# Gaussian distribution with absolute value

I am doing my homework about continuous random variable and Im struggling with this problem :

Given a Gaussian random variable $$T(85,10)$$, find $$c$$ satisfying $$\mathbb{P}[|T| < c] = 0.9$$.

Could you help me with this question? Thanks a lot in advance for your help!

HINT

1. Can you find some transformation $$X = (T-a)/b$$ so that $$X \sim \mathcal{N}(0,1)$$ and $$b>0$$?
2. Then $$T = bX+a$$ and your expression becomes $$\begin{split} 0.9 &= \mathbb{P}[|T| < c] \\ &= \mathbb{P}[-c < T < c] \\ &= \mathbb{P}[-c < bX+a < c] \\ &= \mathbb{P}\left[\frac{-c-a}{b} < X < \frac{c-a}{b}\right] \\ &= \Phi\left(\frac{c-a}{b}\right) - \Phi\left(\frac{-c-a}{b}\right), \end{split}$$ where $$\Phi$$ is the standard normal CDF...
• oh wow, I get it. thank you so much! – Nguyên Chương Jan 16 at 16:15
• oops!! I still dont know how to find c, can you give me a hint ? At first, I thought that : because the probability is 0.9 then c must be greater than 85(because if c < 85 then P[|T|<c] < 0.5) , so -c < -85, then P[|T|<c] would approximately be P[T < c], which is 0.9, and then I can find c, but I'm not very sure that is the solution. – Nguyên Chương Jan 16 at 16:40
• @NguyênChương do the first part, what are the values of $a$ and $b$? – gt6989b Jan 16 at 17:24
• T = 10X + 85, is that right ? – Nguyên Chương Jan 16 at 18:36
• @NguyênChương $b = \sqrt{10}$ and $a$ is correct. Now plug them in, what do you get? – gt6989b Jan 16 at 21:53