4. [20 points] To study the time to failure of a cable, a civil engineer perform

Question

4. [20 points] To study the time to failure of a cable, a civil engineer performs tensile tests by subjecting Gaussian the cable to random loads so that the tension recorded by the measuring instrument follows a distribution with mean 1 and standard deviation 1, both in tons. At a certain last measuremen greater than 2.96 tons. What is the probability that the cable tension in the last 3.17 tons given that the tension was greater than 2.96 tons? instant during tensile measurement, the measuring instrument suddenly broke down. The t was not recorded properly, but the engineer believes that the cable tension was measurement excceded

Explanation / Answer

Solution:

We are given that random variable tension (X) follows a normal or Gaussian distribution.

Mean = 1, SD = 1

We have to find P(X>3.17|X>2.96)

P(X>3.17|X>2.96) = P(2.96<X<3.17) / P(X>2.96)

P(2.96<X<3.17) = P(X<3.17) – P(X<2.96)

For X<3.17

Z = (X – Mean) / SD

Z = (3.17 – 1) / 1 = 2.17

P(Z<2.17) = P(X<3.17) = 0.984996577 (by using z-table or excel)

For X<2.96

Z = (2.96 – 1) / 1 = 1.96

P(Z<1.96) = P(X<2.96) = 0.975002105 (by using z-table or excel)

P(2.96<X<3.17) = P(X<3.17) – P(X<2.96)

P(2.96<X<3.17) = 0.984996577 – 0.975002105

P(2.96<X<3.17) = 0.009994472

Now,

P(X>2.96) = 1 – P(X<2.96)

P(X<2.96) = 0.975002105

P(X>2.96) = 1 – 0.975002105

P(X>2.96) = 0.024997895

P(X>3.17|X>2.96) = P(2.96<X<3.17) / P(X>2.96)

P(X>3.17|X>2.96) = 0.009994472/0.024997895

P(X>3.17|X>2.96) = 0.399812544

Required probability = 0.399812544

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