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< async src="//pagead2.googlesyndication.com/pagead/js/adsbygoogle.js"> Probability

發問:

http://holland.pk/1n3H

最佳解答:

1. (a) Normal table used : http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf (i) The required probability = P(x > 5.2) = P(z > [5.2-5]/0.14) = P(z > 1.429) = 1 - P(z < 1.429) = 1 - 0.9235 = 0.0765 (ii) P(4.85 < x < 5.15) = P([4.85-5]/0.14 < x < [5.15-5]/0.14) = P(-1.071 < x < 1.071) = P(x < 1.071) - P(x < -1.071) = 0.8579 - 0.1421 = 0.7158 (iii) P(z < 2.054) = 98% P(z ≥ 2.054) = 2% P(x ≥ [5+2.054*0.14]) = 2% P(x ≥ 5.29) = 2% Expected minimum weight = 5.29 grams (b) A: arrive within specified time A': not arrive within specified time P(A') = 2% = 0.02 P(A) = 1 - 0.02 = 0.98 (i) The required probability = P(1A'+9A) = 10C1 * 0.02 * 0.98? = 0.1667 (ii) The required probability = P(10A) + P(1A'+9A) = 0.981? + 10C1 * 0.02 * 0.98? = 0.9838

其他解答:5FAD1C75A5AA61DA