Probability more than 25% greater?

Question

The random variable X is distributed N(60,64). The random variable Y is distributed N(52,36).

Find the probability that a random observation from X is more than 25% greater than a random observation of Y.

Answer 1

This will get you half way there.

Suppose we have obtained an observation from Y. Call it $y$. I want to express these probability densities as functions. So let g(x) be the probability density of X and h(y) be the probability density of Y. You can formally write the probability that an observation from X is at least 25% greater than it as

$P(x > 1.25y) = \int_{1.25y}^\infty $ g(x) dx

Now you just have "sum" this over all possible values of $y$, properly weighted by the probability density of $y$.

Answer 2

Outline: We do indeed want $\Pr(X\gt 5Y/4)$ or equivalently $\Pr(W\gt 0)$ where $$W=X-\frac{5}{4}Y.$$ If we assume that $X$ and $Y$ are independent, then $W$ is normally distributed, with mean $E(X)-\frac{5}{4}E(Y)$ and with variance $\text{Var}(X)+\frac{25}{16}\text{Var}(Y)$.

Compute the mean and variance of $W$. The rest should be standard.