Suppose X is N(4,2) and Y is N(10,1). what kind of distribution is 2X+Y? What is its mean and standard deviation if X and Y are independent?
If N(4,2) means a distribution with a mean of 4 and standard deviation of 2, etc. and 2X +Y is a distribution made up of twice as many from group X as group Y,
then the mean of the combined group is
(2*4 + 10)/3 = 6,
and the new standard deviation is
sqrt[(2^2 + 2^2 + 1^2)/3] = sqrt3 = 1.73
My answer was a guess. It is probably wrong regarding the standard deviation. A non-normal "bipolar distribution" is possible when two different populations are mixed.
To determine the distribution of 2X + Y, where X ~ N(4, 2) and Y ~ N(10, 1), we can use the properties of normal distributions.
First, let's consider the mean and standard deviation of 2X + Y when X and Y are independent.
The mean of a linear combination of independent random variables is equal to the sum of their individual means. So, the mean of 2X + Y is:
mean(2X + Y) = 2 * mean(X) + mean(Y)
Now, since X ~ N(4, 2), the mean of X is 4. And since Y ~ N(10, 1), the mean of Y is 10.
mean(2X + Y) = 2 * 4 + 10 = 8 + 10 = 18
Therefore, the mean of 2X + Y is 18.
The standard deviation of a linear combination of independent random variables is calculated as follows:
std(2X + Y) = sqrt((2^2) * var(X) + var(Y))
Now, var(X) represents the variance of X, and since X ~ N(4, 2), the variance is 2.
var(Y) represents the variance of Y, and since Y ~ N(10, 1), the variance is 1.
std(2X + Y) = sqrt((2^2) * 2 + 1) = sqrt(8 + 1) = sqrt(9) = 3
Therefore, the standard deviation of 2X + Y is 3.
In conclusion, the distribution of 2X + Y, when X and Y are independent, would be a normal distribution with a mean of 18 and a standard deviation of 3.