Respuesta :
Answer:
The probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day is [tex]P(Y>190)=\frac{1}{e^{\frac{19}{10}}}\approx 0.1496[/tex]
Step-by-step explanation:
Let Y be the water demand in the early afternoon.
If the random variable Y has density function f (y) and a < b, then the probability that Y falls in the interval [a, b] is
[tex]P(a\leq Y \leq b)=\int\limits^a_b {f(y)} \, dy[/tex]
A random variable Y is said to have an exponential distribution with parameter [tex]\beta > 0[/tex] if and only if the density function of Y is
[tex]f(y)=\left \{ {{\frac{1}{\beta}e^{-\frac{y}{\beta} }, \quad{0\:\leq \:y \:\leq \:\infty} } \atop {0}, \quad elsewhere} \right.[/tex]
If Y is an exponential random variable with parameter β, then
mean = β
To find the probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day, you must:
We are given the mean = β = 100 cubic feet per second
[tex]P(Y>190)=\int\limits^{\infty}_{190} {\frac{1}{100}e^{-y/100} } \, dy[/tex]
Compute the indefinite integral [tex]\int \frac{1}{100}e^{-\frac{y}{100}}dy[/tex]
[tex]\frac{1}{100}\cdot \int \:e^{-\frac{y}{100}}dy\\\\\mathrm{Apply\:u \:substitution}\:u=-\frac{y}{100}\\\\\frac{1}{100}\cdot \int \:-100e^udu\\\\\frac{1}{100}\left(-100\cdot \int \:e^udu\right)\\\\\frac{1}{100}\left(-100e^u\right)\\\\\mathrm{Substitute\:back}\:u=-\frac{y}{100}\\\\\frac{1}{100}\left(-100e^{-\frac{y}{100}}\right)\\\\-e^{-\frac{y}{100}}[/tex]
Compute the boundaries
[tex]\int _{190}^{\infty \:}\frac{1}{100}e^{-\frac{y}{100}}dy=0-\left(-\frac{1}{e^{\frac{19}{10}}}\right)[/tex]
[tex]\int _{190}^{\infty \:}\frac{1}{100}e^{-\frac{y}{100}}dy=\frac{1}{e^{\frac{19}{10}}}\approx 0.1496[/tex]
The probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day is [tex]P(Y>190)=\frac{1}{e^{\frac{19}{10}}}\approx 0.1496[/tex]
