x is a random variable with with a positive mean and for which E[x(x+1)]=80 and Var(x)=50. what is E[x]
Solution:
E[x(x+1)] = 80
E[x2 + x ] = 80
E[x2 ] + E[x] = 80
E[x2 ] = 80 - E[x]
Now, we know that
Var(x) = E[x2 ] - [E(x)]2
50 = 80 -
E[x] - [E(x)]2
[E(x)]2 + E[x ] -30 = 0
The roots of this quadratic equations are 6 and -5
But it is given that the mean of x is positive.i.e. E[x ] is positive.
E[x
] = 6
x is a random variable with with a positive mean and for which E[x(x+1)]=80 and Var(x)=50....
Problem 1. (a) Let X be a Binomial random variable such that E(X) 4 and Var(x) 2. Find the parameters of X (b) Let X be a standard normal random variable. Write down one function f(t) so that the random variable Y-f(X) is normal with mean a and variance b.
in another word, find E[Yt] and var[Yt]
t X be a random variable with mean 0 and variance σ2. Define Yİ = (-1)'X. Is this process stationary?
What is Var[3X]? Let X be a random variable such that Var[X] = 5 and E[X] = 4.
(1 point) For a random variable X, suppose that E[X] = 2 and Var(X) = 3. Then (a) E[(5 + x)2) = (b) Var(2 + 6X) =
Let the random variable, X have the survival function SX(x) =e(−x/80), for x>0.Then the mean of that random variable equals 1/80. True or False? (show work/calculations)
Consider the random variable X such that E(X) = 3 and Var(X) = 2 if the random variable Z = 1+5X. Variance of Z, i.e., σ2 is:
Recall that the variance of a random variable is defined as
Var[X]=E[(X−μ)2], where μ = E[X]. Use the properties of
expectation to show that we can rewrite the variance of a random
variable X as Var [X]=E[X^2]−(E[X])^2
Problem 3. (1 point) Recall that the variance of a random variable is defined as Var X-E(X-μ)21, where μ= E[X]. Use the properties of expectation to show that we can rewrite the variance of a random variable X as u hare i- ElX)L...
Consider a random variable X with the following properties E[X] - 10 and var(X) - 9. Consider a new random variable such that Y-1-5X Calculate the following (a) EY] - (b) var(Y) = 5
A random variable X is known to always be positive and have a standard deviation of 5 and E[x^2] = 125. Another random variable (Y) is known to have a mean twice as large as (X) and E[Y^2] = 500. Find the following: a.) E[X] b.) E[2X + 5] c.) Var(Y) d.) E[(Y-5)^2] e.) Assuming X and Y are independent find Var(2X - Y +5)
Consider a random variable X with the following properties E[X] = 20 and var(X) = 2. Consider a new random variable such that Y = 5 – 5X Calculate the following. (a) E[Y] = = (b) var(Y) = À