Answer:
a) P ( 3 ≤X≤ 5 ) = 0.02619
b) E(X) = 1
Step-by-step explanation:
Given:
- The CDF of a random variable X = { 0 , 1 , 2 , 3 , .... } is given as:
Find:
a.Calculate the probability that 3 ≤X≤ 5
b) Find the expected value of X, E(X), using the fact that. (Hint: You will have to evaluate an infinite sum, but that will be easy to do if you notice that
Solution:
- The CDF gives the probability of (X < x) for any value of x. So to compute the P ( 3 ≤X≤ 5 ) we will set the limits.
- The Expected Value can be determined by sum to infinity of CDF:
E(X) = Σ ( 1 - F(X) )
E(X) = Limit n->∞ [1 - 1 / ( n + 2 ) ]
E(X) = 1
The value of the probability P(A) is 0.40
<h3>How to determine the
probability?</h3>
The given parameters about the probability are
P(A or B) = 0.6
P(B) = 0.3
P(A and B) = 0.1
To calculate the probability P(A), we use the following formula
P(A and B) = P(A) + P(B) - P(A or B)
Substitute the known values in the above equation
0.1 = 0.3 + P(A) - 0.6
Collect the like terms
P(A)= 0.1 - 0.3 + 0.6
Evaluate the expression
P(A)= 0.4
Hence, the value of the probability P(A) is 0.40
Read more about probability at
brainly.com/question/25870256
#SPJ1
Step-by-step explanation:
Y= m x+ C
m (slope ) = 2 - (_3) / 0 _( - 9 )
= 5 / 9
( O, 2 ) Satisfying equation
2= 5/9 ( 0) + C
C= 2
So eq. is
y= 5/ 9 X + 2
Go up on the graph (y axis) and you need to go down 3 on the y axis and go to the right one and repeat. let me know if you want me to draw it out!
Step-by-step explanation:
