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
Answer:
m=2
Step-by-step explanation:
Answer:
Please check the explanation.
Step-by-step explanation:
Finding Domain:
We know that the domain of a function is the set of input or argument values for which the function is real and defined.
From the given graph, it is clear that the starting x-value of the line is x=-2, the closed circle at the starting value of x= -2 means the x-value x=-2 is included.
And the line ends at the x-value x=1 with a closed circle, meaning the ending value of x=1 is also included.
Thus, the domain is:
D: {-2, -1, 0, 1} or D: −2 ≤ x ≤ 1
Finding Range:
We also know that the range of a function is the set of values of the dependent variable for which a function is defined
From the given graph, it is clear that the starting y-value of the line is y=0, the closed circle at the starting value of y = 0 means the y-value y=0 is included.
And the line ends at the y-value y=2 with a closed circle, meaning the ending value of y=2 is also included.
Thus, the range is:
R: {0, 1, 2} or R: 0 ≤ y ≤ 2
Hey There
According to me,
no 1 is
SQUARE
no. 2 is
CONGRUENT
HOPE THIS HELPS
