Probability dice flip Central Limit Theorem problem












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Suppose you flip a fair dice 300 times. Let $X$ be the number of times a $6$ was thrown. What is the probability that $X$ is greater than 60?




How can I start with such a problem?






probability dice central-limit-theorem







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    0












    $begingroup$



    Suppose you flip a fair dice 300 times. Let $X$ be the number of times a $6$ was thrown. What is the probability that $X$ is greater than 60?




    How can I start with such a problem?






    probability dice central-limit-theorem







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$



      Suppose you flip a fair dice 300 times. Let $X$ be the number of times a $6$ was thrown. What is the probability that $X$ is greater than 60?




      How can I start with such a problem?






      probability dice central-limit-theorem







      share|cite|improve this question











      $endgroup$





      Suppose you flip a fair dice 300 times. Let $X$ be the number of times a $6$ was thrown. What is the probability that $X$ is greater than 60?




      How can I start with such a problem?






      probability dice central-limit-theorem




      probability dice central-limit-theorem






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 10 at 10:49









      jvdhooft

      5,65961641




      5,65961641










      asked Jan 30 at 2:36









      Murad DavudovMurad Davudov

      234




      234






















          1 Answer
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          $begingroup$

          Let's take $X_i$ to be $1$ if you got $6$ for the $i$-th dice.



          $E(X_i)=frac{1}{6}, Var(X_i)=frac{1}{6}-frac{1}{36}=frac{5}{36}$



          Now we are looking for $X=sum_{i=1}^{300}X_i$



          $E(X)=E(sum_{i=1}^{300}X_i)=sum_{i=1}^{300}E(X_i)=50$



          $Var(X)=Var(sum_{i=1}^{300}X_i)=sum_{i=1}^{300}Var(X_i)=300cdotfrac{5}{36}=41frac{2}{3}$



          Now we can start playing with the central limit theorem.



          When $nrightarrow infty $ you get $frac{frac{1}{n}sum_{i=1}^{n}X_i-E(X_i)}{sigma/sqrt(n)}rightarrow z$ when $z$ is a normal distribution.



          So $P(X>60)=P(X-E(X)>10)=P(frac{X}{n}-E(X_i)>frac{10}{n})= $



          $=P(frac{frac{x}{n}-E(X_i)}{frac{5}{36}/sqrt(n)}>frac{10}{frac{5}{36}sqrt(n)})= P(frac{frac{x}{n}-E(X_i)}{sigma/sqrt(n)}>4.157)=1- phi(4.157)$



          So I ended up calculating $Var(X)$ for free.



          I didn't justify every step as you suppose to, but if there is any step you don't understand you are welcome to ask.



          And I might have some calculation errors, but the concept should be correct.






          share|cite|improve this answer











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            $begingroup$

            Let's take $X_i$ to be $1$ if you got $6$ for the $i$-th dice.



            $E(X_i)=frac{1}{6}, Var(X_i)=frac{1}{6}-frac{1}{36}=frac{5}{36}$



            Now we are looking for $X=sum_{i=1}^{300}X_i$



            $E(X)=E(sum_{i=1}^{300}X_i)=sum_{i=1}^{300}E(X_i)=50$



            $Var(X)=Var(sum_{i=1}^{300}X_i)=sum_{i=1}^{300}Var(X_i)=300cdotfrac{5}{36}=41frac{2}{3}$



            Now we can start playing with the central limit theorem.



            When $nrightarrow infty $ you get $frac{frac{1}{n}sum_{i=1}^{n}X_i-E(X_i)}{sigma/sqrt(n)}rightarrow z$ when $z$ is a normal distribution.



            So $P(X>60)=P(X-E(X)>10)=P(frac{X}{n}-E(X_i)>frac{10}{n})= $



            $=P(frac{frac{x}{n}-E(X_i)}{frac{5}{36}/sqrt(n)}>frac{10}{frac{5}{36}sqrt(n)})= P(frac{frac{x}{n}-E(X_i)}{sigma/sqrt(n)}>4.157)=1- phi(4.157)$



            So I ended up calculating $Var(X)$ for free.



            I didn't justify every step as you suppose to, but if there is any step you don't understand you are welcome to ask.



            And I might have some calculation errors, but the concept should be correct.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              Let's take $X_i$ to be $1$ if you got $6$ for the $i$-th dice.



