Determining $n$ for which the confidence interval will have level of confidence >= 0.9











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I have come to the point where there is an $Xtext{~}N(a,frac{9}{n})$ and $psi$ is its CDF, I have to determine the smallest n for which $psi(a+1)-psi(a-1)>=0.9$ I have no idea how to approach this, any help appriciated!










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    I have come to the point where there is an $Xtext{~}N(a,frac{9}{n})$ and $psi$ is its CDF, I have to determine the smallest n for which $psi(a+1)-psi(a-1)>=0.9$ I have no idea how to approach this, any help appriciated!










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      I have come to the point where there is an $Xtext{~}N(a,frac{9}{n})$ and $psi$ is its CDF, I have to determine the smallest n for which $psi(a+1)-psi(a-1)>=0.9$ I have no idea how to approach this, any help appriciated!










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      I have come to the point where there is an $Xtext{~}N(a,frac{9}{n})$ and $psi$ is its CDF, I have to determine the smallest n for which $psi(a+1)-psi(a-1)>=0.9$ I have no idea how to approach this, any help appriciated!







      statistics normal-distribution confidence-interval






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      asked Nov 22 at 19:58









      ryszard eggink

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          $psi(a+1)-psi(a-1)=Phi(1/sqrt{(9/n)})-Phi(-1/sqrt{(9/n)})=2Phi(1/sqrt{(9/n)})-1geq 0.9Rightarrow Phi(1/sqrt{(9/n)})geq (1+0.9)/2=0.95 Rightarrow sqrt{n}/3geq Phi^{-1}(0.95)approx 1.64 Rightarrow ngeq (3times 1.64)^2=24.20Rightarrow ngeq 25.$



          Here $Phi$ is cdf of $N(0,1).$






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            Hint. $psi(a+1)-psi(a-1)=mathbf{P}(a - 1 < X < a + 1) = mathbf{P}(-1 < X - a < 1) = ...$ then use a table for standard normal.






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              2 Answers
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              2 Answers
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              $psi(a+1)-psi(a-1)=Phi(1/sqrt{(9/n)})-Phi(-1/sqrt{(9/n)})=2Phi(1/sqrt{(9/n)})-1geq 0.9Rightarrow Phi(1/sqrt{(9/n)})geq (1+0.9)/2=0.95 Rightarrow sqrt{n}/3geq Phi^{-1}(0.95)approx 1.64 Rightarrow ngeq (3times 1.64)^2=24.20Rightarrow ngeq 25.$



              Here $Phi$ is cdf of $N(0,1).$






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                up vote
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                $psi(a+1)-psi(a-1)=Phi(1/sqrt{(9/n)})-Phi(-1/sqrt{(9/n)})=2Phi(1/sqrt{(9/n)})-1geq 0.9Rightarrow Phi(1/sqrt{(9/n)})geq (1+0.9)/2=0.95 Rightarrow sqrt{n}/3geq Phi^{-1}(0.95)approx 1.64 Rightarrow ngeq (3times 1.64)^2=24.20Rightarrow ngeq 25.$



                Here $Phi$ is cdf of $N(0,1).$






                share|cite|improve this answer























                  up vote
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                  up vote
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                  accepted






                  $psi(a+1)-psi(a-1)=Phi(1/sqrt{(9/n)})-Phi(-1/sqrt{(9/n)})=2Phi(1/sqrt{(9/n)})-1geq 0.9Rightarrow Phi(1/sqrt{(9/n)})geq (1+0.9)/2=0.95 Rightarrow sqrt{n}/3geq Phi^{-1}(0.95)approx 1.64 Rightarrow ngeq (3times 1.64)^2=24.20Rightarrow ngeq 25.$



                  Here $Phi$ is cdf of $N(0,1).$






                  share|cite|improve this answer












                  $psi(a+1)-psi(a-1)=Phi(1/sqrt{(9/n)})-Phi(-1/sqrt{(9/n)})=2Phi(1/sqrt{(9/n)})-1geq 0.9Rightarrow Phi(1/sqrt{(9/n)})geq (1+0.9)/2=0.95 Rightarrow sqrt{n}/3geq Phi^{-1}(0.95)approx 1.64 Rightarrow ngeq (3times 1.64)^2=24.20Rightarrow ngeq 25.$



                  Here $Phi$ is cdf of $N(0,1).$







                  share|cite|improve this answer












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                  answered Nov 22 at 20:08









                  John_Wick

                  1,199111




                  1,199111






















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                      Hint. $psi(a+1)-psi(a-1)=mathbf{P}(a - 1 < X < a + 1) = mathbf{P}(-1 < X - a < 1) = ...$ then use a table for standard normal.






                      share|cite|improve this answer

























                        up vote
                        1
                        down vote













                        Hint. $psi(a+1)-psi(a-1)=mathbf{P}(a - 1 < X < a + 1) = mathbf{P}(-1 < X - a < 1) = ...$ then use a table for standard normal.






                        share|cite|improve this answer























                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









                          Hint. $psi(a+1)-psi(a-1)=mathbf{P}(a - 1 < X < a + 1) = mathbf{P}(-1 < X - a < 1) = ...$ then use a table for standard normal.






                          share|cite|improve this answer












                          Hint. $psi(a+1)-psi(a-1)=mathbf{P}(a - 1 < X < a + 1) = mathbf{P}(-1 < X - a < 1) = ...$ then use a table for standard normal.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 22 at 20:06









                          Will M.

                          2,337313




                          2,337313






























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