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
Understanding Discrete Random Variables and Variance
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:
- $E(X)$ is the expected value (mean) of X.
- $E(X^2)$ is the expected value of $X^2$.
Calculating the Expected Value E(X)
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.
Calculating the Expected Value E(X<sup>2</sup>)
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:
- $(-2)^2 = 4$
- $(-1)^2 = 1$
- $(0)^2 = 0$
- $(1)^2 = 1$
- $(2)^2 = 4$
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.
Calculating the Variance of X
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.
Summary of Variance Calculation
To find the variance of a discrete random variable from its probability distribution:
- Calculate the mean $E(X) = \sum x_i P(X=x_i)$.
- Calculate the expected value of the square $E(X^2) = \sum x_i^2 P(X=x_i)$.
- Use the formula $\text{Var}(X) = E(X^2) - [E(X)]^2$.
Applying these steps to the given problem yielded a variance of 1.89.
Revision Table: Key Statistical Concepts
| 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. | $\sigma = \sqrt{\text{Var}(X)}$ |
Additional Information: Properties of Variance
- Variance is always non-negative: $\text{Var}(X) \ge 0$.
- The variance of a constant is zero: $\text{Var}(c) = 0$.
- For a constant 'a' and random variable X: $\text{Var}(aX) = a^2 \text{Var}(X)$.
- For constants 'a' and 'b' and random variable X: $\text{Var}(aX + b) = a^2 \text{Var}(X)$. Adding a constant 'b' shifts the distribution but doesn't change its spread.
Important Questions from Determinants
An even number is the determinant of which of the following matrices?
(A) \(\begin{bmatrix} 1 & -1 \\ -1 & 5 \end{bmatrix}\)
(B) \(\begin{bmatrix} 13 & -1 \\ -1 & 15 \end{bmatrix}\)
(C) \(\begin{bmatrix} 16 & -1 \\ -11 & 15 \end{bmatrix}\)
(D) \(\begin{bmatrix} 6 & -12 \\ 11 & 15 \end{bmatrix}\)
Choose the correct answer from the options given below:
Two pipes A and B together can fill a tank in 40 minutes. Pipe A is twice as fast as pipe B. Pipe A alone can fill the tank in :
There are 6 cards numbered 1 to 6, one number on one card. Two cards are drawn at random without replacement.
Let X denote the sum of the numbers on the two cards drawn.
Then P(X > 3) is:
The angle between two lines whose direction ratios are proportional to \( (\sqrt{3} - 1) \), \( (-\sqrt{3} - 1) \), and -4 is:
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:
