1 reliability of earthen dam 25 points\nan earthen dam with a design life of 50 years is planned in a region…
1 reliability of earthen dam 25 points\nan earthen dam with a design life of 50 years is planned in a region where low - to moderate intensity earthquakes may occur.\nthe earthquake magnitude, measured on richter scale ( m, in energy units: ergs) follows a probability distribution function of the form: $f_m(m)=me^{-\frac{1}{2}m^{2}};\forall mgeq0$\nthe seismometer (earthquake detection instrument) signals an earthquake when m exceeds 3 ergs. the annual occurrence of low - to moderate - intensity earthquakes in this region is assumed to be a bernoulli sequence.\nthe earthen dam is designed to have a 95% probability of not suffering damage due to earthquakes registered by the seismometer within its design life. damage effects between earthquakes are statistically independent.\na.) what is the probability of damage to the earthen dam under a single earthquake registered by the seismometer?\nb.) what would be the probability of at most two damages to the earthen dam during its design life assuming that the occurrences of earthquakes constitute a poisson process?
Answer
Explanation:
Step1: Calculate probability of single - earthquake damage
We first need to find the probability that an earthquake with magnitude $M$ exceeds 3 ergs. The probability density function is $f_M(m)=me^{-\frac{1}{2}m^{2}},m\geq0$. The probability $P(M > 3)$ is given by the integral $\int_{3}^{\infty}me^{-\frac{1}{2}m^{2}}dm$. Let $u = \frac{1}{2}m^{2}$, then $du=mdm$. When $m = 3$, $u=\frac{9}{2}$, and as $m\rightarrow\infty$, $u\rightarrow\infty$. So, $\int_{3}^{\infty}me^{-\frac{1}{2}m^{2}}dm=\int_{\frac{9}{2}}^{\infty}e^{-u}du$. Using the integral formula $\int e^{-u}du=-e^{-u}+C$, we have $\int_{\frac{9}{2}}^{\infty}e^{-u}du=\lim_{b\rightarrow\infty}\int_{\frac{9}{2}}^{b}e^{-u}du=\lim_{b\rightarrow\infty}(-e^{-u})\big|{\frac{9}{2}}^{b}=\lim{b\rightarrow\infty}(-e^{-b}+e^{-\frac{9}{2}})=e^{-\frac{9}{2}}$.
Step2: Define parameters for Poisson process
The design life of the dam is $T = 50$ years. Let $\lambda$ be the average number of damaging earthquakes in 50 years. If the annual probability of a damaging earthquake is $p = e^{-\frac{9}{2}}$, then $\lambda=50p = 50e^{-\frac{9}{2}}$. The probability mass function of a Poisson distribution is $P(X = k)=\frac{e^{-\lambda}\lambda^{k}}{k!}$, where $X$ is the number of events, $\lambda$ is the average number of events in the given time - interval, and $k$ is the actual number of events that occur. We want to find $P(X\leq2)=P(X = 0)+P(X = 1)+P(X = 2)$. $P(X = 0)=\frac{e^{-\lambda}\lambda^{0}}{0!}=e^{-\lambda}$, $P(X = 1)=\frac{e^{-\lambda}\lambda^{1}}{1!}=\lambda e^{-\lambda}$, $P(X = 2)=\frac{e^{-\lambda}\lambda^{2}}{2!}=\frac{\lambda^{2}e^{-\lambda}}{2}$. So, $P(X\leq2)=e^{-\lambda}(1 + \lambda+\frac{\lambda^{2}}{2})$, substituting $\lambda = 50e^{-\frac{9}{2}}$.
Answer:
a. The probability of damage to the earthen dam under a single earthquake registered by the seismometer is $e^{-\frac{9}{2}}$. b. First, $\lambda = 50e^{-\frac{9}{2}}$, and $P(X\leq2)=e^{-50e^{-\frac{9}{2}}}(1 + 50e^{-\frac{9}{2}}+\frac{(50e^{-\frac{9}{2}})^{2}}{2})$