Find the expected value for the random variable. 2. 3. 4 P(x) 0.1 0.2 0.2 0.5
Continuous Probability Distributions
Probability distributions are of two types, which are continuous probability distributions and discrete probability distributions. A continuous probability distribution contains an infinite number of values. For example, if time is infinite: you could count from 0 to a trillion seconds, billion seconds, so on indefinitely. A discrete probability distribution consists of only a countable set of possible values.
Normal Distribution
Suppose we had to design a bathroom weighing scale, how would we decide what should be the range of the weighing machine? Would we take the highest recorded human weight in history and use that as the upper limit for our weighing scale? This may not be a great idea as the sensitivity of the scale would get reduced if the range is too large. At the same time, if we keep the upper limit too low, it may not be usable for a large percentage of the population!
![### Finding the Expected Value for a Random Variable
To calculate the expected value (mean) of a given random variable, you utilize the values of the random variable along with their corresponding probabilities. The expected value \(E(X)\) is calculated as follows:
\[ E(X) = \sum [x \cdot P(x)] \]
where \(x\) are the values of the random variable and \(P(x)\) are the probabilities associated with these values.
**Given Data:**
| \(x\) | 2 | 3 | 4 | 5 |
|-------|-----|-----|-----|-----|
| \(P(x)\) | 0.1 | 0.2 | 0.2 | 0.5 |
**Calculation:**
- For \(x = 2\), \(P(x) = 0.1\), so \(2 \times 0.1 = 0.2\)
- For \(x = 3\), \(P(x) = 0.2\), so \(3 \times 0.2 = 0.6\)
- For \(x = 4\), \(P(x) = 0.2\), so \(4 \times 0.2 = 0.8\)
- For \(x = 5\), \(P(x) = 0.5\), so \(5 \times 0.5 = 2.5\)
**Total Expected Value:**
\[ E(X) = 0.2 + 0.6 + 0.8 + 2.5 = 4.1 \]
**Answer Options:**
- 3.8
- 4.1
- 4.5
- 3.1
The correct answer is **4.1**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7c5fa014-da02-4c57-a781-fd58c7d971af%2F424e725a-23ba-4cbf-b9e9-fa5e133f0969%2Fgv7yry_processed.jpeg&w=3840&q=75)
In this case, we need to find the expected value.
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