Let X be the average of a sample of 16 independent normal random variables with mean 0 and variance 1....












1












$begingroup$


Let $overline{X}$ be the average of a sample of $16$ independent normal random variables with mean $0$ and variance $1$. Determine c such that
$P(| overline{X} | < c) = .5$



I am having a lot of trouble with this question. I know it is related to chi-square but I don't know how to even start.






statistics







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why chi-square? Distribution of $bar X$ is known, which should be enough to find $c$.
    $endgroup$
    – StubbornAtom
    Jan 8 at 17:38
















1












$begingroup$


Let $overline{X}$ be the average of a sample of $16$ independent normal random variables with mean $0$ and variance $1$. Determine c such that
$P(| overline{X} | < c) = .5$



I am having a lot of trouble with this question. I know it is related to chi-square but I don't know how to even start.






statistics







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why chi-square? Distribution of $bar X$ is known, which should be enough to find $c$.
    $endgroup$
    – StubbornAtom
    Jan 8 at 17:38














1












1








1





$begingroup$


Let $overline{X}$ be the average of a sample of $16$ independent normal random variables with mean $0$ and variance $1$. Determine c such that
$P(| overline{X} | < c) = .5$



I am having a lot of trouble with this question. I know it is related to chi-square but I don't know how to even start.






statistics







share|cite|improve this question











$endgroup$




Let $overline{X}$ be the average of a sample of $16$ independent normal random variables with mean $0$ and variance $1$. Determine c such that
$P(| overline{X} | < c) = .5$



I am having a lot of trouble with this question. I know it is related to chi-square but I don't know how to even start.






statistics




statistics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 18:34









Sauhard Sharma

953318




953318










asked Jan 8 at 17:10









George HarrisonGeorge Harrison

133




133












  • $begingroup$
    Why chi-square? Distribution of $bar X$ is known, which should be enough to find $c$.
    $endgroup$
    – StubbornAtom
    Jan 8 at 17:38


















  • $begingroup$
    Why chi-square? Distribution of $bar X$ is known, which should be enough to find $c$.
    $endgroup$
    – StubbornAtom
    Jan 8 at 17:38
















$begingroup$
Why chi-square? Distribution of $bar X$ is known, which should be enough to find $c$.
$endgroup$
– StubbornAtom
Jan 8 at 17:38




$begingroup$
Why chi-square? Distribution of $bar X$ is known, which should be enough to find $c$.
$endgroup$
– StubbornAtom
Jan 8 at 17:38










2 Answers
2






active

oldest

votes


















0












$begingroup$

If $X$ is distributed as $Xsim mathcal N(0,1)$, then $frac1n sumlimits_{i=1}^{16} X_i=overline X$ is distributed as $overline Xsim mathcal Nleft( 0, frac1{16}right)$



Now we have to evaluate how $|overline X|$ is distributed in terms of $overline X$



$P(|overline X| leq c)=P(-c leq overline X leq c)$



$=P(overline X leq c)-P(overline X leq -c)$



$=P(overline X leq c)-left[ 1-P(overline X leq c) right]$



$=2 cdot P(overline X leq c)-1=2cdot F_{overline X}(c)-1$



At the next step we standardize $overline X$ to be able to use the cdf of the standard normal distribution.



$2cdot P(overline X leq c)-1=2cdot Phileft(frac{c-0}{sqrt{frac1{16}}} right)-1=2cdot Phileft(4c right)-1=0.5$



$2cdot Phileft(4c right)=1.5$



$Phileft(4c right)=0.75$



$4c=Phi^{-1}left(0.75right)$



$Phi^{-1}left(pright)$ is the inverse function of $Phileft(zright)$



$4c=0.674Rightarrow c=0.1685$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I am kind of lost on the step where we standardize X. Everything before that makes sense.
    $endgroup$
    – George Harrison
    Jan 8 at 21:25












  • $begingroup$
    @GeorgeHarrison Let $overline X sim mathcal N(mu, sigma^2)$, then $Z=frac{overline X-mu}{sigma}sim N(0,1)$. In your exercise $mu=0$ and $sigma=sqrt{frac1{16}}=frac14$.
    $endgroup$
    – callculus
    Jan 8 at 21:31












  • $begingroup$
    I am not getting where you are getting the mean is 0 and the standard deviation is 0.25.
    $endgroup$
    – George Harrison
    Jan 8 at 21:36










