Consider a distribution with the following probability density function $$f(x) = \begin{cases} 0.5, & 0 < x < 2 \\ 0.0, & Otherwise \end{cases}$$ Given that the mean of the above probability distribution is 1, the variance (rounded off to two decimal places) is _______________.
Calculating Variance for Uniform Distribution
The problem provides a probability density function (PDF) for a continuous random variable $X$:
$f(x) = \begin{cases} 0.5, & 0 < x < 2 \\ 0.0, & \text{Otherwise} \end{cases}$
This describes a uniform distribution on the interval $(0, 2)$. We are given that the mean, $E[X] = 1$, and we need to find the variance, $\text{Var}(X)$.
Variance Formula
The variance is calculated using the formula:
$ \text{Var}(X) = E[X^2] - (E[X])^2 $
Calculating Expected Value of $X^2$
First, we calculate $E[X^2]$:
$ E[X^2] = \int_{-\infty}^{\infty} x^2 f(x) dx $
Substitute the given PDF:
$ E[X^2] = \int_{0}^{2} x^2 (0.5) dx $
$ E[X^2] = 0.5 \int_{0}^{2} x^2 dx $
$ E[X^2] = 0.5 \left[ \frac{x^3}{3} \right]_{0}^{2} $
$ E[X^2] = 0.5 \left( \frac{2^3}{3} - \frac{0^3}{3} \right) $
$ E[X^2] = 0.5 \left( \frac{8}{3} \right) = \frac{4}{3} $
Calculating Variance
Now, use the variance formula with $E[X^2] = \frac{4}{3}$ and the given $E[X] = 1$:
$ \text{Var}(X) = E[X^2] - (E[X])^2 $
$ \text{Var}(X) = \frac{4}{3} - (1)^2 $
$ \text{Var}(X) = \frac{4}{3} - 1 $
$ \text{Var}(X) = \frac{1}{3} $
Final Result
Convert the fraction to a decimal:
$ \text{Var}(X) = \frac{1}{3} \approx 0.3333... $
Rounding to two decimal places, the variance is 0.33.
This value falls between 0.32 and 0.34.
Important Questions from Variance
- Let $X$ and $Y$ be two independent random variables. $X$ follows $Bernoulli(p = 0.3)$ distribution and $Y$ follows $Normal(\mu = 0, \sigma^2 = 100)$ distribution.
Which of the following options is the variance of $(2X - 1)Y$? - For a given data set $\{x_1, x_2, \ldots, x_n\}$, where $n = 100$, it is known that
$$ \frac{1}{2000} \sum_{i=1}^{n} \sum_{j=1}^{n} (x_i - x_j)^2 = 99 $$
Let us denote $\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i$.
The value of $\frac{1}{99} \sum_{i=1}^{n} (x_i - \bar{x})^2$ is __________ . (Answer in integer) - A random variable $X$ has the sample space $\{0,1\}$. The probability $P(X = 0) = 1/4$ and $P(X = 1) = 3/4$.
What is the variance of the random variable?
Hint: $\text{Mean } (\mu) = \sum_{i=1}^{n} x_i p(x_i) ; \text{Variance } (\sigma^2) = \sum_{i=1}^{n} (x_i - \mu)^2 p(x_i)$ The unbiased sample variance for the set of numbers: $S = \{40,45,50,55,60\}$ is_____. (write answer with one decimal place)
A set of 11 (x, y) data points is least-squares fitted to a quadratic polynomial. If the sum of squares of error is 2.4, the variance of error is ________ (round off to 1decimal place).
