Question

Suppose X has an exponential distribution with lambda equals 5. Determine the following. Round the answers to 3 decimal places.a) Upper P left-parenthesis Upper X less-than-or-equal-to 0 right-parenthesis equalsb) Upper P left-parenthesis Upper X greater-than-or-equal-to 2 right-parenthesis equalsc) Upper P left-parenthesis Upper X less-than-or-equal-to 1 right-parenthesis equalsd) Upper P left-parenthesis 1 less-than Upper X less-than 2 right-parenthesis equalse) Find the value of x such that Upper P left-parenthesis Upper X less-than x right-parenthesis equals 0.05.

Answer 1

The value of x such that P(X < x) = 0.05 is approximately 0.01

Exponential Distribution

The exponential distribution is a continuous probability distribution that describes the time between events in a Poisson process. The probability density function (PDF) of the exponential distribution with rate parameter λ is given by:

f(x) = λe^(-λx)

where x is the time between events and λ is the rate parameter.

Cumulative Distribution Function (CDF)

The cumulative distribution function (CDF) of the exponential distribution is given by:

F(x) = 1 - e^(-λx)

The CDF represents the probability that X is less than or equal to a certain value x.

Solving for Probabilities

To solve for probabilities using the CDF, we can simply plug in the desired value of x into the CDF. For example, to find P(X ≤ 2), we would calculate:

P(X ≤ 2) = F(2) = 1 - e^(-5*2) ≈ 0.99324

Solving for Quantiles

To solve for the quantile x such that P(X < x) = p, we can set the CDF equal to p and solve for x:

p = F(x) = 1 - e^(-λx)

Solving for x, we get:

x = -ln(1 - p) / λ

The value of x such that P(X < x) = 0.05 is approximately 0.01.

Here's the revised calculation:

P(X < x) = 0.05

e^(-λ*x) = 0.05

ln(0.05) = -λ*x

x = -ln(0.05) / λ

x = -ln(0.05) / 5

x ≈ 0.01

Therefore, the value of x such that P(X < x) = 0.05 is approximately 0.01

Answer 2

Explained below.

Explanation:

The random variable X follows an exponential distribution with λ = 5.

The probability density function of X is:

`f_(X)(x)=\lambda\cdot e^(-\lambda\cdot x);\ 0<x<\infty`

`P(X\leq x)=1-e^(-\lambda\cdot x)\\\\P(X\geq x)=e^(-\lambda\cdot x)`

(a)

Compute the value of P (X ≤ 0) as follows:

`P(X\leq 0)=1-e^(-\lambda\cdot x)`

`=1-e^(-5* 0)\\\\=1-1\\\\=0`

Thus, the value of P (X ≤ 0) is 0.

(b)

Compute the value of P (X ≥ 2) as follows:

`P(X\geq 2)=e^(-5* 2)=4.54* 10^(-5)\approx 0`

Thus, the value of P (X ≥ 2) is approximately 0.

(c)

Compute the value of P (X ≤ 1) as follows:

`P(X\leq 0)=1-e^(-\lambda\cdot x)`

`=1-e^(-5* 1)\\\\=1-0.00674\\\\=0.99326`

Thus, the value of P (X ≤ 1) is 0.99326.

(d)

Compute the value of P (1 ≤ X ≤ 2) as follows:

`P(1<X<2)=\int\limits^(2)_(1) {5* e^(-5x)} \, dx \\\\=5* [(e^(-5x))/(-5)]^(2)_(1) \\\\=5* [(0-0.00674)/(-5)]\\\\=0.0067`

Thus, the value of P (1 ≤ X ≤ 2) is 0.0067.

(e)

Compute the value of x such that P (X < x) = 0.05 as follows:

`P(X<x)=0.05\\\\\int\limits^(x)_(0) {5* e^(-5x)} \, dx =0.05\\\\5* [(e^(-5x))/(-5)]^(x)_(0) =0.05\\\\1-e^(-5x)=0.05\\\\-e^(-5x)=0.95\\\\-5x=\ln(0.95)\\\\-5x=-0.05\\\\x=0.01\\\\`

Thus, the value of x is 0.01.