A target is located at the point 0 on a horizontal axis. Let x be the landing point of a shot aimed at the target, a continuous variable with density function f(x) = 0.75(1 - x^2) for -1≤ x≤ 1.
What is the mean value of x?

The mean value (expected value) of a continuous random variable \( X \) with probability density function (PDF) \( f(x) \) is given by:

\[
E[X] = \int_{-\infty}^{\infty} x f(x) \, dx
\]

In this case, the random variable \( X \) has the PDF:

\[
f(x) = 0.75(1 - x^2) \quad \text{for } -1 \leq x \leq 1
\]

To find the mean, we need to evaluate:

\[
E[X] = \int_{-1}^{1} x \cdot 0.75(1 - x^2) \, dx
\]

Let's compute this integral step-by-step.

1. **Set up the integral:**

\[
E[X] = 0.75 \int_{-1}^{1} x (1 - x^2) \, dx
\]

2. **Expand the integrand:**

\[
E[X] = 0.75 \int_{-1}^{1} (x - x^3) \, dx
\]

3. **Split the integral:**

\[
E[X] = 0.75 \left( \int_{-1}^{1} x \, dx - \int_{-1}^{1} x^3 \, dx \right)
\]

4. **Evaluate each integral separately:**

   - For \(\int_{-1}^{1} x \, dx\):
   
   \[
   \int_{-1}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{-1}^{1} = \frac{1^2}{2} - \frac{(-1)^2}{2} = \frac{1}{2} - \frac{1}{2} = 0
   \]
   
   - For \(\int_{-1}^{1} x^3 \, dx\):
   
   \[
   \int_{-1}^{1} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{-1}^{1} = \frac{1^4}{4} - \frac{(-1)^4}{4} = \frac{1}{4} - \frac{1}{4} = 0
   \]

5. **Combine the results:**

\[
E[X] = 0.75 \left( 0 - 0 \right) = 0
\]

Thus, the mean value of \( x \) is:

\[
E[X] = 0
\]

Would you like more details or have any questions?

Here are 5 related questions for further exploration:
1. How would the variance of \( x \) be calculated for the given PDF?
2. What is the median value of \( x \) for this distribution?
3. How do you find the mode of the given density function?
4. If we change the interval to \([-2, 2]\), what would be the normalization constant for the PDF?
5. How does the mean value change if the PDF is modified to \( f(x) = c(1 - x^2) \) where \( c \) is a new constant?

**Tip:** The mean value for a symmetric distribution around 0 is often 0, as the positive and negative contributions cancel out.

A target is located at the point 0 on a horizontal axis. Let x be the landing point of a shot aimed at the target, a continuous variable with density function f(x) = 0.75(1 - x^2) for -1≤ x≤ 1. What is the mean value of x?

Find the mean value (expected value) of a continuous random variable with the PDF f(x) = 0.75(1 - x^2) for -1 ≤ x ≤ 1. This detailed solution walks through each step of the calculation, making it suitable for college-level students in probability and statistics.

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