Let X~U[0,9.2], what is E(X)? Give answer to 2 decimal places.

The expected value \( E(X) \) of a uniform distribution \( X \sim U[a, b] \) is given by the formula:

\[
E(X) = \frac{a + b}{2}
\]

For the given distribution \( X \sim U[0, 9.2] \):

\[
E(X) = \frac{0 + 9.2}{2} = \frac{9.2}{2} = 4.6
\]

Thus, \( E(X) = 4.60 \) (rounded to 2 decimal places).

Would you like more details or have any questions? Here are 5 related questions:

1. What is the variance \( \text{Var}(X) \) for \( X \sim U[0, 9.2] \)?
2. How do you calculate the probability that \( X \) is greater than a certain value?
3. What is the cumulative distribution function (CDF) for \( X \sim U[0, 9.2] \)?
4. How would \( E(X) \) change if the interval was \( U[a, b] \) instead?
5. What is the median of \( X \) in the uniform distribution \( U[0, 9.2] \)?

**Tip:** The expected value for any uniform distribution \( U[a, b] \) is always the midpoint of the interval.

Learn how to calculate the expected value \( E(X) \) for a uniform distribution \( X \sim U[0, 9.2] \). Understand the formula and steps involved in determining the expected value. Suitable for advanced high school students and anyone interested in probability theory.

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