Question

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 \(0 \le X \le 100\). The mean is

• A. 5.0
• B. (2)5
• C. 25.0
• D. 50.0

Hint:

Mean of Continuous Distribution. \(E[X] = \int x f(x) dx\). For a Uniform distribution on [a, b], the mean is the midpoint \((a+b)/2\).

Answer 1

The Correct Option is D

The probability density function (pdf) is given as \(f(x) = 0.
01\) for \(0 \le x \le 100\), and \(f(x) = 0\) otherwise.
This is a uniform distribution over the interval [0, 100].
The mean (or expected value) \(E[X]\) of a continuous random variable X with pdf \(f(x)\) is calculated as: $$ E[X] = \int_{-\infty}^{\infty} x f(x) dx $$ For this uniform distribution: $$ E[X] = \int_{0}^{100} x (0.
01) dx $$ $$ E[X] = 0.
01 \int_{0}^{100} x dx $$ $$ E[X] = 0.
01 \left[ \frac{x^2}{2} \right]_{0}^{100} $$ $$ E[X] = 0.
01 \left( \frac{100^2}{2} - \frac{0^2}{2} \right) $$ $$ E[X] = 0.
01 \left( \frac{10000}{2} \right) = 0.
01 \times 5000 = 50 $$ The mean is 50.
0.
Alternatively, for a uniform distribution over the interval [a, b], the mean is simply the midpoint of the interval: \((a+b)/2\).
Here, \(a=0\) and \(b=100\), so the mean is \((0+100)/2 = 50\).