Statistics Homework Solutions 19

Statistics Homework Solutions 19 - 19 SECTION 2.3 ¤ ¡...

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Unformatted text preview: 19 SECTION 2.3 ¤ ¡ ¢¥ ¨ ¤ ¡ ¢ ¨ ¢ ¡ ¨ ¢¥ ¤ ¡ ¢¥ 0 20 ¨ ¤ ¡ ¢ ¨ ¡ ¢ ¨ ¢¥ ¡ ¤ 0 20 ¨ ¢¥ ¢ ¤ ¡ 0 01 P G1 G2 ¤ ¨ ¢ ¡ 0 30 0 51 P Y1 Y2 ¡ ¨ ¨ ¤ ¡ ¤ ¡ ¥ ¡ ¤ ¨ ¨ ¨ P R1 Y 2 P Y2 0 04 0 04 0 01 0 40 ¥ ¤ ¡ ¥ 0 05 ¨ ¥ ¡ ¡ P R1 G2 P R1 0 16 0 50 0 32 ¥ ¤ ¥ ¨ ¢ ¤ ¡ ¡ ¥ ¤ ¡ ¨ ¨ ¡ ¨ ¢ ¤ ¡ ¨ P R1 R2 ¥ ¥ (d) P G2 R1 ¤ 0 04 P G1 G2 ¡ ¤ ¡ 0 16 0 40 P Y 1 G2 ¥ ¥ ¡ ¨ P R1 G2 (c) P same color (e) P R1 Y 2 ¢¥ (b) P G2 ¤ 0 50 P R1 G2 ¡ P R1 R2 P R1 Y 2 0 30 0 04 0 16 ¥ (a) P R1 ¡ 13. (a) That the gauges fail independently. (b) One cause of failure, a fire, will cause both gauges to fail. Therefore, they do not fail independently. (c) Too low. The correct calculation would use P second gauge fails first gauge fails in place of P second gauge fails . Because there is a chance that both gauges fail together in a fire, the condition that the first gauge fails makes it more likely that the second gauge fails as well. Therefore P second gauge fails first gauge fails P second gauge fails . ¥ ¤ ¥ ¥ ¤ ¤ ¡ ¥ ¤ ...
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