The random variable X is normally distributed with mean 5 and standard deviation 25. The random variable Y is defined by Y = 2 + 4X. What are the mean and the standard deviation of Y ? The mean is 20 and the standard deviation is 102. A) The mean is 20 and the standard deviation is 50. B) The mean is 22 and the standard deviation is 102. C) The mean is 22 and the standard deviation is 100. D) The mean is 22 and the standard deviation is 50.

Answer:C) The mean is 22 and the standard deviation is 100. Step-by-step explanation:Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X). If [tex] X\ sim N(\mu_x =5, \sigma_x =25)[/tex]And the random variable Y =2+4X.If we are interested in find the mean and the standard deviation for this new variable we need to apply the concepts of expected value and Variance of random variables.Using the concept of expected value we have:[tex]E(Y) =E(2+4X) = E(2) +E(4X)= 2+ 4E(X) =2+4(5) =2+20=22[/tex]So on this case the [tex]E(Y) =\mu_Y =22[/tex]Using the concept of variance we have:[tex]Var(Y)=Var(2+4x)=Var(2)+Var(4X)+2Cov(2,4X)[/tex]But since 2 is a constant we don't have variance for this and the [tex]Cov(2,4x)=0[/tex] because the covariance of a random variable with a constant is zero. So applying these concepts we have:[tex]Var(Y)= Var(4x)=4^2 Var (X)= 16 \sigma^2_x =16(25^2)=10000[/tex]And then if we need the standard deviation we just need to take the square root:[tex]sd(Y) = \sqrt{10000}=100[/tex]So the best option for this case would be:C) The mean is 22 and the standard deviation is 100.