How to find sampling distribution of sample mean












1












$begingroup$


So suppose $Y$ takes values $0$ and $1$ with probabilities



$Pr(Y=1)=p=0.78$ and $Pr(Y=0)=1-p=0.22$



I calculated the mean of Y, which is $0.78$ and the variance of Y, which is $0.1716$.



I also know that the sampling distribution of the sample mean depends on n. For example, let $n=2$, and I want to calculate the sampling distribution of the sample mean .



$overline Y$=sample mean



On the answers sheet, it states that the sampling distribution of the sample mean is :



$Pr(overline Y=0)=(1-p)^2=0.22^2=0.0484$



$Pr(overline Y=0.5)=2·p(1-p)=2·0.22·0.78=0.3432$



$Pr(overline Y=1)=p^2=0.78^2=0.6084$



Now I have two questions:



1) where is the $overline Y=0, overline Y=0.5,overline Y=1$ coming from ? Meaning, where are the $0,0.5,1$ coming from?



2) Once I know the $0,0.5,1$, how to i calculate the sampling distributions? (meaning, where are the $(1-p)^2,2·p(1-p),p^2$ coming from ?)



Thanks for the help, I would greatly appreciate simple and clear answers!






statistics probability-distributions sampling







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$endgroup$

















    1












    $begingroup$


    So suppose $Y$ takes values $0$ and $1$ with probabilities



    $Pr(Y=1)=p=0.78$ and $Pr(Y=0)=1-p=0.22$



    I calculated the mean of Y, which is $0.78$ and the variance of Y, which is $0.1716$.



    I also know that the sampling distribution of the sample mean depends on n. For example, let $n=2$, and I want to calculate the sampling distribution of the sample mean .



    $overline Y$=sample mean



    On the answers sheet, it states that the sampling distribution of the sample mean is :



    $Pr(overline Y=0)=(1-p)^2=0.22^2=0.0484$



    $Pr(overline Y=0.5)=2·p(1-p)=2·0.22·0.78=0.3432$



    $Pr(overline Y=1)=p^2=0.78^2=0.6084$



    Now I have two questions:



    1) where is the $overline Y=0, overline Y=0.5,overline Y=1$ coming from ? Meaning, where are the $0,0.5,1$ coming from?



    2) Once I know the $0,0.5,1$, how to i calculate the sampling distributions? (meaning, where are the $(1-p)^2,2·p(1-p),p^2$ coming from ?)



    Thanks for the help, I would greatly appreciate simple and clear answers!






    statistics probability-distributions sampling







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      So suppose $Y$ takes values $0$ and $1$ with probabilities



      $Pr(Y=1)=p=0.78$ and $Pr(Y=0)=1-p=0.22$



      I calculated the mean of Y, which is $0.78$ and the variance of Y, which is $0.1716$.



      I also know that the sampling distribution of the sample mean depends on n. For example, let $n=2$, and I want to calculate the sampling distribution of the sample mean .



      $overline Y$=sample mean



      On the answers sheet, it states that the sampling distribution of the sample mean is :



      $Pr(overline Y=0)=(1-p)^2=0.22^2=0.0484$



      $Pr(overline Y=0.5)=2·p(1-p)=2·0.22·0.78=0.3432$



      $Pr(overline Y=1)=p^2=0.78^2=0.6084$



      Now I have two questions:



      1) where is the $overline Y=0, overline Y=0.5,overline Y=1$ coming from ? Meaning, where are the $0,0.5,1$ coming from?



      2) Once I know the $0,0.5,1$, how to i calculate the sampling distributions? (meaning, where are the $(1-p)^2,2·p(1-p),p^2$ coming from ?)



      Thanks for the help, I would greatly appreciate simple and clear answers!






      statistics probability-distributions sampling







      share|cite|improve this question











      $endgroup$




      So suppose $Y$ takes values $0$ and $1$ with probabilities



      $Pr(Y=1)=p=0.78$ and $Pr(Y=0)=1-p=0.22$



      I calculated the mean of Y, which is $0.78$ and the variance of Y, which is $0.1716$.