  • $begingroup$
    In your example $P(overline Xleq c)$ is the same as $P(Zleq 4c)$ due standardizing.
    $endgroup$
    – callculus
    Jan 8 at 21:37










  • $begingroup$
    @GeorgeHarrison The expected value of the mean of n variables is the mean of the expected values. And for $n$ i.i.d random variables we have $Varleft (frac1n sumlimits_{i=1}^n X_i right)=frac1{n^2}cdot Varleft (sumlimits_{i=1}^n X_i right)=Varleft (sumlimits_{i=1}^n X_i right)$ $=frac1{n^2}cdotleft(Var(X_1)+Var(X_2)+ldots+Var(X_n)right)$. Since the $X_i$´s has the same variance we get: $=frac1{n^2}cdot ncdot Var(X_1)=frac1ncdot sigma^2$
    $endgroup$
    – callculus
    Jan 8 at 21:48





















0












$begingroup$

HINT- The sum of independent normal variables also follows the normal distribution with



$$N(sum_{i=1}^n {mu_i},sum_{i=1}^n {sigma_i^2})$$



Also, if a random variable $X$ has mean $mu$ and variance $sigma^2$, then the random variable $Y=kX$ (where $k$ is a constant) has mean $kmu$ and variance $k^2sigma^2$






share|cite|improve this answer









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    2 Answers
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    2 Answers
    2






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    oldest

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    active

    oldest

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    0












    $begingroup$

    If $X$ is distributed as $Xsim mathcal N(0,1)$, then $frac1n sumlimits_{i=1}^{16} X_i=overline X$ is distributed as $overline Xsim mathcal Nleft( 0, frac1{16}right)$



    Now we have to evaluate how $|overline X|$ is distributed in terms of $overline X$



    $P(|overline X| leq c)=P(-c leq overline X leq c)$



    $=P(overline X leq c)-P(overline X leq -c)$



    $=P(overline X leq c)-left[ 1-P(overline X leq c) right]$



    $=2 cdot P(overline X leq c)-1=2cdot F_{overline X}(c)-1$



    At the next step we standardize $overline X$ to be able to use the cdf of the standard normal distribution.



    $2cdot P(overline X leq c)-1=2cdot Phileft(frac{c-0}{sqrt{frac1{16}}} right)-1=2cdot Phileft(4c right)-1=0.5$



    $2cdot Phileft(4c right)=1.5$



    $Phileft(4c right)=0.75$



    $4c=Phi^{-1}left(0.75right)$



    $Phi^{-1}left(pright)$ is the inverse function of $Phileft(zright)$



    $4c=0.674Rightarrow c=0.1685$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I am kind of lost on the step where we standardize X. Everything before that makes sense.
      $endgroup$
      – George Harrison
      Jan 8 at 21:25












    • $begingroup$
      @GeorgeHarrison Let $overline X sim mathcal N(mu, sigma^2)$, then $Z=frac{overline X-mu}{sigma}sim N(0,1)$. In your exercise $mu=0$ and $sigma=sqrt{frac1{16}}=frac14$.
      $endgroup$
      – callculus
      Jan 8 at 21:31












    • $begingroup$
      I am not getting where you are getting the mean is 0 and the standard deviation is 0.25.
      $endgroup$
      – George Harrison
      Jan 8 at 21:36










    • $begingroup$
      In your example $P(overline Xleq c)$ is the same as $P(Zleq 4c)$ due standardizing.
      $endgroup$
      – callculus
      Jan 8 at 21:37










    • $begingroup$
      @GeorgeHarrison The expected value of the mean of n variables is the mean of the expected values. And for $n$ i.i.d random variables we have $Varleft (frac1n sumlimits_{i=1}^n X_i right)=frac1{n^2}cdot Varleft (sumlimits_{i=1}^n X_i right)=Varleft (sumlimits_{i=1}^n X_i right)$ $=frac1{n^2}cdotleft(Var(X_1)+Var(X_2)+ldots+Var(X_n)right)$. Since the $X_i$´s has the same variance we get: $=frac1{n^2}cdot ncdot Var(X_1)=frac1ncdot sigma^2$
      $endgroup$
      – callculus
      Jan 8 at 21:48


















    0












    $begingroup$

    If $X$ is distributed as $Xsim mathcal N(0,1)$, then $frac1n sumlimits_{i=1}^{16} X_i=overline X$ is distributed as $overline Xsim mathcal Nleft( 0, frac1{16}right)$



    Now we have to evaluate how $|overline X|$ is distributed in terms of $overline X$



    $P(|overline X| leq c)=P(-c leq overline X leq c)$



    $=P(overline X leq c)-P(overline X leq -c)$



    $=P(overline X leq c)-left[ 1-P(overline X leq c) right]$



    $=2 cdot P(overline X leq c)-1=2cdot F_{overline X}(c)-1$



    At the next step we standardize $overline X$ to be able to use the cdf of the standard normal distribution.



