Let X be a continuous random variable with density function √ 3e-3x, 0, (a) Verify that f is a density function. (b) Calculate P(-1 < X < 1). (c) Calculate P(x < 5). (d) Caculate P(2 < X < 4|X<5) f(x) = x > 0 else.
Let X be a continuous random variable with density function √ 3e-3x, 0, (a) Verify that f is a density function. (b) Calculate P(-1 < X < 1). (c) Calculate P(x < 5). (d) Caculate P(2 < X < 4|X<5) f(x) = x > 0 else.
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![**Continuous Random Variable with Density Function Example**
Let \( X \) be a continuous random variable with the following density function \( f(x) \):
\[ f(x) =
\begin{cases}
3e^{-3x} & x > 0, \\
0 & \text{else.}
\end{cases}
\]
### (a) Verification of the Density Function:
To verify that \( f \) is a density function, we need to check two conditions:
1. \( f(x) \geq 0 \) for all \( x \).
2. The integral of \( f(x) \) over its entire range is equal to 1, i.e., \(\int_{-\infty}^{\infty} f(x) \, dx = 1 \).
### (b) Calculation of \( P(-1 < X < 1) \):
To find \( P(-1 < X < 1) \), we integrate \( f(x) \) from -1 to 1:
\[ P(-1 < X < 1) = \int_{-1}^{1} f(x) \, dx \]
Since \( f(x) = 0 \) for \( x \leq 0 \):
\[ P(-1 < X < 1) = \int_{0}^{1} 3e^{-3x} \, dx \]
Evaluate the integral:
\[ P(-1 < X < 1) = \left[-e^{-3x}\right]_0^1 = -(e^{-3 \cdot 1} - e^{-3 \cdot 0}) = -(e^{-3} - 1) = 1 - e^{-3} \]
### (c) Calculation of \( P(X < 5) \):
To find \( P(X < 5) \), we integrate \( f(x) \) from 0 to 5:
\[ P(X < 5) = \int_{-\infty}^{5} f(x) \, dx = \int_{0}^{5} 3e^{-3x} \, dx \]
Evaluate the integral:
\[ P(X < 5) = \left[-e^{-3x}\right]_0^5 = -(e^{-3 \cdot 5} - e^{-3 \cdot 0}) = -(e^{-15} - 1) =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff7e7c908-505f-4c35-8ac4-e541f2ab69b2%2F9e930d01-d255-42e2-baaf-37417a21bd60%2Fi69u67r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Continuous Random Variable with Density Function Example**
Let \( X \) be a continuous random variable with the following density function \( f(x) \):
\[ f(x) =
\begin{cases}
3e^{-3x} & x > 0, \\
0 & \text{else.}
\end{cases}
\]
### (a) Verification of the Density Function:
To verify that \( f \) is a density function, we need to check two conditions:
1. \( f(x) \geq 0 \) for all \( x \).
2. The integral of \( f(x) \) over its entire range is equal to 1, i.e., \(\int_{-\infty}^{\infty} f(x) \, dx = 1 \).
### (b) Calculation of \( P(-1 < X < 1) \):
To find \( P(-1 < X < 1) \), we integrate \( f(x) \) from -1 to 1:
\[ P(-1 < X < 1) = \int_{-1}^{1} f(x) \, dx \]
Since \( f(x) = 0 \) for \( x \leq 0 \):
\[ P(-1 < X < 1) = \int_{0}^{1} 3e^{-3x} \, dx \]
Evaluate the integral:
\[ P(-1 < X < 1) = \left[-e^{-3x}\right]_0^1 = -(e^{-3 \cdot 1} - e^{-3 \cdot 0}) = -(e^{-3} - 1) = 1 - e^{-3} \]
### (c) Calculation of \( P(X < 5) \):
To find \( P(X < 5) \), we integrate \( f(x) \) from 0 to 5:
\[ P(X < 5) = \int_{-\infty}^{5} f(x) \, dx = \int_{0}^{5} 3e^{-3x} \, dx \]
Evaluate the integral:
\[ P(X < 5) = \left[-e^{-3x}\right]_0^5 = -(e^{-3 \cdot 5} - e^{-3 \cdot 0}) = -(e^{-15} - 1) =
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