Statistics Homework Solutions 76

A p can be used 2 a p 0 12 the uncertainty is pb c

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Unformatted text preview: an be used. Then X ¥ 0 06 ¨ 0 90 ¡ P long ¨ ¡ P meets spec ¡ σ2 B p 0 05 ¨ 13. (a) P can be used σ2 A p 0 12 ¡ The uncertainty is pB ¨ (c) The difference is estimated with pA ¤ 17. (a) Let X be the number of components that function. Then X Bin 5 0 9 . 5! 5! PX 3 09 3 1 09 5 3 09 4 1 09 5 4 3! 5 3 ! 4! 5 4 ! 5! 0 9 5 1 0 9 5 5 0 9914 5! 5 5 ! ¥¨  ¥¨ ¤ ¤¥¨ ¡ ¤ ¥ ¢ ¤ ¡ ¨ ¡ ¥¨ ¤¥¨ ¡ ¥¨ ¡ ¤¥¨ ¤ ¤ ¥ ¥ ¤ ¢ ¤ ¡ § ¡¥ ¤ ¡ (b) We need to find the smallest value of n so that P X 2 0 10 when X Bin n 0 9 . Consulting Table A.1, we find that if n 3, P X 2 0 271, and if n 4, P X 2 0 052. The smallest value of n is therefore n 4. ¨ ¡¥ ¥ ¡ ¤ ¨ ¤ ¨ ¡ ¨ ¡¥ ¤ ¨ ¡ ¡ 9 ¢¥ ¡ PX PX ¤ ¤ ¢¥ 8 ¡ ¡ PX 10 ¥ ¥ ¤ ¢¥ 7 ¨ ¡ PX ¤ ¤ ¨ ¡  ¥ ¤ 7  § PX Bin 10 0 15 . ¥¨ 19. (a) X ¤...
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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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