      I also know that the sampling distribution of the sample mean depends on n. For example, let $n=2$, and I want to calculate the sampling distribution of the sample mean .



      $overline Y$=sample mean



      On the answers sheet, it states that the sampling distribution of the sample mean is :



      $Pr(overline Y=0)=(1-p)^2=0.22^2=0.0484$



      $Pr(overline Y=0.5)=2·p(1-p)=2·0.22·0.78=0.3432$



      $Pr(overline Y=1)=p^2=0.78^2=0.6084$



      Now I have two questions:



      1) where is the $overline Y=0, overline Y=0.5,overline Y=1$ coming from ? Meaning, where are the $0,0.5,1$ coming from?



      2) Once I know the $0,0.5,1$, how to i calculate the sampling distributions? (meaning, where are the $(1-p)^2,2·p(1-p),p^2$ coming from ?)



      Thanks for the help, I would greatly appreciate simple and clear answers!






      statistics probability-distributions sampling




      statistics probability-distributions sampling






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      share|cite|improve this question













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      edited Dec 29 '18 at 17:58









      callculus

      18.9k31729




      18.9k31729










      asked Dec 29 '18 at 17:36









      BM97BM97

      778




      778






















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          $begingroup$

          You have two (Bernoulli) independent random variables $Y_1$ and $Y_2$ (with probability of success $p=0.78$) and you want to compute the distribution of the sample mean $overline{Y} := frac{Y_1+Y_2}{2}$.



          1) Since the variables $Y_1$ and $Y_2$ can take only the values $0$ and $1$, it is clear that the sample mean can take only the values $0$ (when $Y_1 = Y_2 = 0$), $1$ (when $Y_1 = Y_2 = 1$) or $1/2$ (when $Y_1=0$, $Y_2 = 1$ or viceversa).



          2) $P(overline{Y} = 0) = P(Y_1 =0 text{and} Y_2=0) = P(Y_1=0)cdot P(Y_2=0) = (1-p)^2$ (since $Y_1$ and $Y_2$ are independent). Similarly $P(overline{Y} = 1) = p^2$.
          Finally,
          $$
          P(overline{Y} = 1/2) = P(Y_1 =0 text{and} Y_2=1) + P(Y_1 =1 text{and} Y_2=0) =
          2p(1-p).
          $$






          share|cite|improve this answer









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          • $begingroup$
            Thankyou!!!!! You are awesome. Greatly appreciated-
            $endgroup$
            – BM97
            Dec 29 '18 at 17:51












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          0












          $begingroup$

          You have two (Bernoulli) independent random variables $Y_1$ and $Y_2$ (with probability of success $p=0.78$) and you want to compute the distribution of the sample mean $overline{Y} := frac{Y_1+Y_2}{2}$.



          1) Since the variables $Y_1$ and $Y_2$ can take only the values $0$ and $1$, it is clear that the sample mean can take only the values $0$ (when $Y_1 = Y_2 = 0$), $1$ (when $Y_1 = Y_2 = 1$) or $1/2$ (when $Y_1=0$, $Y_2 = 1$ or viceversa).



          2) $P(overline{Y} = 0) = P(Y_1 =0 text{and} Y_2=0) = P(Y_1=0)cdot P(Y_2=0) = (1-p)^2$ (since $Y_1$ and $Y_2$ are independent). Similarly $P(overline{Y} = 1) = p^2$.
          Finally,
          $$
          P(overline{Y} = 1/2) = P(Y_1 =0 text{and} Y_2=1) + P(Y_1 =1 text{and} Y_2=0) =
          2p(1-p).
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thankyou!!!!! You are awesome. Greatly appreciated-
            $endgroup$
            – BM97
            Dec 29 '18 at 17:51
















          0












          $begingroup$

          You have two (Bernoulli) independent random variables $Y_1$ and $Y_2$ (with probability of success $p=0.78$) and you want to compute the distribution of the sample mean $overline{Y} := frac{Y_1+Y_2}{2}$.



