Question

1 Find the value of C to that the function probability density function defined as follow, also calculate the man and variance /4

X -1 0 1
f(x) 3c 3c 6c

Answer 1

`c = (1)/(12)`

The mean of the distribution is 0.25.

The variance of the distribution is of 0.6875.

Explanation:

Probability density function:

For it to be a probability function, the sum of the probabilities must be 1. The probabilities are 3c, 3c and 6c, so:

`3c + 3c + 6c = 1`

`12c = 1`

`c = (1)/(12)`

So the probability distribution is:

`P(X = -1) = 3c = 3(1)/(12) = (1)/(4) = 0.25`

`P(X = 0) = 3c = 3(1)/(12) = (1)/(4) = 0.25`

`P(X = 1) = 6c = 6(1)/(12) = (1)/(2) = 0.5`

Mean:

Sum of each outcome multiplied by its probability. So

`E(X) = -1(0.25) + 0(0.25) + 1(0.5) = -0.25 + 0.5 = 0.25`

The mean of the distribution is 0.25.

Variance:

Sum of the difference squared between each value and the mean, multiplied by its probability. So

`V^2(X) = 0.25(-1-0.25)^2 + 0.25(0 - 0.25)^2 + 0.5(1 - 0.25)^2 = 0.6875`

The variance of the distribution is of 0.6875.