Respuesta :
Answer:
a) [tex] \frac{1+3k}{3} + \frac{1+2k}{3} +\frac{0.5+5k}{3}= 1[/tex]
We can multiply both sides by 3 and we got:
[tex] 1+3k+ 1+2k + 0.5+5k = 3 [/tex]
[tex] 10 k = 3-1-0.5-1=0.5[/tex]
[tex] k = 0.05[/tex]
And if we replace we got:
X 1 2 3
P(X) 0.383 0.367 0.25
b) [tex] E(X)= \sum_{i=1}^n X_i P(X_i) = 1*0.383 + 2*0.367 +3*0.25= 1.867[/tex]
[tex] 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[/tex]
[tex] Var(X) = E(X^2) -[E(X)]^2 = 4.101- [1.867]^2 =0.615[/tex]
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:
[tex] \frac{1+3k}{3} + \frac{1+2k}{3} +\frac{0.5+5k}{3}= 1[/tex]
We can multiply both sides by 3 and we got:
[tex] 1+3k+ 1+2k + 0.5+5k = 3 [/tex]
[tex] 10 k = 3-1-0.5-1=0.5[/tex]
[tex] k = 0.05[/tex]
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:
[tex] E(X)= \sum_{i=1}^n X_i P(X_i) = 1*0.383 + 2*0.367 +3*0.25= 1.867[/tex]
For the variance we need to find the second moment first like this:
[tex] 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[/tex]
And we can find the variance with this formula:
[tex] Var(X) = E(X^2) -[E(X)]^2 = 4.101- [1.867]^2 =0.615[/tex]
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
The value of k, mean, and variance are 0.05, 1.867, and 0.615. And the cumulative distribution function is shown below.
X 1 2 3
F(X) 0.383 0.75 1
What is probability?
Probability means possibility. It deals with the occurrence of a random event. The value of probability can only be from 0 to 1. Its basic meaning is something is likely to happen. It is the ratio of the favorable event to the total number of events.
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. We know that the sum of probability of all outcomes must be 1.
[tex]\rm \dfrac{1+3k}{3} + \dfrac{1+2k}{3} + \dfrac{0.5+5k}{3} = 1\\[/tex]
On solving the equation we have the value of k will be
[tex]\rm 10 k = 0.5\\\\k \ \ \ = 0.05[/tex]
Then we have
x 1 2 3
P(x) 0.383 0.367 0.25
b. Mean will be
[tex]\rm E(X) = \Sigma ^{n}_{i = 1} X_i P(X_i)\\\\E(X) = 1*0.383 + 2*0.367 + 3 * 0.25\\\\E(X) = 1.867[/tex]
Variance will be
[tex]\rm E(X^2) = \Sigma ^{n}_{i = 1} X_i P(X_i)\\\\E(X^2) = 1^2*0.383 + 2^2*0.367 + 3^2 * 0.25\\\\E(X^2) = 4.1041[/tex]
Then
[tex]\rm Var (X) = E(X^2) -[E(X)]^2 = 4.101 - 1.867^2 = 0.615[/tex]
c. The cumulative distribution on this case is given by
X 1 2 3
F(X) 0.383 0.75 1
More about the probability link is given below.
https://brainly.com/question/795909
