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HW3 Solution

# HW3 Solution - ME335-Engineering Statistics-Fall 2008 HW 3...

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ME335-Engineering Statistics-Fall 2008 HW # 3 Solution 4.3 (a) - ( ) 1, f x dx x x x x 4 0 1 x ke dx x - x x , 4 1 k x , k=1/4 (b)P{2<X<8}= 8 2 ( ) f x dx x = 8 4 2 1 4 x e dx - = 1 2 2 e e - - - =0.4712 (c) P{X>7}= 7 4 4 7 7 1 ( ) 0.1738 4 x f x dx e dx e x - - = = = (d) P{X<1}= 1 1 1 4 4 0 1 ( ) 1 0.2212 4 x f x dx e dx e - - - = = - = (e) P{X>C}= 1 4 4 1 ( ) 0.5 4 C C C f x dx e dx e x - - = = (Obviously C>0) 4 ln( ) ln(0.05) 4 C C e - = = - , ln(0.05) 4ln(20) 11.98 C = - = (f) P{X<d}= d 4 4 0 1 ( ) 1 0.025 4 x d d f x dx e dx e - - - = = - = , d 0.1 W 4.9 From the additive property of the normal distribution, we know that, if you take several normal random variables, Xi, with i=1,2,…,n, and form weighted sums and/or differences among them, the result will always be another normal random variable, i.e. , Y is a normal random variable and its mean and variance are: , . For this question, here Y= A+B+C, A,B and C are independently normal random variables, so Y is a normal random variable, and its mean and variance are as follows: , Entering the formula Normal Dist(12.2, 12, 0.10488) to JMP to calculate the result.

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