Question

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.

Answer 1

Answer:

a) $\frac{1+3k}{3} + \frac{1+2k}{3} +\frac{0.5+5k}{3}= 1$

We can multiply both sides by 3 and we got:

$1+3k+ 1+2k + 0.5+5k = 3$

$10 k = 3-1-0.5-1=0.5$

$k = 0.05$

And if we replace we got:

X            1             2               3

P(X)    0.383      0.367        0.25

b) $E(X)= \sum_{i=1}^n X_i P(X_i) = 1*0.383 + 2*0.367 +3*0.25= 1.867$

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i)=1^2*0.383 + 2^2*0.367 +3^2*0.25= 4.101$

$Var(X) = E(X^2) -[E(X)]^2 = 4.101- [1.867]^2 =0.615$

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:

$\frac{1+3k}{3} + \frac{1+2k}{3} +\frac{0.5+5k}{3}= 1$

We can multiply both sides by 3 and we got:

$1+3k+ 1+2k + 0.5+5k = 3$

$10 k = 3-1-0.5-1=0.5$

$k = 0.05$

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:

$E(X)= \sum_{i=1}^n X_i P(X_i) = 1*0.383 + 2*0.367 +3*0.25= 1.867$

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

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i)=1^2*0.383 + 2^2*0.367 +3^2*0.25= 4.101$

And we can find the variance with this formula:

$Var(X) = E(X^2) -[E(X)]^2 = 4.101- [1.867]^2 =0.615$

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