By Chien W., Hung Q.

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**Sample text**

Unlike in the discrete probability case, there is no simple analytical formula for obtaining the probability for the normal distribution. Instead, in evaluating such a probability, we often consider the transformation Z = (X − µ)/σ. It can be shown that when X ∼ N(µ, σ 2 ), then Z = X−µ ∼ N(0, 1), σ which is a normal distribution with mean 0 and variance 1. The particular N(0, 1) distribution is called the standard normal distribution. 1) when X ∼ N(70, 62 ). 1 under the probability density curve (note that the total area under any density curve is 1).

A and B) is just the event that the value shown is a 2. Thus, P(AB) = 1/6. 6) is 3/6 + 3/6 − 1/6 = 5/6, which is equal to P(A or B) = 5/6 that we obtained earlier. 2 DEPENDENT EVENTS AND CONDITIONAL PROBABILITY 9 or generally with n events E1 , . . 7) i=1 where the notation ki=1 ai stands for the summation of the terms ai over the index i from 1 to k, n n P i=1 Ei means P(E1 ∩ E2 ∩ · · · ∩ En ) i=1 Ei means P(E1 ∪ E2 ∪ · · · ∪ En ), and P or just P(E1 E2 · · · En ). 7) is also called the principle of inclusion and exclusion.

3. 7528. 4115 are obtained from the standard normal probability table given in Appendix A. The first probability P(−1 < Z < 0) is obtained based on the fact that the standard normal distribution is symmetric about its mean 0. 9 Likelihood ratio Consider a blood donation center in which every donor is required to have a screening blood test for a particular kind of disease. Suppose a person has had the test and the result is positive. In this situation, two propositions or explanations may be considered: H0 : the person is a carrier of the disease; H1 : the person is not a carrier of the disease.

### 3-D numerical study on the bending of symmetric composite laminates by Chien W., Hung Q.

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