A random variable X has a mean of 120 and a standard deviation of 15 a random variable Y has a mean of 100 and a standard deviation of 9. If X and Y are independent approximately what is the standard deviation of X plus Y.
Solution:
Given, X and Y are independent random variables
A random variable X has a mean of 120 and a standard deviation of 15.
A random variable Y has a mean of 100 and a standard deviation of 9.
We have to find the standard deviation of X plus Y.
Let S be the sum of standard deviation of X and Y.
The standard deviation of S will be the square root of the sum of the variances of X and Y.
We know that variance is the square of standard deviation.
So, variance of X = (standard deviation of X)2
Variance of X = (15)2 = 225
variance of Y = (standard deviation of Y)2
Variance of Y = (9)2 = 81
SD(X +Y) = √SD(X)2 + SD(Y)2
Now, S = \(\sqrt{225+81}\\=\sqrt{306}\)
S = 17.5
Therefore, the standard deviation of X plus Y = 17.5
A random variable X has a mean of 120 and a standard deviation of 15 a random variable Y has a mean of 100 and a standard deviation of 9. If X and Y are independent approximately what is the standard deviation of X plus Y.
Summary:
A random variable X has a mean of 120 and a standard deviation of 15 a random variable Y has a mean of 100 and a standard deviation of 9. if X and Y are independent approximately the standard deviation of X plus Y is 17.5
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