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 mistake is in step 3:
I: −2m =−12 − 2n
II: 2m = 8 + 8n
the student set both right sides of the equations equal (−12 − 2n = 8 + 8n), but the left sides aren't equal: first equation has -2m, second has +2m so the signs of one of the equations must be inverted first
One and one-fourth equals to 5 over 4
5/4 times 3/3 equals to 15/12
subtract 5/12 from 15/12
the answer is 10/12
simplest form is 5/6
hope it helped
hit the brainiest button!
If the parabola has y = -4 at both x = 2 and x = 3, then since a parabola is symmetric, its axis of symmetry must be between x = 2 and x = 3, or at x = 5/2. Our general equation can then be:
y = a(x - 5/2)^2 + k
Substitute (1, -2): -2 = a(-3/2)^2 + k
-2 = 9a/4 + k
Substitute (2, -4): -4 = a(-1/2)^2 + k
-4 = a/4 + k
Subtracting: 2 = 2a, so a = 1. Substituting back gives k = -17/4.
So the equation is y = (x - 5/2)^2 - 17/4
Expanding: y = x^2 - 5x + 25/4 - 17/4
y = x^2 - 5x + 2 (This is the standard form.)
