What is the expected value (mean) of X?
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Q: How do u find expected value
A: How do u find expected value
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![**Probability Distribution and Expected Value**
Consider a random variable with the following probability distribution:
| X | P(X) |
|---|------|
| 0 | 0.25 |
| 1 | 0.5 |
| 2 | 0.25 |
**Question: What is the expected value (mean) of X?**
Choices:
- O 1.0
- O 2.0
- O 0
- O Can't be determined
To determine the expected value (mean) of a discrete random variable X, we use the following formula:
\[ E(X) = \sum [X \cdot P(X)] \]
Let's calculate:
\[ E(X) = (0 \cdot 0.25) + (1 \cdot 0.5) + (2 \cdot 0.25) \]
\[ E(X) = 0 + 0.5 + 0.5 \]
\[ E(X) = 1.0 \]
Thus, the expected value (mean) of X is 1.0. The correct answer is:
- O 1.0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbc42dbd0-41a9-480b-bad1-d3b723150fc3%2F02709f03-8e5d-4acc-a308-7dbe465e1a84%2Fyj2d4cj_processed.jpeg&w=3840&q=75)
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- let X be a random variable and X-B(9,0.49) find the mean (expected value) of the random variable XThe Weibull distribution is widely used in statistical problems relating to aging of solid insulating materials subjected to aging and stress. Use this distribution as a model for time (in hours) to failure of solid insulating specimens subjected to AC voltage. The values of the parameters depend on the voltage and temperature; suppose a = 2.4 and ß = 220. (a) What is the probability that a specimen's lifetime is at most 250? Less than 250? More than 300? (Round your answers to five decimal places.) at most 250 less than 250 more than 300 0.74310 0.74310 0.12183 (b) What is the probability that a specimen's lifetime is between 100 and 250? (Round your answer to four decimal places.) 6032 (c) What value (in hr) is such that exactly 50% of all specimens have lifetimes exceeding that value? (Round your answer to three decimal places.) hr
