X 0 1 2 3 F(X) 0.5 0.6 0.8 1.0
Finding the Mean of a Discrete Random Variable
The mean, or expected value, of a discrete random variable $X$, denoted as $E[X]$, is calculated by summing the product of each possible value of $X$ and its corresponding probability $P(X=x)$. The formula is:
$E[X] = \sum_{i} x_i P(X=x_i)$
Deriving Probability Mass Function (PMF) from Cumulative Distribution Function (CDF)
The cumulative probability $F(X)$ (or CDF) gives $P(X \le x)$. We need the probability mass function $P(X=x)$ to calculate the mean. We can find $P(X=x)$ using the CDF values:
- $P(X=0) = F(0) = 0.5$
- $P(X=1) = F(1) - F(0) = 0.6 - 0.5 = 0.1$
- $P(X=2) = F(2) - F(1) = 0.8 - 0.6 = 0.2$
- $P(X=3) = F(3) - F(2) = 1.0 - 0.8 = 0.2$
Let's summarize the PMF in a table:
| X | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| P(X=x) | 0.5 | 0.1 | 0.2 | 0.2 |
Calculating the Mean Value
Now, we apply the formula for the expected value using the derived PMF:
$E[X] = (0 \times 0.5) + (1 \times 0.1) + (2 \times 0.2) + (3 \times 0.2)$
$E[X] = 0 + 0.1 + 0.4 + 0.6$
$E[X] = 1.1$
The mean value of X is 1.1.
Important Questions from Random Variables
Consider the following probability density function for a random variable x.
(a) \(f_{1}(x)=1\ ;\ -1\le x\le 1\)
(b) \(f_{2}(x)=x\ ;\ 0\le x\le 1\)
(c) \(f_{3}(x)=(1-x)\ ;\ -1\le x\le 1\)
Arrange the above functions in terms of the increasing value of mean of random variable x.
Considering all the symbols with their usual meanings, match the following :
List - I List - II (a) \(f_{X}(x)\) (i) \(\displaystyle\int_{-\infty}^{\infty}f_{X,Y}(x,y)\,dx\) (b) \(\displaystyle\int_{-\infty}^{\infty}f_{X}(x)\,dx\) (ii) \(\displaystyle\int_{-\infty}^{a}f_{X}(x)\,dx\) (c) \(F_{X}(a)\) (iii) \(\displaystyle\int_{-\infty}^{\infty}f_{X,Y}(x,y)\,dy\) (d) \(f_{Y}(y)\) (iv) 1 Codes :
The density function of a random variable is given by \(p(x)=Ke^{-\frac{x^{2}}{2}}\) for \(-\infty \lt x \lt \infty\). The value of K should be :
If the odds in favour of any random event A are 5 ∶ 6, then the odds against the event are:
If random variable X follows binomial distribution with parameter n and p with mean 15 and variance 10, then the value of mode is
