Statistics Homework Solutions 75

# Statistics Homework Solutions 75 - 75 SECTION 4.2 Â 0 2188...

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Unformatted text preview: 75 SECTION 4.2 Â¡ 0 2188 Â¨ Â¥ Â¢ 05 Â¤ 1 8! 05 8! 8 8 ! Â¡ 87 Â¥ 05 Â¤ Â¥Â¨ Â¡ 1 8 Â¤Â¥Â¨ 7 1 Â¡ Â¡ Â¥Â¨ Â¡ Â¤ Â¤ Â¡Â¥ Â¡Â¥ Â¥Â¨ Â¤ Â¥Â¨ Â¥ Â¤ Â¤ Â¡ Â¥ Â¡ Â¤ 6 PX 7 PX 8 8! 8! 05 6 1 05 8 6 05 6! 8 6 ! 7! 8 7 ! 0 10938 0 03125 0 00391 0 1445 05 Â¥Â¨ Â¥Â¨ Â¤Â¥Â¨ Â¤ Â¥ Â¡ Â¢ Â¤ Â¡ Â¤ Â¢Â¥ Â¡ Â¥Â¨ 88 Â¢Â¥ Â¤Â¥Â¨ Â¡ Â¨ Â¤ Â¤ Â¢ Â¡ Â¥ Â¨ Â¤ Â¡ Â¢ Â¨ 8! 05 1! 8 1 ! 1 Â¤Â¥Â¨ Â¤ Â¥ Â¡ 80 Â¡ Â¥Â¨ Â¤ Â¡ Â¤ Â¥ Â¡ Â¡ Â¡ Â¤ Â¡ Â¤Â¥Â¨ Â¡ 81 Â¨ Â¤ Â¡ Â¥ Â¡ Â¥ Â¨ Â¡ Â§ Â¤ Â¡ Â¡ Â¤ Â¡ Â¤ Â¡ 8! 05 0 1 05 0! 8 0 ! 1 0 00391 0 03125 0 9648 1 Â¥Â¨ 1 PX Â¡ 2 0 Â¥ Â¤ Â§ Â¤ Â¡ Â¨ PX PX Â¡ Â¡ Â¤ Â¡ Â¤ Â¡ 1 1 Â¥ Â¥ Â¡ 2 83 (d) P X 05 Â¡ PX 1 0 0039 Â¨ 6 3 88 (c) P X 8! 05 3! 8 3 ! 05 1 Â¡ 3 8 Â¤ Â¤ (b) P X 8! 05 8! 8 8 !  Â¤ Â¦Â¥ 8 Bin 8 0 5 . Â¥Â¨ Â¤ Â¦Â¥ Â¨ Let X be the number of bits out of the eight that are equal to 1. Then X (a) P X Â¡ Â¡ Â¨ Â¡ Let X be the number of defective parts in the sample from vendor A, and let Y be the number of defective parts in the sample from vendor B. Let pA be the probability that a part from vendor A is defective, and let p B be the probability that a part from vendor B is defective. Then X Bin 100 p A and Y Bin 200 pB . The observed value of X is 12 and the observed value of Y is 10. Â¤ Â¥ Â¨ Â¡ Â¡ Â¥ 0 05 1 0 05 200 0 032. 0 015. Â¨ Â¤ Â¨ Â¡ Â¥ Â¨ Â¨ Â¡ Â¤ Â¨ Â¤ Â¡ 0 12 1 0 12 100 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¡ pB 1 p B 200 0 05 and n with 200, Ïƒ pB Â¥ Â¡ Replacing pB with pB 0 05, Ïƒ pB  10 200 Â¥ Y 200 pA 1 p A 100 0 12 and n with 100, Ïƒ pA Replacing pA with pA (b) pB 0 12, Ïƒ pA 12 100 Â¤ X 100  (a) pA Â¥ 11. 1 3856 Â¡Â¥Â¨ 9. 12 0 2 0 8 Â¨ (e) ÏƒX ...
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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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