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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:

$P(W>5000)=1-\Phi(\frac{5000-3250}{\sqrt{625000}})=0.0134$.

However, apparently the solution is given by this: enter image description here

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?

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    $\begingroup$ There is a distinction between exactly a 75/25 split (deterministic), and a probability .75 of one and .25 of the other. In the latter case you have to account for the variance of the Bernoulli split (as per @drhab's Answer. $\endgroup$
    – BruceET
    Feb 27, 2018 at 9:26

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To make things clear suppose that $X\sim N(\mu,\sigma^2)$ and $Y\sim N(\mu,\sigma^2)$ so both the same.

Then there is no split up and it would be just a matter of finding $P(W>5000)$ where $W\sim N(\mu,\sigma^2)$.

But it seems you would go for finding $P(0.25X+0.75Y>5000)$.

Now observe that the variance of $0.25X+0.75Y$ is not $\sigma^2$ (as it should) but is $\frac{10}{16}\sigma^2$.

The correct route is working with $W=BX+(1-B)Y$ where $B\sim\text{Bernoulli}(0.25)$ and where $B,X,Y$ are independent.

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  • $\begingroup$ could you please detail "Now observe that the variance .. is not $\sigma^2$ .." $\endgroup$
    – G Cab
    Feb 27, 2018 at 9:54
  • $\begingroup$ If $X$ and $Y$ are independent then so are $aX$ and $bX$ so that $Var(aX+bY)=Var(aX)+Var(bX)=a^2Var(X)+b^2Var(Y)$. Here $a=0.25$ and $b=0.75$. $\endgroup$
    – drhab
    Feb 27, 2018 at 9:58
  • $\begingroup$ I see it now, thanks: the weighted average of two variables with same variance produces a decrease of the variance. $\endgroup$
    – G Cab
    Feb 27, 2018 at 10:27
  • $\begingroup$ Indeed. You are welcome. $\endgroup$
    – drhab
    Feb 27, 2018 at 10:29

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