Gaussian distribution with absolute value












0












$begingroup$


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!






probability absolute-value







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    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!






    probability absolute-value







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      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!






      probability absolute-value







      share|cite|improve this question











      $endgroup$




      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!






      probability absolute-value




      probability absolute-value






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 16 at 16:10









      gt6989b

      34k22455




      34k22455










      asked Jan 16 at 16:02









      Nguyên ChươngNguyên Chương

      51




      51






















          1 Answer
          1






          active

          oldest

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          0












          $begingroup$

          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
            $$
            begin{split}
            0.9 &= mathbb{P}[|T| < c] \
            &= mathbb{P}[-c < T < c] \
            &= mathbb{P}[-c < bX+a < c] \
            &= mathbb{P}left[frac{-c-a}{b} < X < frac{c-a}{b}right] \
            &= Phileft(frac{c-a}{b}right) - Phileft(frac{-c-a}{b}right),
            end{split}
            $$
            where $Phi$ is the standard normal CDF...






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          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
            $$
            begin{split}
            0.9 &= mathbb{P}[|T| < c] \
            &= mathbb{P}[-c < T < c] \
            &= mathbb{P}[-c < bX+a < c] \
            &= mathbb{P}left[frac{-c-a}{b} < X < frac{c-a}{b}right] \
            &= Phileft(frac{c-a}{b}right) - Phileft(frac{-c-a}{b}right),
            end{split}
            $$
            where $Phi$ is the standard normal CDF...






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
















          0












          $begingroup$

          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
            $$
            begin{split}
            0.9 &= mathbb{P}[|T| < c] \
            &= mathbb{P}[-c < T < c] \
            &= mathbb{P}[-c < bX+a < c] \
            &= mathbb{P}left[frac{-c-a}{b} < X < frac{c-a}{b}right] \
            &= Phileft(frac{c-a}{b}right) - Phileft(frac{-c-a}{b}right),
            end{split}
            $$
            where $Phi$ is the standard normal CDF...






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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














          0












          0








          0





          $begingroup$

          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
            $$
            begin{split}
            0.9 &= mathbb{P}[|T| < c] \
            &= mathbb{P}[-c < T < c] \
            &= mathbb{P}[-c < bX+a < c] \
            &= mathbb{P}left[frac{-c-a}{b} < X < frac{c-a}{b}right] \
            &= Phileft(frac{c-a}{b}right) - Phileft(frac{-c-a}{b}right),
            end{split}
            $$
            where $Phi$ is the standard normal CDF...






          share|cite|improve this answer









          $endgroup$



          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
            $$
            begin{split}
            0.9 &= mathbb{P}[|T| < c] \
            &= mathbb{P}[-c < T < c] \
            &= mathbb{P}[-c < bX+a < c] \
            &= mathbb{P}left[frac{-c-a}{b} < X < frac{c-a}{b}right] \
            &= Phileft(frac{c-a}{b}right) - Phileft(frac{-c-a}{b}right),
            end{split}
            $$
            where $Phi$ is the standard normal CDF...







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 16 at 16:07









          gt6989bgt6989b

          34k22455




          34k22455












          • $begingroup$
            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


















          • $begingroup$
            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
















          $begingroup$
          oh wow, I get it. thank you so much!
          $endgroup$
          – Nguyên Chương
          Jan 16 at 16:15




          $begingroup$
          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




          $begingroup$
          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




          $begingroup$
          @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




          $begingroup$
          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




          $begingroup$
          @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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