Find the mean variance and standard deviation  for the following
f(x)= x.(bx to the power -1). the interval is (0,2)


To find the **mean**, **variance**, and **standard deviation** for the given probability density function \( f(x) = x \cdot b x^{-1} \) on the interval \( (0, 2) \), we need to first analyze the function and apply the standard definitions of these statistical measures. Let's proceed step by step.

### Step 1: Set up the Probability Density Function
The given function is:

\[
f(x) = x \cdot b x^{-1} = b
\]

This implies that \( f(x) = b \), a constant. The constant probability density function over the interval \( (0, 2) \) means that we have a uniform distribution. The value of \( b \) can be determined using the fact that the total probability over the interval must equal 1. So:

\[
\int_0^2 f(x) \, dx = 1
\]

Substituting \( f(x) = b \) into this equation:

\[
\int_0^2 b \, dx = 1
\]

\[
b \cdot (2 - 0) = 1
\]

\[
b = \frac{1}{2}
\]

Thus, \( f(x) = \frac{1}{2} \) over the interval \( (0, 2) \).

### Step 2: Calculate the Mean
The **mean** \( \mu \) of a continuous random variable is given by:

\[
\mu = \int_0^2 x \cdot f(x) \, dx
\]

Substitute \( f(x) = \frac{1}{2} \):

\[
\mu = \int_0^2 x \cdot \frac{1}{2} \, dx
\]

\[
\mu = \frac{1}{2} \int_0^2 x \, dx
\]

Now, calculate the integral:

\[
\int_0^2 x \, dx = \left[ \frac{x^2}{2} \right]_0^2 = \frac{2^2}{2} - \frac{0^2}{2} = 2
\]

Thus, the mean is:

\[
\mu = \frac{1}{2} \times 2 = 1
\]

### Step 3: Calculate the Variance
The **variance** \( \sigma^2 \) is given by:

\[
\sigma^2 = \int_0^2 (x - \mu)^2 \cdot f(x) \, dx
\]

Substituting \( f(x) = \frac{1}{2} \) and \( \mu = 1 \):

\[
\sigma^2 = \frac{1}{2} \int_0^2 (x - 1)^2 \, dx
\]

First, expand \( (x - 1)^2 \):

\[
(x - 1)^2 = x^2 - 2x + 1
\]

Now, integrate term by term:

\[
\int_0^2 (x^2 - 2x + 1) \, dx = \int_0^2 x^2 \, dx - 2 \int_0^2 x \, dx + \int_0^2 1 \, dx
\]

Each of these integrals is straightforward:

\[
\int_0^2 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^2 = \frac{8}{3}
\]
\[
\int_0^2 x \, dx = 2
\]
\[
\int_0^2 1 \, dx = 2
\]

Substituting these results:

\[
\sigma^2 = \frac{1}{2} \left( \frac{8}{3} - 2 \times 2 + 2 \right)
\]

\[
\sigma^2 = \frac{1}{2} \left( \frac{8}{3} - 4 + 2 \right) = \frac{1}{2} \left( \frac{8}{3} - 2 \right)
\]

\[
\sigma^2 = \frac{1}{2} \left( \frac{8 - 6}{3} \right) = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}
\]

### Step 4: Calculate the Standard Deviation
The **standard deviation** \( \sigma \) is simply the square root of the variance:

\[
\sigma = \sqrt{\sigma^2} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} \approx 0.577
\]

### Final Results:
- **Mean** \( \mu = 1 \)
- **Variance** \( \sigma^2 = \frac{1}{3} \)
- **Standard Deviation** \( \sigma = \frac{1}{\sqrt{3}} \approx 0.577 \)

Would you like any additional details or explanations?

Here are some related questions you might find useful:
1. How do you find the mean for a non-uniform distribution?
2. What does the variance of a distribution tell you about its spread?
3. Can variance ever be negative? Why or why not?
4. How do the mean and median compare in a uniform distribution?
5. What is the relationship between variance and standard deviation?

**Tip:** When calculating variance, always remember that it represents the average of the squared deviations from the mean.

Find the mean, variance, and standard deviation for the following: f(x) = x * bx^(-1). The interval is (0, 2).

This page provides a detailed solution to finding the mean, variance, and standard deviation for the probability density function f(x) = x * bx^(-1) on the interval (0, 2). Learn how to calculate these statistical measures through integration, including the step-by-step process for determining the uniform distribution constant. Suitable for advanced high school or early college students.

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