Gaussian distribution with mean and variance of probability

Question

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

Answer 1

The tension, $t$, is distributed normally with known mean and standard deviation, $t \sim \mathcal{N}(0,1)$.

The probabilities can be obtained with the cumulative distribution function,

$$ \Phi(x) = \frac{1}{\sigma \sqrt{2\pi}} \int_{-\infty}^x \! \exp{\frac{-(t-\mu)^2}{2 \sigma^2}} \, \mathrm{d}t$$

where $\mu = 0$ is the mean, $\sigma = 1$ is the standard deviation, and $x$ is the corresponding value. Since you want to know the probabilities for larger values, you have to use the negative of your values:

$$p_{2.96} = \Phi_{0,1}(-2.96) \approx 0.0015$$ $$p_{3.17} = \Phi_{0,1}(-3.17) \approx 0.00076$$

Hence, the probability of $t$ being greater than $3.17$ given $t > 2.96$ is

$$ p_{3.17} / p_{2.96} \approx 0.5$$