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Given a normal distribution $X$~$N(60,9^2)$ with a random variable $A$ and a normal distribution $Y$~$N(50,7^2)$ with a random variable $B$, how do I go about finding the probability $P(B>A)$?

(Given that A and B are independent events).

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If ther are dependent you cannot do this. If they are independent then $C=B-A$ has normal distribution with mean $50-60$ and variance $9^{2}+7^{2}$. You can compute $P(C>0)$ by integrating the density function from $0$ to $\infty$.

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  • $\begingroup$ So we get to something like this? $f(x) = \begin{cases} 0, & \text{if $A≥x≥B$} \\ c, & \text{if $A≤x≤B$} \end{cases}$ $\endgroup$ – sdds Apr 17 at 8:01
  • $\begingroup$ The answer is $\int_0^{\infty} \frac 1 {\sqrt {260 \pi}} e^{-(x+10)^{2}/{260}}dx$. $\endgroup$ – Kavi Rama Murthy Apr 17 at 8:04
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Hint: If $A$ and $B$ are independent, you have that $A-B \sim \mathcal{N}(10,7^2+9^2)$ and $P(B>A) = P(A-B<0)$.

If they are dependent and are jointly normally distributed, you need to use the joint distribution.

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