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3. Assume that the time required to pour a concrete floor for a structure (D) has a triangular distribution between 18 and 22
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Page 1 ® Triangular Distribution : f(x) a18 19 c=20 21 b=22 We know that, for triangular distribution xea acxsc f (x) 2x-a) (1/2 x Page-2 2(22 - 2D x (222) (22-18) (22-20) 1/2 0,25x1 P(0321) = 0.125 p (D520) = 1 / 2 x f(x) x base a=18 c=20=X b=22 1 xprob (D =19) Page 3 1 x basex f(x) = 1/2x1 x 22-0) (b-a) ((-a) 2(19-18) 22 10 re 21 = 1/2X1 X (92-18) (*20-18 base PP(D=19) =

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