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

Answer:

$c = \frac{1}{12}$

The mean of the distribution is 0.25.

The variance of the distribution is of 0.6875.

Step-by-step 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:

$3c + 3c + 6c = 1$

$12c = 1$

$c = \frac{1}{12}$

So the probability distribution is:

$P(X = -1) = 3c = 3\frac{1}{12} = \frac{1}{4} = 0.25$

$P(X = 0) = 3c = 3\frac{1}{12} = \frac{1}{4} = 0.25$

$P(X = 1) = 6c = 6\frac{1}{12} = \frac{1}{2} = 0.5$

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.