          1) Since the variables $Y_1$ and $Y_2$ can take only the values $0$ and $1$, it is clear that the sample mean can take only the values $0$ (when $Y_1 = Y_2 = 0$), $1$ (when $Y_1 = Y_2 = 1$) or $1/2$ (when $Y_1=0$, $Y_2 = 1$ or viceversa).



          2) $P(overline{Y} = 0) = P(Y_1 =0 text{and} Y_2=0) = P(Y_1=0)cdot P(Y_2=0) = (1-p)^2$ (since $Y_1$ and $Y_2$ are independent). Similarly $P(overline{Y} = 1) = p^2$.
          Finally,
          $$
          P(overline{Y} = 1/2) = P(Y_1 =0 text{and} Y_2=1) + P(Y_1 =1 text{and} Y_2=0) =
          2p(1-p).
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thankyou!!!!! You are awesome. Greatly appreciated-
            $endgroup$
            – BM97
            Dec 29 '18 at 17:51














          0












          0








          0





          $begingroup$

          You have two (Bernoulli) independent random variables $Y_1$ and $Y_2$ (with probability of success $p=0.78$) and you want to compute the distribution of the sample mean $overline{Y} := frac{Y_1+Y_2}{2}$.



          1) Since the variables $Y_1$ and $Y_2$ can take only the values $0$ and $1$, it is clear that the sample mean can take only the values $0$ (when $Y_1 = Y_2 = 0$), $1$ (when $Y_1 = Y_2 = 1$) or $1/2$ (when $Y_1=0$, $Y_2 = 1$ or viceversa).



          2) $P(overline{Y} = 0) = P(Y_1 =0 text{and} Y_2=0) = P(Y_1=0)cdot P(Y_2=0) = (1-p)^2$ (since $Y_1$ and $Y_2$ are independent). Similarly $P(overline{Y} = 1) = p^2$.
          Finally,
          $$
          P(overline{Y} = 1/2) = P(Y_1 =0 text{and} Y_2=1) + P(Y_1 =1 text{and} Y_2=0) =
          2p(1-p).
          $$






          share|cite|improve this answer









          $endgroup$



          You have two (Bernoulli) independent random variables $Y_1$ and $Y_2$ (with probability of success $p=0.78$) and you want to compute the distribution of the sample mean $overline{Y} := frac{Y_1+Y_2}{2}$.



          1) Since the variables $Y_1$ and $Y_2$ can take only the values $0$ and $1$, it is clear that the sample mean can take only the values $0$ (when $Y_1 = Y_2 = 0$), $1$ (when $Y_1 = Y_2 = 1$) or $1/2$ (when $Y_1=0$, $Y_2 = 1$ or viceversa).



          2) $P(overline{Y} = 0) = P(Y_1 =0 text{and} Y_2=0) = P(Y_1=0)cdot P(Y_2=0) = (1-p)^2$ (since $Y_1$ and $Y_2$ are independent). Similarly $P(overline{Y} = 1) = p^2$.
          Finally,
          $$
          P(overline{Y} = 1/2) = P(Y_1 =0 text{and} Y_2=1) + P(Y_1 =1 text{and} Y_2=0) =
          2p(1-p).
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 29 '18 at 17:46









          RigelRigel

          11.4k11320




          11.4k11320












          • $begingroup$
            Thankyou!!!!! You are awesome. Greatly appreciated-
            $endgroup$
            – BM97
            Dec 29 '18 at 17:51


















          • $begingroup$
            Thankyou!!!!! You are awesome. Greatly appreciated-
            $endgroup$
            – BM97
            Dec 29 '18 at 17:51
















          $begingroup$
          Thankyou!!!!! You are awesome. Greatly appreciated-
          $endgroup$
          – BM97
          Dec 29 '18 at 17:51




          $begingroup$
          Thankyou!!!!! You are awesome. Greatly appreciated-
          $endgroup$
          – BM97
          Dec 29 '18 at 17:51


















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