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.