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
<span>6.8 x 10^-2
The scientific notation of 70, 030, 000. To find the scientific notation of this value </span><span><span>
1. </span>We first move the period which separates the whole number from the decimal number which is located after the numbers of the given value.</span>
<span><span>2. </span>We move it in the very recent order number which is seventy million, seven and zero.</span> <span><span>
3. </span>It becomes 7.003</span>
<span><span>4. </span>Thus we count how many moves we did from the tens to the ten million order place.</span> <span><span>
5. </span>7.003 x 10^7<span>
</span></span>
Answer:
come back
Step-by-step explanation:
It's D, it's the only one where it has a -6 at the end other than A, but if you look at the A when distributed there isn't a 23x<span />
