Rescaling a normal with Y = 3X - 5

Question

Let X \sim N(	ext{mean } 50,\ 	ext{variance } 16) and define Y = 3X - 5. What are the mean and variance of Y?

Accepted Answer

For a linear transform Y = aX + b, the mean transforms as a\mu + b and the variance as a^2\sigma^2 (the shift b leaves the variance untouched). Here a = 3, b = -5: \mathbb{E}[Y] = 3(50) - 5 = 150 - 5 = 145, \qquad \operatorname{Var}(Y) = 3^2 	imes 16 = 9 	imes 16 = 144. Standard deviation \sqrt{144} = 12 = 3 	imes 4, i.e. the scale factor times the original sd. The subtraction of 5 shifted the center but did nothing to the spread. Takeaway. Y = 3X - 5 has mean 3\mu - 5 = 145 and variance 3^2\sigma^2 = 144; multipliers hit variance squared, additive shifts do not touch it.

Rescaling a normal with Y = 3X - 5

Your answer

Discussion

Learn the concepts

Related questions