MECH 262 Study Guide - Final Guide: Poisson Distribution, Normal Distribution, Binomial Distribution
23 May 2016
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6. 12 probability of an undergraduate biology student to be a woman: 6. 14 probability of all three components being defective: 4 all 6 > 3600 hours; p=0. 95, n=6, r = 6. We want probability of 1 or 2 failures. p=0. 2, n=2, r=1 and r=2. The probability of 1 or 2 is then p(1)+p(2) = 0. 36. 3 (a) f(x) satisfies the requirement of a probability distribution function because: P x f x dx f x dx. 0 (a) f(x) satisfies the requirement of a probability distribution function because: 6. 24 binomial distribution can be used because of the satisfactory/ unsatisfactory outcome of the process. 1 p n r n r n r n r. In this case: p=0. 95 (a) all four parts be satisfactory: n=4. 8145% (b) for at least two parts to be satisfactory, we should calculate the probability that 2,3 and 4 parts be satisfactory: Probability of having at least two satisfactory parts:
