A random variable X has the following probability distribution: The variance of X will be: X | -2 | -1 | 0 | 1 | 2
--------------------------------------------
P(X) | 0.2 | 0.1 | 0.3 | 0.2 | 0.2
1.89
A random variable is a variable whose value is a numerical outcome of a random phenomenon. A discrete random variable is one that can only take a finite or countable number of values.
The probability distribution of a discrete random variable lists all possible values the variable can take, along with their associated probabilities. For this problem, the probability distribution of variable X is given as:
| X | -2 | -1 | 0 | 1 | 2 |
|---|---|---|---|---|---|
| P(X) | 0.2 | 0.1 | 0.3 | 0.2 | 0.2 |
The variance of a discrete random variable X, denoted as Var(X) or $\sigma^2$, measures the spread or dispersion of the values of X around its expected value (mean). It is calculated using the formula:
$\text{Var}(X) = E(X^2) - [E(X)]^2$
Where:
The expected value $E(X)$ is the weighted average of the possible values of X, where the weights are the corresponding probabilities. The formula is:
$E(X) = \sum [x_i \cdot P(X=x_i)]$
Using the given data:
$E(X) = (-2)(0.2) + (-1)(0.1) + (0)(0.3) + (1)(0.2) + (2)(0.2)$
$E(X) = -0.4 - 0.1 + 0 + 0.2 + 0.4$
$E(X) = -0.5 + 0.6$
$E(X) = 0.1$
So, the expected value of X is 0.1.
Next, we need to calculate $E(X^2)$. This is the expected value of the square of the random variable. First, we square each possible value of X, and then multiply by its corresponding probability and sum the results:
$E(X^2) = \sum [x_i^2 \cdot P(X=x_i)]$
Let's calculate $x_i^2$ for each value of X:
Now, calculate $E(X^2)$:
$E(X^2) = (4)(0.2) + (1)(0.1) + (0)(0.3) + (1)(0.2) + (4)(0.2)$
$E(X^2) = 0.8 + 0.1 + 0 + 0.2 + 0.8$
$E(X^2) = 0.9 + 1.0$
$E(X^2) = 1.9$
So, the expected value of $X^2$ is 1.9.
Now we have both $E(X)$ and $E(X^2)$. We can use the variance formula:
$\text{Var}(X) = E(X^2) - [E(X)]^2$
Substitute the calculated values:
$\text{Var}(X) = 1.9 - (0.1)^2$
$\text{Var}(X) = 1.9 - 0.01$
$\text{Var}(X) = 1.89$
The variance of the random variable X is 1.89.
To find the variance of a discrete random variable from its probability distribution:
Applying these steps to the given problem yielded a variance of 1.89.
| Concept | Definition | Formula (Discrete RV) |
|---|---|---|
| Discrete Random Variable | A variable taking a countable number of distinct values. | X = $x_1, x_2, ..., x_n$ |
| Probability Distribution | List of possible values of X and their probabilities $P(X=x_i)$. Sum of $P(X=x_i)$ must equal 1. | $\sum P(X=x_i) = 1$ |
| Expected Value (Mean) | Average value of the random variable over many trials. | $E(X) = \sum x_i P(X=x_i)$ |
| Variance | Measure of the spread of the distribution. Average of the squared differences from the mean. | $\text{Var}(X) = E(X^2) - [E(X)]^2$ |
| Standard Deviation | Square root of the variance. Indicates typical deviation from the mean. | $\sigma = \sqrt{\text{Var}(X)}$ |
If A is a square matrix of order 4 and |A|= 4, then |2A| will be:
For a square matrix \( A_{n \times n} \):
(A) \( |\text{adj} A| = |A|^{n-1} \)
(B) \( |A| = |\text{adj} A|^{n-1} \)
(C) \( A (\text{adj} A) = |A| I \)
(D) \( |A^{-1}| = \frac{1}{|A|} \)
Choose the correct answer from the options given below:
Given the determinant:
\[ \Delta = \begin{vmatrix} 1 & \cos x & 1 \\ -\cos x & 1 & \cos x \\ -1 & -\cos x & 1 \end{vmatrix} \]
Which of the following statements are correct?
(A) \( \Delta = 2(1 - \cos^2 x) \)
(B) \( \Delta = 2(2 - \sin^2 x) \)
(C) Minimum value of \( \Delta \) is 2
(D) Maximum value of \( \Delta \) is 4
Choose the correct answer from the options given below:
The angle between two lines whose direction ratios are proportional to \( (\sqrt{3} - 1) \), \( (-\sqrt{3} - 1) \), and -4 is:
If \( B \) is a non-singular \( 4 \times 4 \) matrix and \( A \) is its adjoint such that \( |A| = 125 \), then \( |B| \) is: