Respuesta :
Answer:
[tex]P(X<0.5)=0.22[/tex]
[tex]P(X>2)=0.36[/tex]
The value 6 hours exceeded with probability 0.05.
Step-by-step explanation:
We are given that The waiting time for service at a hospital emergency department (in hours) follows a distribution with probability density function
[tex]f(x)=0.5e^{(-0.5x)}[/tex] for x>0
a. P(X < 0.5)
[tex]P(X< 0.5)=\int\limits^{0.5}_0 {f(x)} \, dx[/tex]
[tex]P(X< 0.5)=\int\limits^{0.5}_0 {0.5e^{-0.5x}} \, dx[/tex]
[tex]P(X<0.5)=(0.5)[\frac{e^{-0.5x}}{-0.5}]^{0.5}_{0}[/tex]
[tex]P(X<0.5)=1-e^{-0.25}[/tex]
[tex]P(X<0.5)=0.22[/tex]
b. P(X > 2)
[tex]P(X>2)=\int\limits^{\infty}_2 {f(x)} \, dx[/tex]
[tex]P(X>2)=\int\limits^{\infty}_2 {0.5e^{-0.5x}} \, dx[/tex]
[tex]P(X>2)=(0.5)[\frac{e^{-0.5x}}{-0.5}]^{\infty}_{2}[/tex]
[tex]P(X>2)=e^{-1}[/tex]
[tex]P(X>2)=0.36[/tex]
c. Value x (in hours) exceeded with probability 0.05.
P(X>x)=0.05
[tex]\int\limits^{\infty}_x {0.5e^{-0.5x}} \, dx=0.05[/tex]
[tex](0.5)[\frac{e^{-0.5x}}{-0.5}]^{\infty}_{x}=0.05[/tex]
[tex]e^{-0.5x}=0.05[/tex]
[tex]-0.5x=log(0.05)[/tex]
[tex]x=\frac{-log(0.05)}{-0.5}[/tex]
[tex]x=5.991[/tex]
[tex]x\sim 6[/tex]
The value 6 hours exceeded with probability 0.05.
The value of P(X < 0.5) is 0.22.
The value of P(x>2) is 0.32.
The value of x is 5.991 hours exceeded probability 0.05.
Given that,
The waiting time for service at a hospital emergency department (in hours) follows a distribution with probability density function;
[tex]\rm f(x) = 0.5^{(-0.5)} , \ for\ 0<x[/tex]
We have to determine,
The value of P(X < 0.5) and P(X > 2).
The value x (in hours) exceeded probability 0.05.
According to the question,
[tex]\rm f(x) = 0.5^{(-0.5)} , \ for\ 0<x[/tex]
1. The value of P(X < 0.5) is determined by integration of the function,
[tex]\rm P(x<0.5)= \int\limits^{0.5}_5 {f(x)} \, dx \\\\ P(x<0.5)= \int\limits^{0.5}_5 {0.5\times e^{-0.5x}} \, dx \\\\ P(x<0.5)= 0.5\int\limits^{0.5}_5 {e^{-0.5x}} \, dx } \\\\ P (x<0.5) = (0.5) \times \left[\dfrac{e^{-0.5x}}{-0.5}\right]^{0.5}_0\\\\P (x<0.5) = (0.5) \times \left[\dfrac{e^{-0.5(0.5-0)}}{-0.5}\right]\\\\P (x<0.5) = (0.5) \times \left[\dfrac{e^{-0.25}}{-0.5}\right]\\\\P (x<0.5) = 1-e^{-0.25}\\\\P (x<0.5) = 0.22\\[/tex]
The value of P(X < 0.5) is 0.22.
2. The value of P(X > 2 ) is determined by integration of the function,
[tex]\rm P(x>2)= \int\limits^{\infty}_2{f(x)} \, dx \\\\ P(x>2)= \int\limits^{\infty}_2{0.5\times e^{-0.5x}} \, dx \\\\ P(x>2)= 0.5\int\limits^{\infty}_2{e^{-0.5x}} \, dx } \\\\ P (x>2) = (0.5) \times \left[\dfrac{e^{-0.5x}}{-0.5}\right]^{\infty}_2\\\\P (x>2) = (0.5) \times \left[\dfrac{e^{-0.5(\infty - 2)}}{-0.5}\right]\\\\P (x>2) = 1-e^{1}\\\\P (x>2) = 0.36\\[/tex]
The value of P(x>2) is 0.32.
3. The value of x (in hours) exceeded probability 0.05.
[tex]\rm \int\limits^\infty_x {0.5 e^{-0.5x}} \, dx = 0.05\\\\0.5 \int\limits^\infty_x {e^{-0.5x}} \, dx = 0.05\\\\ 0.5 {e^{-0.5x}} = 0.05\\\\ Taking \ log \ on \ both \ sides\\\\- 0.5x= log(0.05)\\\\x = \dfrac{log(0.05)}{-0.5}\\\\x = \dfrac{-2.99}{-0.5}\\\\x = 5.991[/tex]
The value of x is 5.991 hours exceeded probability 0.05.
For more details refer to the link given below.
https://brainly.com/question/18849758