    $2cdot P(overline X leq c)-1=2cdot Phileft(frac{c-0}{sqrt{frac1{16}}} right)-1=2cdot Phileft(4c right)-1=0.5$



    $2cdot Phileft(4c right)=1.5$



    $Phileft(4c right)=0.75$



    $4c=Phi^{-1}left(0.75right)$



    $Phi^{-1}left(pright)$ is the inverse function of $Phileft(zright)$



    $4c=0.674Rightarrow c=0.1685$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I am kind of lost on the step where we standardize X. Everything before that makes sense.
      $endgroup$
      – George Harrison
      Jan 8 at 21:25












    • $begingroup$
      @GeorgeHarrison Let $overline X sim mathcal N(mu, sigma^2)$, then $Z=frac{overline X-mu}{sigma}sim N(0,1)$. In your exercise $mu=0$ and $sigma=sqrt{frac1{16}}=frac14$.
      $endgroup$
      – callculus
      Jan 8 at 21:31












    • $begingroup$
      I am not getting where you are getting the mean is 0 and the standard deviation is 0.25.
      $endgroup$
      – George Harrison
      Jan 8 at 21:36










    • $begingroup$
      In your example $P(overline Xleq c)$ is the same as $P(Zleq 4c)$ due standardizing.
      $endgroup$
      – callculus
      Jan 8 at 21:37










    • $begingroup$
      @GeorgeHarrison The expected value of the mean of n variables is the mean of the expected values. And for $n$ i.i.d random variables we have $Varleft (frac1n sumlimits_{i=1}^n X_i right)=frac1{n^2}cdot Varleft (sumlimits_{i=1}^n X_i right)=Varleft (sumlimits_{i=1}^n X_i right)$ $=frac1{n^2}cdotleft(Var(X_1)+Var(X_2)+ldots+Var(X_n)right)$. Since the $X_i$´s has the same variance we get: $=frac1{n^2}cdot ncdot Var(X_1)=frac1ncdot sigma^2$
      $endgroup$
      – callculus
      Jan 8 at 21:48
















    0












    0








    0





    $begingroup$

    If $X$ is distributed as $Xsim mathcal N(0,1)$, then $frac1n sumlimits_{i=1}^{16} X_i=overline X$ is distributed as $overline Xsim mathcal Nleft( 0, frac1{16}right)$



    Now we have to evaluate how $|overline X|$ is distributed in terms of $overline X$



    $P(|overline X| leq c)=P(-c leq overline X leq c)$



    $=P(overline X leq c)-P(overline X leq -c)$



    $=P(overline X leq c)-left[ 1-P(overline X leq c) right]$



    $=2 cdot P(overline X leq c)-1=2cdot F_{overline X}(c)-1$



    At the next step we standardize $overline X$ to be able to use the cdf of the standard normal distribution.



    $2cdot P(overline X leq c)-1=2cdot Phileft(frac{c-0}{sqrt{frac1{16}}} right)-1=2cdot Phileft(4c right)-1=0.5$



    $2cdot Phileft(4c right)=1.5$



    $Phileft(4c right)=0.75$



    $4c=Phi^{-1}left(0.75right)$



    $Phi^{-1}left(pright)$ is the inverse function of $Phileft(zright)$



    $4c=0.674Rightarrow c=0.1685$






    share|cite|improve this answer









    $endgroup$



    If $X$ is distributed as $Xsim mathcal N(0,1)$, then $frac1n sumlimits_{i=1}^{16} X_i=overline X$ is distributed as $overline Xsim mathcal Nleft( 0, frac1{16}right)$



    Now we have to evaluate how $|overline X|$ is distributed in terms of $overline X$



    $P(|overline X| leq c)=P(-c leq overline X leq c)$



    $=P(overline X leq c)-P(overline X leq -c)$



    $=P(overline X leq c)-left[ 1-P(overline X leq c) right]$



    $=2 cdot P(overline X leq c)-1=2cdot F_{overline X}(c)-1$



    At the next step we standardize $overline X$ to be able to use the cdf of the standard normal distribution.



