What is the variance of the random variable?
Hint: $\text{Mean } (\mu) = \sum_{i=1}^{n} x_i p(x_i) ; \text{Variance } (\sigma^2) = \sum_{i=1}^{n} (x_i - \mu)^2 p(x_i)$
Variance Calculation for Random Variable X
The problem asks for the variance of a random variable $X$ with a given sample space and probabilities.
Step 1: Calculate Mean ($\mu$)
The mean (or expected value) $\mu$ is calculated using the formula $\mu = \sum x_i p(x_i)$.
The random variable $X$ takes values $0$ and $1$ with probabilities $P(X=0) = \frac{1}{4}$ and $P(X=1) = \frac{3}{4}$.
$ \mu = (0 \times P(X = 0)) + (1 \times P(X = 1)) $
Substitute the probabilities:
$ \mu = (0 \times \frac{1}{4}) + (1 \times \frac{3}{4}) $
$ \mu = 0 + \frac{3}{4} = \frac{3}{4} $
Step 2: Calculate Variance ($\sigma^2$)
The variance $\sigma^2$ is calculated using the formula $\sigma^2 = \sum (x_i - \mu)^2 p(x_i)$.
$ \sigma^2 = (0 - \mu)^2 P(X=0) + (1 - \mu)^2 P(X=1) $
Substitute the values of $\mu$, $P(X=0)$, and $P(X=1)$:
$ \sigma^2 = (0 - \frac{3}{4})^2 \times \frac{1}{4} + (1 - \frac{3}{4})^2 \times \frac{3}{4} $
Calculate the terms:
$ \sigma^2 = (-\frac{3}{4})^2 \times \frac{1}{4} + (\frac{1}{4})^2 \times \frac{3}{4} $
$ \sigma^2 = (\frac{9}{16} \times \frac{1}{4}) + (\frac{1}{16} \times \frac{3}{4}) $
$ \sigma^2 = \frac{9}{64} + \frac{3}{64} $
$ \sigma^2 = \frac{12}{64} $
Step 3: Simplify Variance
Simplify the resulting fraction:
$ \sigma^2 = \frac{12}{64} = \frac{3 \times 4}{16 \times 4} = \frac{3}{16} $
The variance of the random variable $X$ is $\frac{3}{16}$.
Important Questions from Variance
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 _______________.
- 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) 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).
