Suppose X has an exponential distribution with λ = 2. Determine the following:
(a)P(x < 0)=””>
(b)P(X > 2)
(c)P(X <>
(d) P(1) < x=””><>
(e)Find the value of x such that P(X < x)=””>
The correct answer and explanation is :
Given that $X$ follows an exponential distribution with rate parameter $\lambda = 2$, we use the probability density function (PDF) of the exponential distribution:
$$
f(x) = \lambda e^{-\lambda x}, \quad x \geq 0
$$
where $\lambda = 2$. Therefore, the PDF becomes:
$$
f(x) = 2 e^{-2x}, \quad x \geq 0
$$
Now, we will calculate the required probabilities.
(a) $P(X < 0)$
For an exponential distribution, the probability of $X$ being less than zero is zero because $X$ can only take non-negative values. Mathematically:
$$
P(X < 0) = 0
$$
(b) $P(X > 2)$
The probability that $X > 2$ can be calculated using the survival function of the exponential distribution:
$$
P(X > 2) = 1 – P(X \leq 2)
$$
The cumulative distribution function (CDF) is given by:
$$
P(X \leq x) = 1 – e^{-\lambda x}
$$
Substituting $\lambda = 2$ and $x = 2$:
$$
P(X \leq 2) = 1 – e^{-2 \cdot 2} = 1 – e^{-4}
$$
Using a calculator:
$$
P(X \leq 2) \approx 1 – 0.0183 = 0.9817
$$
So:
$$
P(X > 2) = 1 – 0.9817 = 0.0183
$$
(c) $P(X < 1)$
Using the CDF again for $x = 1$:
$$
P(X \leq 1) = 1 – e^{-2 \cdot 1} = 1 – e^{-2}
$$
$$
P(X \leq 1) \approx 1 – 0.1353 = 0.8647
$$
Thus:
$$
P(X < 1) = 0.8647
$$
(d) $P(1 < X < 3)$
We can calculate this as:
$$
P(1 < X < 3) = P(X < 3) – P(X < 1)
$$
From earlier, $P(X < 3)$ is:
$$
P(X \leq 3) = 1 – e^{-2 \cdot 3} = 1 – e^{-6} \approx 1 – 0.0025 = 0.9975
$$
Thus:
$$
P(1 < X < 3) = 0.9975 – 0.8647 = 0.1328
$$
(e) Find the value of $x$ such that $P(X < x) = 0.5$
We want to solve for $x$ such that:
$$
P(X < x) = 0.5
$$
Using the CDF:
$$
1 – e^{-2x} = 0.5
$$
Solving for $x$:
$$
e^{-2x} = 0.5
$$
Taking the natural logarithm of both sides:
$$
-2x = \ln(0.5) = -\ln(2)
$$
$$
x = \frac{\ln(2)}{2} \approx \frac{0.6931}{2} = 0.3466
$$
Thus, $x \approx 0.3466$.
Summary
- $P(X < 0) = 0$
- $P(X > 2) \approx 0.0183$
- $P(X < 1) \approx 0.8647$
- $P(1 < X < 3) \approx 0.1328$
- $x \approx 0.3466$ such that $P(X < x) = 0.5$