The image shows the probability density function (PDF) of a random variable $X$ over the interval [0, 2], and the probability that $X > 1.78$ or $X < 0.04$ needs to be computed.

Step-by-Step Solution:

1. The PDF is constant from $X = 0$ to $X = 2$, forming a rectangular shape. This suggests that the distribution is uniform over this interval.

2. In a uniform distribution, the total area under the curve (which represents the entire probability) is 1. The height of the PDF, $f(x)$, is calculated by: $f(x) = \frac{1}{\text{length of the interval}} = \frac{1}{2 - 0} = \frac{1}{2}$ This means the height of the PDF is $\frac{1}{2}$.

3. The probability for any event is the area under the PDF over the corresponding interval. We need to calculate: $P(X > 1.78 \text{ or } X < 0.04)$

   This can be broken into two parts:

   • $P(X > 1.78)$ (the area from $X = 1.78$ to $X = 2$)
   • $P(X < 0.04)$ (the area from $X = 0$ to $X = 0.04$)

4. For $X > 1.78$: $P(X > 1.78) = \text{Height} \times \text{Width} = \frac{1}{2} \times (2 - 1.78) = \frac{1}{2} \times 0.22 = 0.11$

5. For $X < 0.04$: $P(X < 0.04) = \frac{1}{2} \times 0.04 = 0.02$

6. The total probability is the sum of these two probabilities: $P(X > 1.78 \text{ or } X < 0.04) = 0.11 + 0.02 = 0.13$

Thus, the probability that $X > 1.78$ or $X < 0.04$ is 0.13.

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Related Questions:

1. How do you calculate the probability for other ranges using this uniform distribution?
2. What is the cumulative distribution function (CDF) for this PDF?
3. How would the probability change if the interval was [0, 3] instead of [0, 2]?
4. How is the PDF normalized to ensure that the total probability is 1?
5. What is the expected value and variance of a uniformly distributed random variable on [0, 2]?

Tip:

When dealing with uniform distributions, the key is to recognize the constant height and to compute probabilities as simple areas under the curve over the interval of interest.