This question asks us to find the mean of a continuous random variable, denoted as X, which represents temperature. We are given the range of temperature as [0, 100] degrees Celsius and its probability density function (PDF), $f(x) = 0.01$, for $0 \le X \le 100$. This type of PDF, where the probability is constant over a specific interval, represents a uniform distribution.

Mean Calculation Formula for Continuous Variables

The mean, or expected value, $E(X)$ of a continuous random variable X with a probability density function $f(x)$ over the interval $[a, b]$ is calculated using the integral:

$$ E(X) = \int_{a}^{b} x \cdot f(x) \,dx $$

Applying the PDF and Range

In this specific problem:

• The lower limit of the range is $a = 0$.
• The upper limit of the range is $b = 100$.
• The probability density function is $f(x) = 0.01$.

Substituting these values into the formula, we get:

$$ E(X) = \int_{0}^{100} x \cdot (0.01) \,dx $$

Step-by-Step Integration

To find the mean, we perform the integration:

1. Factor out the constant $0.01$:

   $$ E(X) = 0.01 \int_{0}^{100} x \,dx $$

2. Integrate $x$ with respect to $x$. The integral of $x$ is $\frac{x^2}{2}$:

   $$ E(X) = 0.01 \left[ \frac{x^2}{2} \right]_{0}^{100} $$

3. Evaluate the definite integral using the limits $0$ and $100$:

   $$ E(X) = 0.01 \left( \frac{(100)^2}{2} - \frac{(0)^2}{2} \right) $$

4. Calculate the values:

   $$ E(X) = 0.01 \left( \frac{10000}{2} - 0 \right) $$

   $$ E(X) = 0.01 (5000) $$

5. Calculate the final result:

   $$ E(X) = 50.0 $$

Verification of PDF

Before concluding, it's good practice to ensure the PDF is valid by checking if its integral over the entire range equals 1:

$$ \int_{0}^{100} f(x) \,dx = \int_{0}^{100} 0.01 \,dx $$

$$ = 0.01 [x]_{0}^{100} = 0.01 (100 - 0) = 0.01 \times 100 = 1 $$

Since the integral equals 1, the PDF is valid.

Final Mean Value

The calculated mean of the continuous random variable X, representing the temperature, is 50.0 degrees Celsius.