(Random Sampling + Approximate Probability + Normal Approximation)












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Suppose that 16% of the population of the U.S. is left-handed. If a random sample of 130 people from the U.S. is chosen, approximate the probability that at least 20 are left-handed. Use the normal approximation to the binomial with a correction for continuity. Round your answer to at least three decimal places. Do not round any intermediate steps.



======================



MY CALCULATIONS:



0.16 x 130 = 20.8



Answer = 20.800?






probability statistics random







share|cite|improve this question









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    0












    $begingroup$


    Suppose that 16% of the population of the U.S. is left-handed. If a random sample of 130 people from the U.S. is chosen, approximate the probability that at least 20 are left-handed. Use the normal approximation to the binomial with a correction for continuity. Round your answer to at least three decimal places. Do not round any intermediate steps.



    ======================



    MY CALCULATIONS:



    0.16 x 130 = 20.8



    Answer = 20.800?






    probability statistics random







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Suppose that 16% of the population of the U.S. is left-handed. If a random sample of 130 people from the U.S. is chosen, approximate the probability that at least 20 are left-handed. Use the normal approximation to the binomial with a correction for continuity. Round your answer to at least three decimal places. Do not round any intermediate steps.



      ======================



      MY CALCULATIONS:



      0.16 x 130 = 20.8



      Answer = 20.800?






      probability statistics random







      share|cite|improve this question









      $endgroup$




      Suppose that 16% of the population of the U.S. is left-handed. If a random sample of 130 people from the U.S. is chosen, approximate the probability that at least 20 are left-handed. Use the normal approximation to the binomial with a correction for continuity. Round your answer to at least three decimal places. Do not round any intermediate steps.



      ======================



      MY CALCULATIONS:



      0.16 x 130 = 20.8



      Answer = 20.800?






      probability statistics random




      probability statistics random






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked May 8 '15 at 8:12









      DinosaursDinosaurs

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          First you have to use the cdf of the binomial distribution:



          $P(X geq 20)=sum_{x=20}^{130}{130 choose x}cdot left(0.16right)^xcdotleft(0.84right)^{130-x} $



          Applying converse probability



          $P(X geq 20)=1-P(X leq 19)=1-sum_{x=0}^{19} {130 choose x}cdot left(0.16right)^xcdotleft(0.84right)^{1000-x} $



          Approximation of the Binomial distribution by the Normal distribution:



          $$P(X geq 20)=1-Phileft(frac{19+0.5-0.16cdot 130}{sqrt{130cdot 0.16 cdot 0.84}} right)$$



          0.5 is the continuity correction factor. And $Phi(cdot)$ is the cdf of the Standard Normal distribution. You can do the approximation, because $ncdot p cdot (1-p) >9$ (rule of thumb).



          I think you can proceed from here.






          share|cite|improve this answer









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            $begingroup$

            First you have to use the cdf of the binomial distribution:



            $P(X geq 20)=sum_{x=20}^{130}{130 choose x}cdot left(0.16right)^xcdotleft(0.84right)^{130-x} $



            Applying converse probability



            $P(X geq 20)=1-P(X leq 19)=1-sum_{x=0}^{19} {130 choose x}cdot left(0.16right)^xcdotleft(0.84right)^{1000-x} $



            Approximation of the Binomial distribution by the Normal distribution:



            $$P(X geq 20)=1-Phileft(frac{19+0.5-0.16cdot 130}{sqrt{130cdot 0.16 cdot 0.84}} right)$$



            0.5 is the continuity correction factor. And $Phi(cdot)$ is the cdf of the Standard Normal distribution. You can do the approximation, because $ncdot p cdot (1-p) >9$ (rule of thumb).



            I think you can proceed from here.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              First you have to use the cdf of the binomial distribution:



              $P(X geq 20)=sum_{x=20}^{130}{130 choose x}cdot left(0.16right)^xcdotleft(0.84right)^{130-x} $



              Applying converse probability



              $P(X geq 20)=1-P(X leq 19)=1-sum_{x=0}^{19} {130 choose x}cdot left(0.16right)^xcdotleft(0.84right)^{1000-x} $



              Approximation of the Binomial distribution by the Normal distribution:



              $$P(X geq 20)=1-Phileft(frac{19+0.5-0.16cdot 130}{sqrt{130cdot 0.16 cdot 0.84}} right)$$



              0.5 is the continuity correction factor. And $Phi(cdot)$ is the cdf of the Standard Normal distribution. You can do the approximation, because $ncdot p cdot (1-p) >9$ (rule of thumb).



              I think you can proceed from here.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                First you have to use the cdf of the binomial distribution:



                $P(X geq 20)=sum_{x=20}^{130}{130 choose x}cdot left(0.16right)^xcdotleft(0.84right)^{130-x} $



                Applying converse probability



                $P(X geq 20)=1-P(X leq 19)=1-sum_{x=0}^{19} {130 choose x}cdot left(0.16right)^xcdotleft(0.84right)^{1000-x} $



                Approximation of the Binomial distribution by the Normal distribution:



                $$P(X geq 20)=1-Phileft(frac{19+0.5-0.16cdot 130}{sqrt{130cdot 0.16 cdot 0.84}} right)$$



                0.5 is the continuity correction factor. And $Phi(cdot)$ is the cdf of the Standard Normal distribution. You can do the approximation, because $ncdot p cdot (1-p) >9$ (rule of thumb).



                I think you can proceed from here.






                share|cite|improve this answer









                $endgroup$



                First you have to use the cdf of the binomial distribution:



                $P(X geq 20)=sum_{x=20}^{130}{130 choose x}cdot left(0.16right)^xcdotleft(0.84right)^{130-x} $



                Applying converse probability



                $P(X geq 20)=1-P(X leq 19)=1-sum_{x=0}^{19} {130 choose x}cdot left(0.16right)^xcdotleft(0.84right)^{1000-x} $



                Approximation of the Binomial distribution by the Normal distribution:



                $$P(X geq 20)=1-Phileft(frac{19+0.5-0.16cdot 130}{sqrt{130cdot 0.16 cdot 0.84}} right)$$



                0.5 is the continuity correction factor. And $Phi(cdot)$ is the cdf of the Standard Normal distribution. You can do the approximation, because $ncdot p cdot (1-p) >9$ (rule of thumb).



                I think you can proceed from here.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered May 8 '15 at 10:23









                callculuscallculus

                18.5k31428




                18.5k31428






























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