For a discrete probability distribution, you add up x*P(x) over all possible values of x, where P(x) is the probability that the random variable X takes the value x.
For a continuous distribution you need to integrate x*P(x) with respect to x.
No. The mean is the expected value of the random variable but you can also have expected values of functions of the random variable. If you define X as the random variable representing the result of a single throw of a fair die, the expected value of X is 3.5, the mean of the probability distribution of X. However, you play a game where you pay someone a certain amount of money for each throw of the die and the other person pays you your "winnings" which depend on the outcome of the throw. The variable, "your winnings", will also have an expected value. As will your opponent's winnings.
The formula to find the value of X would be Y-2X. This would equal to y-9 times 2 X.
No. The expected value is the mean!
The expected value is the average of a probability distribution. It is the value that can be expected to occur on the average, in the long run.
x is a variable. You do not have to know what its value is. In that equation you have to find the value of x, so if you already know what x stands for then you know the answer
Follow these steps:Find all the values that the random variable (RV) can take, x.For each x, find the probability that the RV takes than value, p(x).Multiply them: x*p(x).Sum these over all possible values of x.The above sum is the expected value of the RV, X.
The expected value of a Martingale system is the last observed value.
If X takes the value 1 with probability p and 0 with probability (1-p), and there are n independent trials then E(X) = np
If a random variable X is distributed normally with probability distribution function p(x), then the expected value of X is E(X) = integral of x*p(x)dx evaluated over the whole of the real line.
The deviation is the observed value less the expected value.
The formula for expected value under certainty is simply the value of the certain outcome itself, as there is no variability involved. Mathematically, it can be expressed as ( EV = x ), where ( x ) is the guaranteed outcome. In scenarios involving multiple outcomes with probabilities, the expected value is calculated as ( EV = \sum (p_i \cdot x_i) ), but under certainty, ( p_i ) for the certain outcome is 1, rendering the expected value equal to that outcome.
The expected value of a random variable ( x ) is a measure of the central tendency and is calculated as the weighted average of all possible values, where each value is weighted by its probability of occurrence. Mathematically, it is expressed as ( E(x) = \sum (x_i \cdot P(x_i)) ) for discrete variables, or as ( E(x) = \int x \cdot f(x) , dx ) for continuous variables, where ( f(x) ) is the probability density function. The expected value provides insight into the long-term average outcome of a random variable in a probability distribution.
No. The mean is the expected value of the random variable but you can also have expected values of functions of the random variable. If you define X as the random variable representing the result of a single throw of a fair die, the expected value of X is 3.5, the mean of the probability distribution of X. However, you play a game where you pay someone a certain amount of money for each throw of the die and the other person pays you your "winnings" which depend on the outcome of the throw. The variable, "your winnings", will also have an expected value. As will your opponent's winnings.
In a binomial distribution, the expected value (mean) is calculated using the formula ( E(X) = n \times p ), where ( n ) is the sample size and ( p ) is the probability of success. For your experiment, with ( n = 100 ) and ( p = 0.5 ), the expected value is ( E(X) = 100 \times 0.5 = 50 ). Thus, the expected value of this binomial distribution is 50.
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E(X+Y)=E(X)+E(Y)where E(X)=population mean
180-x...... hahaahahaaa