The probability distribution for a discrete random variable X is given below. The expectation value of X is (up to one decimal place)______.X 1 2 3 4 P(X) 0.3 0.4 0.2 0.1
The expectation value, often denoted as E[X] or $\mu$, represents the weighted average of all possible values a discrete random variable can take. It's calculated by summing the product of each value and its corresponding probability.
The provided probability distribution for the discrete random variable X is:
| X | 1 | 2 | 3 | 4 |
| P(X) | 0.3 | 0.4 | 0.2 | 0.1 |
The formula for the expectation value of a discrete random variable X is:
$ E[X] = \sum_{i} x_i P(X=x_i) $
Using the given distribution, we calculate the expectation value:
$ E[X] = (1 \times 0.3) + (2 \times 0.4) + (3 \times 0.2) + (4 \times 0.1) $
$ E[X] = 0.3 + 0.8 + 0.6 + 0.4 $
$ E[X] = 2.1 $
The expectation value of X, calculated to one decimal place, is 2.1.
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?