Question

if x has an exponential distribution with mean 2 find p(x less than 1/x less than 2)​

Answer 1

Answer:

Mathematically, it says that P(X > x + k|X > x) = P(X > k). If T represents the waiting time between events, and if T ∼ Exp(λ), then the number of events X per unit time follows the Poisson distribution with mean λ. The probability density function of P(X=k)=λke−λk!

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Answer 2

Answer:

The value of P(X < 1 | X < 2) is e^(0.5).

Step-by-step explanation:

Given:

Mean (1/λ) = 2

To find:

P(X < 1 | X < 2)

Solution:

The mean of the exponential distribution is given to be 2.

Mean = 1/λ = 2

λ = 1/2

λ = 0.5

We have to find P(X < 1 | X < 2)

P(X < 1 | X < 2) = P(X < 1 ∩ X < 2)/P(X < 2)

$P(X < 1 | X < 2) = \frac{P(X < 1 )}{P(X < 2)}$

$P(X < 1 | X < 2) = \frac{1-P(X \geq 1 )}{1-P(X \geq 2)}$

$P(X < 1 | X < 2) = \frac{1-F(1)}{1-F(2)}$

We know that F(x) = 1 - e^(-λx)

$P(X < 1 | X < 2) = \frac{1-(1-e^{-0.5*1})}{1-(1-e^{-0.5*2})}$

$P(X < 1 | X < 2) = \frac{1-1+e^{-0.5}}{1-1+e^{-1}}$

$P(X < 1 | X < 2) = \frac{e^{-0.5}}{e^{-1}}$

$P(X < 1 | X < 2) = e^{-0.5+1}$

$P(X < 1 | X < 2) = e^{0.5}$

Therefore, the value of P(X < 1 | X < 2) is $e^{0.5}$.

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