
Write the formula for \[E(X)\] and $Var(X)$
Answer
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Hint: First, we know the random variance of the random variable X is the mean or expected value of the square deviation from the mean of X.
Using their definition, we can arrive at a simpler formula for variance, both for continuous and discrete variables.
These formulas are mostly used in the statistics for the distribution method.
Complete step-by-step answer:
We know that the variance measures how far a set of numbers is spread out, from their average value.
Also, we are aware that variance is the square of standard deviation.
The variance of random variable X is represented by $Var(X)$
Using this definition for the variance for the random variable X, we can write the variance as $Var(X) = E[{(X - E[X])^2}]$, where E(X) represents the expected value or mean for the random variable X.
We can expand the equation as, $Var(X) = E[{(X - E[X])^2}] \Rightarrow E[{X^2} - 2X.E[X] + E{[X]^2}]$
By the use of ${(a - b)^2}$formula,
Now since giving the expectation values inside the equation we get, $Var(X) = E[{X^2}] - 2E[X].E[X] + E{[X]^2}$
Further solving this we get, $Var(X) = E[{X^2}] - E{[X]^2}$ (since $E[X]$or $m$)
Because the expected value of X is usually written as $E[X]$or$m$.
Thus, we get $Var(X) = E[{X^2}] - E{[X]^2} \Rightarrow Var(X) = E[{X^2}] - {m^2}$
For finding the expected value of X, the discrete random variable is known as X, is a weighted average of the possible values that X can take that each value from weighted probability from according to that event occurring.
The formula is $E[X] = Sf(x) \times P(X = x)$which is the expected value E(X) in a discrete random variable.
The expected value can be also expressed as $E(X) = \sum\limits_{i = 1}^n {{X_i}} P({X_i})$
Thus, the formula can be rewritten as $\sum\limits_{i = 1}^n {{X_i}} P({X_i}) = Sf(x) \times P(X = x)$
and variance of X is $Var(X) = E[{X^2}] - {m^2}$
Note: So, the expected value is the sum of each possible outcome into times of the probability of the outcome occurring in the expected value.
We also say that the variance of X equal to the difference of the mean of the square of X and the square of the mean of X for the variance of the X.
The variance value is interrelated to the given expected value for the distribution.
For the continuous random variable for expected value is $\mu = E(X) \Rightarrow \int\limits_{ - \infty }^\infty {x{f_X}} (x)dx$and variance is $Var(X) = E{(X - \mu )^2}$
Using their definition, we can arrive at a simpler formula for variance, both for continuous and discrete variables.
These formulas are mostly used in the statistics for the distribution method.
Complete step-by-step answer:
We know that the variance measures how far a set of numbers is spread out, from their average value.
Also, we are aware that variance is the square of standard deviation.
The variance of random variable X is represented by $Var(X)$
Using this definition for the variance for the random variable X, we can write the variance as $Var(X) = E[{(X - E[X])^2}]$, where E(X) represents the expected value or mean for the random variable X.
We can expand the equation as, $Var(X) = E[{(X - E[X])^2}] \Rightarrow E[{X^2} - 2X.E[X] + E{[X]^2}]$
By the use of ${(a - b)^2}$formula,
Now since giving the expectation values inside the equation we get, $Var(X) = E[{X^2}] - 2E[X].E[X] + E{[X]^2}$
Further solving this we get, $Var(X) = E[{X^2}] - E{[X]^2}$ (since $E[X]$or $m$)
Because the expected value of X is usually written as $E[X]$or$m$.
Thus, we get $Var(X) = E[{X^2}] - E{[X]^2} \Rightarrow Var(X) = E[{X^2}] - {m^2}$
For finding the expected value of X, the discrete random variable is known as X, is a weighted average of the possible values that X can take that each value from weighted probability from according to that event occurring.
The formula is $E[X] = Sf(x) \times P(X = x)$which is the expected value E(X) in a discrete random variable.
The expected value can be also expressed as $E(X) = \sum\limits_{i = 1}^n {{X_i}} P({X_i})$
Thus, the formula can be rewritten as $\sum\limits_{i = 1}^n {{X_i}} P({X_i}) = Sf(x) \times P(X = x)$
and variance of X is $Var(X) = E[{X^2}] - {m^2}$
Note: So, the expected value is the sum of each possible outcome into times of the probability of the outcome occurring in the expected value.
We also say that the variance of X equal to the difference of the mean of the square of X and the square of the mean of X for the variance of the X.
The variance value is interrelated to the given expected value for the distribution.
For the continuous random variable for expected value is $\mu = E(X) \Rightarrow \int\limits_{ - \infty }^\infty {x{f_X}} (x)dx$and variance is $Var(X) = E{(X - \mu )^2}$
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