X 0 1 2 3 F(X) 0.5 0.6 0.8 1.0
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:
$E[X] = \sum_{i} x_i P(X=x_i)$
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:
Let's summarize the PMF in a table:
| X | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| P(X=x) | 0.5 | 0.1 | 0.2 | 0.2 |
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.
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
Let $X$ and $Y$ be continuous random variables with probability density functions $P_X(x)$ and $P_Y(y)$, respectively. Further, let $Y = X^2$ and $P_X(x) = \begin{cases} 1, & x\in (0,1] \\ 0, & \text{otherwise} \end{cases}$
Which one of the following options is correct?