This is the answer.
It is the expected value of the distribution. It also happens to be the mode and median.It is the expected value of the distribution. It also happens to be the mode and median.It is the expected value of the distribution. It also happens to be the mode and median.It is the expected value of the distribution. It also happens to be the mode and median.
Depending on whether you subtract actual value from expected value or other way around, a positive or negative percent error, will tell you on which side of the expected value that your actual value is. For example, suppose your expected value is 24, and your actual value is 24.3 then if you do the following calculation to figure percent error:[percent error] = (actual value - expected value)/(actual value) - 1 --> then convert to percent.So you have (24.3 - 24)/24 -1 = .0125 --> 1.25%, which tells me the actual is higher than the expected. If instead, you subtracted the actual from the expected, then you would get a negative 1.25%, but your actual is still greater than the expected. My preference is to subtract the expected from the actual. That way a positive error tells you the actual is greater than expected, and a negative percent error tells you that the actual is less than the expected.
The expected value of the standard normal distribution is equal to the total amount of the value. It is usually equal to it when the value works out to be the same.
Expected value of a random variable requires that the random variable can be repeated in experiment indefinitely. If the random variable can only be repeated finite times, e.g. once, there is an inadequacy of the expected value principle for a decision maker.
The expected value if you roll a die three times is 3.5.
This is the answer.
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 expected value is the arithmetic mean. It may not always be a value that is realised. Consider rolling a fair normal die. The mean or expected value of the outcome is 3.5 but a normal die will never ever turn up 3.5 since it has only integer values.
Consider a binomial distribution with 10 trials What is the expected value of this distribution if the probability of success on a single trial is 0.5?
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.
The expected value of a Martingale system is the last observed value.
The expected values were 2 of each value. This differs significantly from what was expected. You could show that the die is most likely not fair by using a chi squared test for goodness of fit.
It is the expected value of the distribution. It also happens to be the mode and median.It is the expected value of the distribution. It also happens to be the mode and median.It is the expected value of the distribution. It also happens to be the mode and median.It is the expected value of the distribution. It also happens to be the mode and median.
For a population the mean and the expected value are just two names for the same thing. For a sample the mean is the same as the average and no expected value exists.
The expected value is 7.