# 5 I WILu a head of 3UW ip and a TUV of 20. SUPPOs and jected to a dead load of
ID: 3303737 • Letter: #
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# 5
I WILu a head of 3UW ip and a TUV of 20. SUPPOs and jected to a dead load of 100 dip and a live load of 150 Wip () Calculate the probability of failure of the foundation under dead load only. dead (b) Calculate the probability of failure of the foundation under the combined action of and live loads. (c) r the probability of failure of the foundation needs to be limited to 0.001, and the dead load of 100 kip cannot be what is changed, the maximum amount of live load that can be applied to the foundation? 5. The compressive strength data concrete is given the table below of cylinders in . Find the probability of realizing a compressive strength between 3 and 6klbin assuming a normal distribution and b. lognormal distribution Table 1. Compressive Strength of concrete Cylinders (klbinº) 6.5 1.4 5.9 6.3 2.1 7.2 3.4 8.2 4.1 6.9 6.2 8.9 6.8 6.1 6.8 7.7 5.6 7.5 6.4 5.8 5.3 9.1 5.7 8.3 7.8 3.7 4.3 4.9 5.1 3.7 6.7 4.3 2.6 7.3 4.5 7.9 6.4 6.6 5.4 6. Matlab Skills a. Create a dataset of 1000 points normally distributed with a mean of 500 and standard deviation of 100 using randn b. Visualize the data as a histogram c. Visualize the data as a CDF Report the actual mean and standard deviation for the generated dataset e, Compute the likelihood of the value being below 475. . Compute the likelihood of the value being below 625.Explanation / Answer
5. a, Using technology, mean, xbar=5.790, and standard deviation, s=1.808. Assume the variable X denote realizing compressive strength, therefore, using normal distribution,
Compute z scores for X1=3 and X2=6 using the following formula, Z=(X-xbar)/s, where, X is the raw score.
Z1=(3-5.790)/1.808=-1.54, Z2=(6-5.790)/1.808=0.12
Thw two Z scores are of opposite signs, therefore, find the area between each Z scores and the mean and add teh areas.
Therefore, P[3<X<6]=0.4382+0.0478=0.486
b. The mean and standard deviation of lognormal distribution are as follows:
mean=log xbar-1/2 log [(s/xbar)^2+1]=log 5.790-1/2 log[(1.808/5.790)^2+1]=0.74
standard deviation=sqrt[log (s/xbar)^2+1]=sqrt[log (1.808/5.790)^2+1]=0.20
Assume Y to be lognormally distributed variable.
P(Y<=6)=phi[{log(6)-0.74}/1.5]=phi (0.03)
P(Y<=3)=phi[{log(3)-0.74}/1.5]=phi (-0.18)
P[3<Y<6]=0.0120+0.0714=0.0834
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