The random variable x takes on the values 1, 2, or 3 with probabilities (1 + 3k)/3, (1 + 2k)/3, and (0.5 +5k)/3, respectively.(a) Find the appropriate value of k.(b) Find the mean and variance of x.(c) Find the cumulative distribution function.

a)

We can multiply both sides by 3 and we got:

And if we replace we got:

X            1             2               3

P(X)    0.383      0.367        0.25

b)

c) X            1                         2                              3

F(X)       0.383      0.367+0.383=0.75        0.25+0.75 = 1

Step-by-step explanation:

For this case we have the following probability mass function given:

X            1             2               3

P(X)  (1+3k)/3    (1+2k)/3   (0.5+5k)/3

Part a

In order to satisfy the definition of probability distribution the sum of all the probabilities needs to be 1 and each of the individual probabilities needs to be higher or equal than 0, using this we can do that:

We can multiply both sides by 3 and we got:

And if we replace we got:

X            1             2               3

P(X)    0.383      0.367        0.25

And we satisfy all the conditions.

Part b

For this case we can find the mean using this formula:

For the variance we need to find the second moment first like this:

And we can find the variance with this formula:

Part c

The cumulative distribution on this case is given by:

X            1                         2                              3

F(X)    0.383      0.367+0.383=0.75        0.25+0.75 = 1