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:
This is absolute value equation so collect like terms and find X
15+5=|x|
20=|x|
since it's absolute value it's going to be 20 or -20
Step-by-step explanation:
it filled up half the circle (up to the center point) - if we had a full circle. but a little bit is cut off (below AB).
what we see is that the shaded area is the sum of the area of the triangle AOB and 2 equally sized circle segment areas left and right of AOB.
since we are dealing with a half-circle, we have 180° in total. 120° are taken by AOB, so, that leaves us with 180-120 = 60° for both circle segments (so, one has an angle of 30°).
and 2×30° = 1×60°, so we can calculate the area of one 60° segment instead of two 30° segments.
AOB is an isoceles triangle (the legs are equally long, and therefore also the 2 side angles are equal).
the area of this triangle AOB is
1/2 × a × b × sin(C) = 1/2 × 3 × 3 × sin(120) =
= 3.897114317... m²
a circle segment area of 60° is 60/360 = 1/6 of the full circle area (as a full circle = 360°).
so, it's area is
pi×r² × 1/6 = pi×3²/6 = pi×3/2 = 4.71238898... m²
so, the total area of the shaded area is
3.897114317... m² + 4.71238898... m² =
= 8.609503297... m²
No, it would leave a decimal
It is different because you are doing a different method to complete the problem. It is the same because it ends up with the same answer, provided you complete the process correctly.
