Problem statement: Seventy-five percent of claims have a normal distribution with a mean of 3,000 and a variance of 1,000,000. The remaining 25% have a normal distribution with a mean of 4,000 and a variance of 1,000,000. Determine the probability that a randomly selected claim exceeds 5,000.
My attempt at a solution:
So let $X\sim N(3000, 10^6)$ and let $Y\sim N(4000, 10^6).$ I assume that $X$ and $Y$ are independent.
So if I define the random variable $W=0.75X+0.25Y$, $W\sim N(3250, 625000)$ since it's just a linear combination of independent normal random variables. This implies that:
Clearly this is not the same answer. I'm not entirely sure which is right; if I've gone wrong somewhere in my solution, where was it?