Question

Exercises Suppose the force acting on a column that helps to support a building is normally distributed with mean 15 kips and standard deviation 1.25 kips. What is the probability that the force is at most 18 kips? What is the probability that the force is between 10 and 12 kips? What is the probability that the force differs from 15 kips by at most 2 standard deviations? How would you characterize the largest 5% of all values of the force?

Answer 1

Solution:- Given i U= 15 kips &= 1.25 kips SP (X<18) = P(x-4 L 18 - 15 ) 1. 25 . 1.25 2 p (2 < 2:4) - 0.9918 PCX518) PC10<x< 12) = |( 10-15 < x-x. 12-15) -1S - 12 - 15 1. 25 1. 25 =pl-5 (1. 25 < 2 - 3 l 1. 25 ) = p (-4<2 <-2-4) = P(Z <=2.4) - P(22-4) = PC10<x<12) - 0.0082 - 0 0.0082

At most je Less 2 std than devi from mean a sta devi from mean. (15) - X = x = x = x = o . (1.25) 15- 2 15- 2 15-2.5 12.5 P(X512.5) = plx-et < 12:5215) - P (2 2-2.5) - 2 .5 1. 25 - P (22-2) IPC x < 12.5 0.0228 - The 2 distribution of the St. is P( 272) = 5%. 1-P (2 <z)=0.05 P( 2 22) = 1-0.05. P (2< 1.645) = 0.95 2= 1.645 Using 2-score formula x = zotu x = 1.645X 1.25 + 15 x= 17.06 kips