# Conditioning on combination of normal variables

I am working on a problem that goes like this

$X_i$ ($i=1$ to $6$) are independent random variables with distributions

$X_1=X_2=X_3=X_4=X_5=X_6 \sim{}$ Normal(Mean${}=30$,Variance${}=25$)

I have to find the probability that

$X_1+X_2+X_3+X_4+X_5<180$ and

$X_1+X_2+X_3+X_4+X_5+X_6>180.$

I know that the probability of first condition being met is $0.9963$, while that of the second is $0.5$, but that does not give probability of both conditions being met simultaneously. Is there some way to alter the conditions somehow to make the problem solvable? It is known that the variables $X_i$ are independent of each other.

• Yes, all variables are independent of each other – Sarthak Nigam Aug 6 '17 at 20:35
• Notational comment: if you want to say that $X_1$ and $X_2$ are independent with the same distribution, you should not write $X_1=X_2$, which typically means $P(X_1=X_2)=1$. – angryavian Aug 6 '17 at 20:53
• Your question is equivalent to finding $P(Y<180, Y+X_6 > 180)$ where $Y \sim N(150, 125)$ and $X_6 \sim N(30,25)$. You can write $$P(Y < 180, Y+X_6 > 180) = \int_{-\infty}^{180} p_Y(y) \int_{180-y}^\infty p_{X_6}(x) \mathop{dx} \mathop{dy}$$ but I do not know if there is an easier way to compute this probability. – angryavian Aug 6 '17 at 20:53
• @angryavian Thanks for the info on notation. How do you denote equality in random variables then? Also, could you explain how do you arrive at that expression? – Sarthak Nigam Aug 7 '17 at 7:23