Respuesta :
Answer:
a)
X | 1 3 5 7
f(X) | 0.4 0.2 0.2 0.2
b) [tex] P(4 <X \leq 7)= P(X\leq 7) -P(X<4) = P(X\leq 7) -P(X \leq 3) = (0.4+0.2+0.2+0.2) -(0.4+0.2)= 0.4 [/tex]
Step-by-step explanation:
For this case we have defined the cumulative distribution function like this:
[tex] F(X) = 0, x<1[/tex]
[tex] F(X) = 0.4, 1 \leq x <3 [/tex]
[tex] F(X) = 0.6, 3 \leq x <5 [/tex]
[tex] F(X) = 0.8, 5 \leq x <7 [/tex]
[tex] F(X) = 1, x \geq 7 [/tex]
And we know that the general definition for the distribution function is given by:
[tex] F(x) = P(X \leq x) = \sum_{i\leq k} f(i)[/tex]
Where f represent the density function.
Part a
For this case we need to find the density function, so we can find the values for the density for each value of X = 1,2,3,4,5,6,7,... since X is a discrete random variable.
[tex] f(1) = P(X=1) = P(X \leq 1) - P(X=0) = F(1) -F(0) = 0.4-0=0.4[/tex]
[tex] f(2) = P(X=2) = P(X \leq 2) - P(X=0)- P(X=1) = F(2) -F(1) = 0.4-0.4=0[/tex]
[tex] f(3) = P(X=3) = P(X \leq 3) - P(X=0)- P(X=1) -P(X=2) = F(3) -F(2) = 0.6-0.4=0.2[/tex]
[tex] f(4) = P(X=4) = P(X \leq 4) - P(X=0)- P(X=1) -P(X=2)-P(X=3) = F(4) -F(3) = 0.6-0.6=0[/tex]
[tex] f(5) = P(X=5) = P(X \leq 5) - P(X=0)- P(X=1) -P(X=2)-P(X=3)-P(X=4) = F(5) -F(4) = 0.8-0.6=0.2[/tex]
[tex] f(6) = P(X=6) = P(X \leq 6) - P(X=0)- P(X=1) -P(X=2)-P(X=3)-P(X=4)-P(X=5) = F(6) -F(5) = 0.8-0.8=0[/tex]
[tex] f(7) = P(X=7) = P(X \leq 7) - P(X=0)- P(X=1) -P(X=2)-P(X=3)-P(X=4)-P(X=5)-P(X=6) = F(7) -F(6) = 1-0.8=0.2[/tex]
And for any value higher than 7 we have that:
[tex] x_i \in [8,9,10,...][/tex]
[tex] f(x_i) = F(X_i) -F(X_i -1) = 1-1=0[/tex]
So then we have our density function defined like this:
X | 1 3 5 7
f(X) | 0.4 0.2 0.2 0.2
Part b
For this case we want to find this probability [tex] P(4 <X \leq 7) [/tex]
And since the random variable is discrete we can write this like that:
[tex] P(4 <X \leq 7)= P(X\leq 7) -P(X<4) = P(X\leq 7) -P(X \leq 3) = (0.4+0.2+0.2+0.2) -(0.4+0.2)= 0.4 [/tex]
