Let X be a continuous random variable denoting the temperature measured. The range of temperature is [0, 100] degree Celsius and let the probability density function of X be f(x) = 0.01 for 0 ≤ X ≤ 100. The mean of X is ______
50.0
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:
- Factor out the constant $0.01$:
$$ E(X) = 0.01 \int_{0}^{100} x \,dx $$
- Integrate $x$ with respect to $x$. The integral of $x$ is $\frac{x^2}{2}$:
$$ E(X) = 0.01 \left[ \frac{x^2}{2} \right]_{0}^{100} $$
- Evaluate the definite integral using the limits $0$ and $100$:
$$ E(X) = 0.01 \left( \frac{(100)^2}{2} - \frac{(0)^2}{2} \right) $$
- Calculate the values:
$$ E(X) = 0.01 \left( \frac{10000}{2} - 0 \right) $$
$$ E(X) = 0.01 (5000) $$
- 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.
Important Questions from Continuous Distributions
Suppose X is a continuous random variable with probability density function
\(f(x)=\frac{1}{\pi} \frac{1}{1+(x+1)^2}\), -∞ < x < ∞.
Define
\(Y=\left\{\begin{array}{cc} \frac{X}{|X|}, & \text { if } X \neq 0 \\ 0, & \text { if } X=0 \end{array}\right.\)
Then which of the following statements are true?
Let X1, X2, ..., Xn be a random sample from an absolutely continuous distribution with the probability density function
\(f(x \mid \theta)=\left\{\begin{array}{cl} e^{\theta-x}, & \text { if } x \geq \theta \\ 0, & \text { if } x<\theta \end{array},\right.\)
where θ ∈ ℝ is unknown. Define \(\bar{X}=\frac{1}{n} \sum_{i=1}^n X_i\) and X(1) = min{X1, ..., Xn}. Then
which of the following statements are true?
Suppose that X is a continuous random variable with probability density function given by:
f(x) = \(\left\{ {\begin{array}{c} {\frac{x}{8},}&{x \in \left[ {0,2} \right)}\\ {\frac{1}{4},}&{x \in \left[ {2,4} \right)}\\ { - \frac{x}{8} + \frac{3}{4},}&{x \in \left[ {4,6} \right)} \end{array}}\right.\)
Find the mean of X.
The variable x takes a value between 0 and 10 with uniform probability distribution. The variable y takes a value between 0 and 20 with uniform probability distribution. The probability of the sum of variables (x + y) being greater than 20 is _________
Probability density function of a random variable X is given below
\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {0.25}&{if\;1 \le x \le 5}\\ 0&{otherwise} \end{array}} \right.\)
P (X ≤ 4) is
