Respuesta :
Answer:
a) X 1 2 3 4 5
P(X) 0.7 0.15 0.10 0.03 0.02
b) [tex] P(X \leq 2) = P(X=1) +P(X=2) = 0.7+0.15=0.85[/tex]
c) [tex] P(X >3) = 1-P(X \leq 3) = 1-[P(X=1) +P(X=2)+P(X=3)]=1-[0.7+0.15+0.1]= 0.05[/tex]
d) [tex] E(X) = \sum_{i=1}^n X_i P(X_i) = 1*0.7 +2*0.15+ 3*0.1+4*0.03+ 5*0.02= 1.52[/tex]
e) [tex] E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 1*0.7 +4*0.15+ 9*0.1+16*0.03+ 25*0.02=3.18 [/tex]
[tex] Var(X) = E(X^2) -[E(X)]^2= 3.18- (1.52)^2 = 0.8996[/tex]
[tex]\sigma= \sqrt{Var(X)}= \sqrt{0.8996}= 0.933[/tex]
Step-by-step explanation:
Part a
From the information given we define the probability distribution like this:
X 1 2 3 4 5
P(X) 0.7 0.15 0.10 0.03 0.02
And we see that the sum of the probabilities is 1 so then we have a probability distribution
Part b
We want to find this probability:
[tex] P(X \leq 2) = P(X=1) +P(X=2) = 0.7+0.15=0.85[/tex]
Part c
We want to find this probability [tex] P(X>3)[/tex]
And for this case we can use the complement rule and we got:
[tex] P(X >3) = 1-P(X \leq 3) = 1-[P(X=1) +P(X=2)+P(X=3)]=1-[0.7+0.15+0.1]= 0.05[/tex]
Part d
We can find the expected value with this formula:
[tex] E(X) = \sum_{i=1}^n X_i P(X_i) = 1*0.7 +2*0.15+ 3*0.1+4*0.03+ 5*0.02= 1.52[/tex]
Part e
For this case we need to find first the second moment given by:
[tex] E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 1*0.7 +4*0.15+ 9*0.1+16*0.03+ 25*0.02=3.18 [/tex]
And we can find the variance with the following formula:
[tex] Var(X) = E(X^2) -[E(X)]^2= 3.18- (1.52)^2 = 0.8996[/tex]
And we can find the deviation taking the square root of the variance:
[tex]\sigma= \sqrt{Var(X)}= \sqrt{0.8996}= 0.933[/tex]