              $E(X_i)=frac{1}{6}, Var(X_i)=frac{1}{6}-frac{1}{36}=frac{5}{36}$



              Now we are looking for $X=sum_{i=1}^{300}X_i$



              $E(X)=E(sum_{i=1}^{300}X_i)=sum_{i=1}^{300}E(X_i)=50$



              $Var(X)=Var(sum_{i=1}^{300}X_i)=sum_{i=1}^{300}Var(X_i)=300cdotfrac{5}{36}=41frac{2}{3}$



              Now we can start playing with the central limit theorem.



              When $nrightarrow infty $ you get $frac{frac{1}{n}sum_{i=1}^{n}X_i-E(X_i)}{sigma/sqrt(n)}rightarrow z$ when $z$ is a normal distribution.



              So $P(X>60)=P(X-E(X)>10)=P(frac{X}{n}-E(X_i)>frac{10}{n})= $



              $=P(frac{frac{x}{n}-E(X_i)}{frac{5}{36}/sqrt(n)}>frac{10}{frac{5}{36}sqrt(n)})= P(frac{frac{x}{n}-E(X_i)}{sigma/sqrt(n)}>4.157)=1- phi(4.157)$



              So I ended up calculating $Var(X)$ for free.



              I didn't justify every step as you suppose to, but if there is any step you don't understand you are welcome to ask.



              And I might have some calculation errors, but the concept should be correct.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                Let's take $X_i$ to be $1$ if you got $6$ for the $i$-th dice.



                $E(X_i)=frac{1}{6}, Var(X_i)=frac{1}{6}-frac{1}{36}=frac{5}{36}$



                Now we are looking for $X=sum_{i=1}^{300}X_i$



                $E(X)=E(sum_{i=1}^{300}X_i)=sum_{i=1}^{300}E(X_i)=50$



                $Var(X)=Var(sum_{i=1}^{300}X_i)=sum_{i=1}^{300}Var(X_i)=300cdotfrac{5}{36}=41frac{2}{3}$



                Now we can start playing with the central limit theorem.



                When $nrightarrow infty $ you get $frac{frac{1}{n}sum_{i=1}^{n}X_i-E(X_i)}{sigma/sqrt(n)}rightarrow z$ when $z$ is a normal distribution.



                So $P(X>60)=P(X-E(X)>10)=P(frac{X}{n}-E(X_i)>frac{10}{n})= $



                $=P(frac{frac{x}{n}-E(X_i)}{frac{5}{36}/sqrt(n)}>frac{10}{frac{5}{36}sqrt(n)})= P(frac{frac{x}{n}-E(X_i)}{sigma/sqrt(n)}>4.157)=1- phi(4.157)$



                So I ended up calculating $Var(X)$ for free.



                I didn't justify every step as you suppose to, but if there is any step you don't understand you are welcome to ask.



                And I might have some calculation errors, but the concept should be correct.






                share|cite|improve this answer











                $endgroup$



                Let's take $X_i$ to be $1$ if you got $6$ for the $i$-th dice.



                $E(X_i)=frac{1}{6}, Var(X_i)=frac{1}{6}-frac{1}{36}=frac{5}{36}$



                Now we are looking for $X=sum_{i=1}^{300}X_i$



                $E(X)=E(sum_{i=1}^{300}X_i)=sum_{i=1}^{300}E(X_i)=50$



                $Var(X)=Var(sum_{i=1}^{300}X_i)=sum_{i=1}^{300}Var(X_i)=300cdotfrac{5}{36}=41frac{2}{3}$



                Now we can start playing with the central limit theorem.



                When $nrightarrow infty $ you get $frac{frac{1}{n}sum_{i=1}^{n}X_i-E(X_i)}{sigma/sqrt(n)}rightarrow z$ when $z$ is a normal distribution.



                So $P(X>60)=P(X-E(X)>10)=P(frac{X}{n}-E(X_i)>frac{10}{n})= $



                $=P(frac{frac{x}{n}-E(X_i)}{frac{5}{36}/sqrt(n)}>frac{10}{frac{5}{36}sqrt(n)})= P(frac{frac{x}{n}-E(X_i)}{sigma/sqrt(n)}>4.157)=1- phi(4.157)$



                So I ended up calculating $Var(X)$ for free.



                I didn't justify every step as you suppose to, but if there is any step you don't understand you are welcome to ask.



                And I might have some calculation errors, but the concept should be correct.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Feb 10 at 10:51









                jvdhooft

                5,65961641




                5,65961641










                answered Jan 30 at 8:59









                ShaqShaq

                3049




                3049






























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