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The problem involves a uniform distribution between \( X = 0 \) and \( X = 2 \), and you're asked to find \( P(X \geq 1.82) \).

### Step-by-Step Solution:

1. **Understanding Uniform Distribution**: 
   - A uniform distribution means the probability is evenly distributed across the interval. The probability density function (PDF) of a uniform distribution between \( a \) and \( b \) is:
     \[
     f(x) = \frac{1}{b - a}
     \]
     Here, \( a = 0 \) and \( b = 2 \), so:
     \[
     f(x) = \frac{1}{2 - 0} = \frac{1}{2}
     \]

2. **Calculate the Required Probability**:
   - You're asked to find \( P(X \geq 1.82) \). This is the area under the PDF curve from \( X = 1.82 \) to \( X = 2 \), which is a simple proportion of the total interval.
   - The length of the interval from 1.82 to 2 is:
     \[
     2 - 1.82 = 0.18
     \]
   - The probability of being in this interval is:
     \[
     P(X \geq 1.82) = \text{length of interval} \times \text{height of PDF} = 0.18 \times \frac{1}{2} = 0.09
     \]

Thus, \( P(X \geq 1.82) = 0.09 \).

Would you like further details on this? 

Here are 5 related questions for further practice:

1. What is \( P(X \geq 1.5) \) for the same uniform distribution?
2. How would the probability change if the interval were \( [0, 3] \) instead of \( [0, 2] \)?
3. What is the probability that \( X \) is between 0.5 and 1.5?
4. How do you calculate the expected value of a uniform distribution?
5. What is the variance of \( X \) in a uniform distribution between \( a \) and \( b \)?

**Tip**: The area under the curve in a uniform distribution always represents probability!

Calculate the probability P(X ≥ 1.82) for a uniform distribution between 0 and 2. This solution involves understanding probability concepts and applying the uniform distribution formula. Suitable for students in Grades 10-12.

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