Assume that X is a continuous random variable with the probability density function (pdf) f(x) =
10e−10x ,0 <x<∞. Also recall in the class that E(g(X)) =∞ −∞g(x)f(x)dx.
(a) Verify that f(x) is indeed a pdf
(b) Find P(0 <X <10),P(X >5)
(c) Find µ= E[X]
(d) Find E[(X−µ)2] where µ is found from the previous question

Let's go through each part of this problem step-by-step:

### (a) Verify that \( f(x) \) is a PDF
To confirm that \( f(x) = 10e^{-10x} \) is a probability density function (PDF), two criteria must be met:

1. **Non-negativity**: \( f(x) \geq 0 \) for all \( x \) in the domain.
2. **Integration to 1**: The integral over the entire range of \( f(x) \) must equal 1:
   \[
   \int_{0}^{\infty} f(x) \, dx = 1.
   \]

Since \( f(x) = 10e^{-10x} \), we see that \( f(x) \geq 0 \) for \( x \in (0, \infty) \), which satisfies the non-negativity condition.

Now, let's check if the integral over the range of \( x \) equals 1:
\[
\int_{0}^{\infty} 10e^{-10x} \, dx.
\]
To solve this, we use the fact that \( \int e^{-ax} \, dx = \frac{1}{a} \) for a positive constant \( a \). Here, \( a = 10 \):
\[
\int_{0}^{\infty} 10e^{-10x} \, dx = 10 \cdot \frac{1}{10} = 1.
\]
Thus, \( f(x) \) is indeed a valid PDF.

### (b) Find \( P(0 < X < 10) \) and \( P(X > 5) \)
To find probabilities involving \( X \), we integrate the PDF over the relevant range.

#### For \( P(0 < X < 10) \):
\[
P(0 < X < 10) = \int_{0}^{10} 10e^{-10x} \, dx.
\]
Using the same integration technique as before:
\[
P(0 < X < 10) = \left[ -e^{-10x} \right]_{0}^{10} = -e^{-100} + e^{0} = 1 - e^{-100}.
\]
Since \( e^{-100} \) is extremely small, \( P(0 < X < 10) \approx 1 \).

#### For \( P(X > 5) \):
\[
P(X > 5) = \int_{5}^{\infty} 10e^{-10x} \, dx = \left[ -e^{-10x} \right]_{5}^{\infty} = 0 + e^{-50} = e^{-50}.
\]
So, \( P(X > 5) = e^{-50} \), which is also very small.

### (c) Find \( \mu = E[X] \)
The expected value \( \mu = E[X] \) is given by
\[
E[X] = \int_{0}^{\infty} x f(x) \, dx = \int_{0}^{\infty} x \cdot 10e^{-10x} \, dx.
\]
To solve this integral, we use integration by parts. Let \( u = x \) and \( dv = 10e^{-10x} \, dx \). Then \( du = dx \) and \( v = -e^{-10x} \), so
\[
E[X] = \left[ -x e^{-10x} \right]_{0}^{\infty} + \int_{0}^{\infty} e^{-10x} \, dx.
\]
Evaluating the first term at \( 0 \) and \( \infty \), and solving the remaining integral, we find
\[
E[X] = \frac{1}{10} = 0.1.
\]

### (d) Find \( E[(X - \mu)^2] \)
To find the variance, \( \sigma^2 = E[(X - \mu)^2] \), we use the formula
\[
\text{Var}(X) = E[X^2] - (E[X])^2.
\]
First, calculate \( E[X^2] \):
\[
E[X^2] = \int_{0}^{\infty} x^2 \cdot 10e^{-10x} \, dx.
\]
Using integration techniques, we can solve this integral to obtain

Assume that X is a continuous random variable with the probability density function (pdf) f(x) = 10e^(-10x), 0 < x < ∞. Also recall in the class that E(g(X)) = ∫g(x)f(x)dx.
(a) Verify that f(x) is indeed a pdf
(b) Find P(0 < X < 10), P(X > 5)
(c) Find µ = E[X]
(d) Find E[(X − µ)^2] where µ is found from the previous question

Explore a continuous random variable X with pdf f(x) = 10e^(-10x) over (0, ∞). This guide verifies the PDF, calculates probabilities P(0 < X < 10) and P(X > 5), finds the expectation E[X], and calculates the variance E[(X − µ)^2]. Suitable for undergraduate students studying probability.

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