Find new standard deviation, when givin three standard deviations

Question

I hope someone can help me with this:

"A company sells paint in the colors: blue, black and red. The annual demand for the three variants follows a normal distribution. Blue paint, mean= 2500 and std. = 600 Black paint, mean= 1000 and std. = 500 Red paint, mean= 1300 std. = 750

The demand is of the paint is independent. The company gets the paint in white, and paints it (red, blue or black) when a customer gives an order.

What is the std. and mean for the white paint?

I know that for the std. its: square root(600^2+750^2+500^2) - but which formula is this? I need to make a reference to this formula...

thanks

Answer 1

If the three colors are equally popular, then $W = X+Y+Z,$ where (independently) $X \sim \mathsf{Norm}(\mu_x = 2500, \sigma_x = 600),$ $Y \sim \mathsf{Norm}(\mu_y = 1000, \sigma_y = 500),$ and $Z \sim \mathsf{Norm}(\mu_z = 1300, \sigma_z = 750).$

Then formulas for means of sums of random variables and variances of sums of independent random variables apply:

$$E(W) = E(X+Y+Z) = E(X)+E(Y)+E(Z)$$ and $$V(W) = V(X+Y+Z) = V(X)+V(Y)+V(Z).$$

You can use these formulas to answer the question.

However, I cannot resist pointing out that the specifications of this problem are impossible in real life. For example, the probability of negative demand for Black paint is $P(Y < 0) = 0.0228.$ Similarly, for Red paint $P(Z < 0) \approx 0.04.$

pnorm(0, 1000, 500)
[1] 0.02275013

Usually, when people make up drill problems involving normal distributions, they take care to make the mean at least three times the standard deviation so that the probability of impossible negative events is negligible and can be ignored. For example, $P(X < 0) \approx .000015.$