Suppose a random variable X has the probability density function f(x) = 1.5 x² for -1
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![### Calculating the Standard Deviation of a Random Variable
**Problem Statement:**
Suppose a random variable \(X\) has the probability density function \( f(x) = 1.5x^2 \) for \( -1 < x < 1 \), and \( f(x) = 0 \) elsewhere. Obtain the standard deviation of \(X\). Round your answer to 2 decimal places. **SHOW WORK**.
**Solution Steps:**
1. **Define the Variance Formula:**
The variance \( \sigma^2 \) of a random variable \( X \) is given by:
\[
\sigma^2 = \int_{-1}^{1} x^2 f(x) \, dx - \mu^2
\]
where \( \mu \) is the mean of the random variable \(X\), defined as:
\[
\mu = \int_{-1}^{1} x f(x) \, dx
\]
2. **Calculate the Mean ( \( \mu \) ):**
\[
\mu = \int_{-1}^{1} x (1.5x^2) \, dx
\]
This simplifies to:
\[
\mu = 1.5 \int_{-1}^{1} x^3 \, dx
\]
Evaluating this integral, we get:
\[
\mu = 1.5 \left[ \frac{x^4}{4} \right]_{-1}^{1} = 1.5 \left( \frac{1^4}{4} - \frac{(-1)^4}{4} \right)
\]
Since \(\frac{1^4}{4} - \frac{1^4}{4} = 0\), it follows that:
\[
\mu = 1.5 \times 0 = 0
\]
3. **Calculate \( \sigma^2 \):**
\[
\sigma^2 = \int_{-1}^{1} x^2 (1.5 x^2) \, dx - \mu^2 = 1.5 \int_{-1}^{1} x^4 \, dx
\]
Evaluating this integral, we get:
\[](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5bad4e48-dad8-4710-a64b-b24b80d1efcf%2Fb642b57f-00a4-44bf-924d-828ee035f30d%2Fa9ykxd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Calculating the Standard Deviation of a Random Variable
**Problem Statement:**
Suppose a random variable \(X\) has the probability density function \( f(x) = 1.5x^2 \) for \( -1 < x < 1 \), and \( f(x) = 0 \) elsewhere. Obtain the standard deviation of \(X\). Round your answer to 2 decimal places. **SHOW WORK**.
**Solution Steps:**
1. **Define the Variance Formula:**
The variance \( \sigma^2 \) of a random variable \( X \) is given by:
\[
\sigma^2 = \int_{-1}^{1} x^2 f(x) \, dx - \mu^2
\]
where \( \mu \) is the mean of the random variable \(X\), defined as:
\[
\mu = \int_{-1}^{1} x f(x) \, dx
\]
2. **Calculate the Mean ( \( \mu \) ):**
\[
\mu = \int_{-1}^{1} x (1.5x^2) \, dx
\]
This simplifies to:
\[
\mu = 1.5 \int_{-1}^{1} x^3 \, dx
\]
Evaluating this integral, we get:
\[
\mu = 1.5 \left[ \frac{x^4}{4} \right]_{-1}^{1} = 1.5 \left( \frac{1^4}{4} - \frac{(-1)^4}{4} \right)
\]
Since \(\frac{1^4}{4} - \frac{1^4}{4} = 0\), it follows that:
\[
\mu = 1.5 \times 0 = 0
\]
3. **Calculate \( \sigma^2 \):**
\[
\sigma^2 = \int_{-1}^{1} x^2 (1.5 x^2) \, dx - \mu^2 = 1.5 \int_{-1}^{1} x^4 \, dx
\]
Evaluating this integral, we get:
\[
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