Respuesta :
Answer:
X 1 2 3 4 5 6
P(X) 0.115 a 0.422 0.312 0.034 0.015
If we want a probability distribution we need to satisfy two important conditions:
1) [tex]P(X_i) \geq 0, \forall i=1,2,3,4,5,6[/tex]
2) [tex]\sum_{i=1}^n P(X_i) =1[/tex]
And using the second condition we have this:
[tex] 0.115+a+0.422+0.312 +0.034 +0.015=1[/tex]
With a the probability associated with the value of X=2 and solving for a we got:
[tex] a = 1-0.115-0.422-0.312-0.034-0.015= 0.102[/tex]
So then for this case [tex] P(X=2) = 0.102[/tex]
Step-by-step explanation:
Let X the random variable that represent the number of classes that a student takes
For this case we have the following probability distribution given:
X 1 2 3 4 5 6
P(X) 0.115 a 0.422 0.312 0.034 0.015
If we want a probability distribution we need to satisfy two important conditions:
1) [tex]P(X_i) \geq 0, \forall i=1,2,3,4,5,6[/tex]
2) [tex]\sum_{i=1}^n P(X_i) =1[/tex]
And using the second condition we have this:
[tex] 0.115+a+0.422+0.312 +0.034 +0.015=1[/tex]
With a the probability associated with the value of X=2 and solving for a we got:
[tex] a = 1-0.115-0.422-0.312-0.034-0.015= 0.102[/tex]
So then for this case [tex] P(X=2) = 0.102[/tex]
