÂµY 7 Â¡ NB 3 0 4 . P Y Â¨ 5. (a) Y x1 Â¡ (iv). P X 9. Let X be the number of hours that have elapsed when the fault is detected. Then X Geom 0 8 . Â¤ Â¡Â¥ 1. Since 1 p Â¤ Â¦Â¥ Â¨ 08 1 08 Â¡Â¥ Â¨ Â¤Â¨ 08 08 1 0 04. Â¨ Â¨ Â¡ 1 Â¡ Â¡Â¥ Â¤ Â¡ Â¢Â¥ Â¥ Â¡ Â¤ Â¡ Â¡ Â¤ Â¢Â¥ Â¡ Â¡ Â¥ Â¡ Â¨ Â¡ Â¤ Â¥ Â¡ Â¡ Â¡ Â¨ Â¡ Â¡Â¥ Â¤ Â¤ Â¡Â¥ Â¤ Â¤ Â¨ Â¡ Â¤ Â¥Â¨ Â¨ Â¡ Â¤Â¨ Â¡ Â¡Â¥ Â¡ Â¡ Â¡ Â¡Â¥ Â¨ Â¡Â¥ Â¤ Â¨ Â¡ Â¡ Â¤ Â¡ Â¡ Â¡ Â¤ Â¡ Â¡Â¥ Â¤ Â¡ Â¨ Â£ Â¡ Â¡ Â¨ Â¡ Â¤ Â¨ 03 2 Â¡ Â¡ Â¤ Â¡ Â¡Â¥ Â¡ Â¤ Â¥  Â¤ Â¡ 12 1 Â¥ Â¡ Â¨ Â¡Â¥ 4 3 10 08 Â¥ Â¡Â¥ Â¤ Â£Â¤ Â¡ (b) ÂµX Â¡ 2 Â¥Â¨ H 10 3 4 . P X Â¢ 11. (a) X 10 3 42 10 4 Â¤ 1 25 3 2 Â¤ Â¦Â¥ Â¨ 1 08 Â¡ (c) ÂµX 08 1 Â¥Â¨ PX PX 0 08 Â¢ 3 2 08 1 Â¤ 2 3 Â¤Â¥ Â¨ 3X PX 2 PX Â¡ PX 2 08 Â¥Â¨ PX 2 Â¡ 3 1 1. 0 992 Â¨ Â¤ Â¤ 1 0 8 1 0 8 2 0 032, and PX 2 1 PX 1 PX 0 032 Therefore P X 3 X 2 0 8. 0 04 Now P X PX 2 1, this quantity is maximized when x Â¥Â¨ Â¡ 2 Â¡ 3X Â¥ PX (b) P X 3 for x Â§ (a) P X p Â¡ p1 Â¡ x Â¡ 7. ...
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