Question

A theoretical justiMication based on a certain material failure mechanism underlies the assumption that ductile strength X of a material has a lognormal distribution. If the parameters are μ=5 and σ=0.1 , Find: (a)μ x and σx (b)P (X >120) (c) P(110 ≤ X ≤ 130) (d)T hemedianductilestrength (e)T heexpectednumberhavingstrengthatleast120,iften different samples of an alloy steel of this type were subjected to a strength test. (f) The minimum acceptable strength, If the smallest 5% of strength values were unacceptable.

Answer 1

Given μ = 5, σ= 0, 1, a) 2-е 2-149. 157 =е = 14.953 b)P(X > 120) = P| Z > c)P(110 <X <130)- P(X <130) P (110) 120) . ΡΙΖ> 120-14912] = P(Z >-1.9499)= 0.97441 14.953 130-149.157 14.953 13 310 1901 110-149.157 14.953 -P(Z <-1.281 15)-P(Z <-2.6187) = 0, 10007-0. 00441 = 0.09566 d)median = exp(A) = exp (5) = 148.413 e)P(X > 120) PZ> f)P(X < x)= 0.05 we find x value p|7 x-149.157 From normal table, P (Z <-1.645)-0.05 Therfore149.157 3 4304 120-149.157 PfZ >-1.9499)= 0.97441 14.953 = 0.05 -1.645→ x = 124.5593 14.953