Mean of a discrete random variable can be interpreted as the average outcome if the experiment is repeated many times. Expected value or average of the distribution is analogous to mean of the distribution.

The mean can be found using summation from nothing to nothing x times Upper P (x) , i.e ∑x•P(x).

Example : If two outcomes 100 & 50 occur with probabilities 0.5 each. Expected value (Average) (Mean) : ∑x•P(x) = (0.5)(100) + (0.5)(50) = 50 + 25 = 75

The mean may not be a possible value of the random variable.

Example : Mean of possible no.s on a die = ( 1 + 2 + 3 + 4 + 5 + 6 ) / 6 = 21/6 = 3.5, which is not a possible value of the random variable 'no. on a die'

4 0

3 years ago

Solve the system of equations. −5y−10x=45 −3y+10x=−5

Fynjy0 [20]

(-2,-5)
The first goal with equations like these is to get one of the variables to go away. You can do that by adding/subtracting one equation from the other. Then you solve for the one variable. Then you plug that answer into one of the original equations and solve.

4 0

3 years ago

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