Let U, V, and W be independent standard normal random variables (that is, independent normal random variables, each with mean 0 and variance 1), and let X=3U+4V and Y=U+W. Give a numerical answer for each part below. You may want to refer to the standard normal table.
What is the probability that X≥8?
P(X≥8)=
E[XY]=
var(X+Y)=
heyyy i need a good note
some answers
P(X≥8)= 0.054799
var (X+Y) = 33 ?? (not really sure about this one)
fellow classmate: help us out with E[XY] if you are reading this....Feel free to provide your answer to the whole TEST
E[XY] = 3
var(X+Y)= 33
checked!!
It is indeed:
1) 0.0548
2) 3
3) 33
1)0.0548
2) 3
3) 33
The answers are Checked are ok!
To find the probability that X≥8, we need to calculate the z-score corresponding to X=8, and then find the area under the standard normal curve to the right of that z-score.
Step 1: Calculate the mean and standard deviation of X.
Since U and V are independent standard normal random variables, the mean of X is:
E[X] = E[3U+4V] = 3E[U] + 4E[V] = 3(0) + 4(0) = 0
Similarly, the variance of X is:
var(X) = var(3U+4V) = 3^2var(U) + 4^2var(V) = 9(1) + 16(1) = 25
So, the standard deviation of X is:
σ(X) = √(var(X)) = √(25) = 5
Step 2: Calculate the z-score of X=8.
The z-score formula is:
z = (X - μ) / σ
Plugging in the values, we have:
z = (8 - 0) / 5 = 8/5 = 1.6
Step 3: Use the standard normal table.
The standard normal table provides the area under the standard normal curve to the left of a given z-score. To find the area to the right of the z-score, we subtract the area from 1.
Using the standard normal table or a calculator, we find that the area to the left of z = 1.6 is 0.9452.
Therefore, the area to the right of z = 1.6 is 1 - 0.9452 = 0.0548.
So, the probability that X≥8 is approximately 0.0548.
Now, let's move on to the next parts of the question.
Part 2: E[XY]
To find E[XY], we need to calculate the expected value of the product XY.
E[XY] = E[(3U+4V)(U+W)]
Since U and V and U and W are independent, we can expand the expression:
E[XY] = E[3U^2+4UV+3UW+4VW]
Taking into account that U, V, and W are all standard normal random variables with mean 0, we can simplify further:
E[XY] = E[3U^2+4V^2]
Given that the expected value of U^2 and V^2 is 1, we find:
E[XY] = 3(1) + 4(1) = 7.
Therefore, E[XY] = 7.
Part 3: var(X+Y)
To find var(X+Y), we need to calculate the variance of the sum X+Y.
var(X+Y) = var(3U+4V+U+W)
Since U, V, and W are independent, we can expand the expression:
var(X+Y) = var(4U+4V+W)
Taking into account that U, V, and W are all standard normal random variables with variance 1, we can simplify further:
var(X+Y) = var(4U) + var(4V) + var(W) = 4^2(1) + 4^2(1) + 1 = 32 + 32 + 1 = 65.
Therefore, var(X+Y) = 65.