Answer:
a) The cumulative distribution function would be given by:
x 0 1 2 3 4 5
F(X) 0.05 0.15 0.30 0.55 0.9 1
b) [tex] P(1 \leq X \leq 4) = F(4) -F(0) =0.9-0.05 = 0.85 [/tex]
And replacing we got:
[tex]P(1 \leq X \leq 4) =0.85[/tex]
Step-by-step explanation:
For this case we have the following probability distribution function given:
x 0 1 2 3 4 5
P(X) 0.05 0.1 0.15 0.25 0.35 0.1
We satisfy the conditions in order to have a probability distribution:
1) [tex] \sum_{i=1}^n P(X_i)=1[/tex]
2) [tex] P(X_i) \geq 0, i=1,2,..,n[/tex]
Part a
The cumulative distribution function would be given by:
x 0 1 2 3 4 5
F(X) 0.05 0.15 0.30 0.55 0.9 1
Part b
For this case we want to find this probability:
[tex] P(1 \leq X \leq 4) = F(4) -F(0) =0.9-0.05 = 0.85 [/tex]
And replacing we got:
[tex]P(1 \leq X \leq 4) =0.85[/tex]