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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!

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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...
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  • $\begingroup$ oh wow, I get it. thank you so much! $\endgroup$ – Nguyên Chương Jan 16 at 16:15
  • $\begingroup$ 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. $\endgroup$ – Nguyên Chương Jan 16 at 16:40
  • $\begingroup$ @NguyênChương do the first part, what are the values of $a$ and $b$? $\endgroup$ – gt6989b Jan 16 at 17:24
  • $\begingroup$ T = 10X + 85, is that right ? $\endgroup$ – Nguyên Chương Jan 16 at 18:36
  • $\begingroup$ @NguyênChương $b = \sqrt{10}$ and $a$ is correct. Now plug them in, what do you get? $\endgroup$ – gt6989b Jan 16 at 21:53

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