Let X and Y be independent and normally distributed, X with mean 0 a.pdf

•

0 likes•4 views

Let X and Y be independent normally distributed random variables, with X having a mean of 0 and variance of 1, and Y having a mean of 1. The probability that X is greater than Y is 1/3. To find the standard deviation of Y, we first determine that the joint distribution of X-Y has a mean of -1 and variance equal to the sum of the variances of X and Y. Setting the probability that X-Y is greater than its mean equal to 1/3 allows us to solve for the standard deviation of Y as 2.06.

Education

Let X and Y be independent and normally distributed, X with mean 0 and variance 1, Y with
mean 1. Suppose P(X>Y) = 1/3. Find the standard deviation of Y.
Solution
for the joint distribution x-y the new mean would be E(x) - E(y) = -1 and new
variance would be Var(x)+ Var(y) = 1+sigma(y)^2 ......so final standard deviation would be
sqrt(1+ sigma^2) = k ......now P(X-Y > 0) = 1/3 ......P([X-Y - (-1)]/[k] > 1/k ) = 1/3 ......1 -
F(1/k) = 1/3 // F is the CDF .......F(1/k) = 2/3 ........1/k = 0.435 (from table of cdf of normal
functions) .........k = 2.298 .........var(Y) = k^2 -1 = 4.28 ...........SD(y) = sqrt(var(Y)) = 2.06

More from info665359

Let X be an r.v. defined on a sample space S into the real line R . .pdf

info665359

let X be a raondom variable that represents the number of infants in a group of 2000 who die before reaching their first birthdays. In the US, the probability that a child dies during his or her first year of life is .0085. What is the probability that at most 5 infants out of 2000 die in their first year of life? Solution (a) What is the mean number of infants who would die in a group of this size? Since this is a binomial distribution the mean = n*p = 17 (b) What is the probability that at most five infants out of 2000 die in their first year of life? The conditions are sufficient to approximate this random variable with either a poisson or normal distribution. Poisson: P(X<=5) = P(X=0) +.

let X be a raondom variable that represents the number of infants in.pdf

info665359

Let X be a random variable with cumulative distribution funtion0.pdf

info665359

let X be a random variable having Poisson density parameter Calculate E[1/(X+1)] Solution As we know that, X~poisson(); E[1/(x+1)] =[1/(x+1)](^x)[e^(-)]/x! from x=0 to x=infinity =(1/)[^(x+1)][e^(-)]/(x+1)! from x=0 to x=infinity =(1/){[^(x+1)][e^(-)]/(x+1)! from x=-1 to x=infinity - e^(-)} As we know, [^(x+1)][e^(-)]/(x+1)! from x=-1 to x=infinity will become 1 because it is the sum of all supports of x, therefore, =(1/)[1-e^(-)] =1/ - e^(-).

let X be a random variable having Poisson density parameter Calcula.pdf

info665359

Let X be a random number between 0 and 1 produced by the idealized random number generator described in Figure 10.4 (below). Find the following probabilities: 14. P(X ? 0.4) 15. P(X < 0.4) 16. P(0.3 ? X ? 0.5) Statistics and Probability Find the following probabilities:14.P(X ? 0.4) Solution a) P(X = 0.4 ) ˜ 4 b) P(x < 0.4) ˜ 4 c) P(0.2 = X = 0.6) ˜ 4 The table is wrong. You have committed a grave error building it! The sum of all the heights must be equal to 1!.

Let X be a random number between 0 and 1 produced by the idealized r.pdf

info665359

Let X be a binomial random variable with parameters m and 13. Find .pdf

info665359

Let X be a discrete random variable that takes on two values -1 and.pdf

info665359

Let X and Y have the joint normal distribution described in equation.pdf

info665359

Let X and Y be uniformly distributed in the interval [0,1]. Compute .pdf

info665359

Let V be a vector space containing polynomials in the form f(x)=ao+a.pdf

info665359

Let V be the vector space of ordered pairs of complex numbers over the real field [RR] . Show that V is of dimension 4. Solution Let S={(1,0),(0,1),(i,0),(0,i)} be the set . We wish to prove S is basis of V over the real field R.Consequently we can prove that dim(V)=4, because number of elements in set S is 4. Vectors in S are linearly independent. If a(1,0)+b(0,1)+c(i,0)+d(0,i)=0 a+ic=0 => a=c=0 b+id=0 => b=d=0 a=b=c=d=0 only solution. Thus vectors in S are linearly independent. Let (x+iy,w+iz) in V then (x+iy,w+iz)=x(1,0)+y(i,0)+w(0,1)+z(0,i) (x+iy,w+iz) is an arbitrary element of V. Thus every element of V can be expressed as linear combination of vectors in S. Thus S is basis for V. Number of elements in S are 4 ,so dimension of V be 4. Hence proved..

Let V be the vector space of ordered pairs of complex numbers over t.pdf

info665359

Let X (t) be a birth and death process with possible states 0, 1, ...pdf

info665359

More from info665359 (12)