    $2cdot P(overline X leq c)-1=2cdot Phileft(frac{c-0}{sqrt{frac1{16}}} right)-1=2cdot Phileft(4c right)-1=0.5$



    $2cdot Phileft(4c right)=1.5$



    $Phileft(4c right)=0.75$



    $4c=Phi^{-1}left(0.75right)$



    $Phi^{-1}left(pright)$ is the inverse function of $Phileft(zright)$



    $4c=0.674Rightarrow c=0.1685$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 8 at 19:06









    callculuscallculus

    17.9k31427




    17.9k31427












    • $begingroup$
      I am kind of lost on the step where we standardize X. Everything before that makes sense.
      $endgroup$
      – George Harrison
      Jan 8 at 21:25












    • $begingroup$
      @GeorgeHarrison Let $overline X sim mathcal N(mu, sigma^2)$, then $Z=frac{overline X-mu}{sigma}sim N(0,1)$. In your exercise $mu=0$ and $sigma=sqrt{frac1{16}}=frac14$.
      $endgroup$
      – callculus
      Jan 8 at 21:31












    • $begingroup$
      I am not getting where you are getting the mean is 0 and the standard deviation is 0.25.
      $endgroup$
      – George Harrison
      Jan 8 at 21:36










    • $begingroup$
      In your example $P(overline Xleq c)$ is the same as $P(Zleq 4c)$ due standardizing.
      $endgroup$
      – callculus
      Jan 8 at 21:37










    • $begingroup$
      @GeorgeHarrison The expected value of the mean of n variables is the mean of the expected values. And for $n$ i.i.d random variables we have $Varleft (frac1n sumlimits_{i=1}^n X_i right)=frac1{n^2}cdot Varleft (sumlimits_{i=1}^n X_i right)=Varleft (sumlimits_{i=1}^n X_i right)$ $=frac1{n^2}cdotleft(Var(X_1)+Var(X_2)+ldots+Var(X_n)right)$. Since the $X_i$´s has the same variance we get: $=frac1{n^2}cdot ncdot Var(X_1)=frac1ncdot sigma^2$
      $endgroup$
      – callculus
      Jan 8 at 21:48




















    • $begingroup$
      I am kind of lost on the step where we standardize X. Everything before that makes sense.
      $endgroup$
      – George Harrison
      Jan 8 at 21:25












    • $begingroup$
      @GeorgeHarrison Let $overline X sim mathcal N(mu, sigma^2)$, then $Z=frac{overline X-mu}{sigma}sim N(0,1)$. In your exercise $mu=0$ and $sigma=sqrt{frac1{16}}=frac14$.
      $endgroup$
      – callculus
      Jan 8 at 21:31












    • $begingroup$
      I am not getting where you are getting the mean is 0 and the standard deviation is 0.25.
      $endgroup$
      – George Harrison
      Jan 8 at 21:36










    • $begingroup$
      In your example $P(overline Xleq c)$ is the same as $P(Zleq 4c)$ due standardizing.
      $endgroup$
      – callculus
      Jan 8 at 21:37










    • $begingroup$
      @GeorgeHarrison The expected value of the mean of n variables is the mean of the expected values. And for $n$ i.i.d random variables we have $Varleft (frac1n sumlimits_{i=1}^n X_i right)=frac1{n^2}cdot Varleft (sumlimits_{i=1}^n X_i right)=Varleft (sumlimits_{i=1}^n X_i right)$ $=frac1{n^2}cdotleft(Var(X_1)+Var(X_2)+ldots+Var(X_n)right)$. Since the $X_i$´s has the same variance we get: $=frac1{n^2}cdot ncdot Var(X_1)=frac1ncdot sigma^2$
      $endgroup$
      – callculus
      Jan 8 at 21:48


















    $begingroup$
    I am kind of lost on the step where we standardize X. Everything before that makes sense.
    $endgroup$
    – George Harrison
    Jan 8 at 21:25






    $begingroup$
    I am kind of lost on the step where we standardize X. Everything before that makes sense.
    $endgroup$
    – George Harrison
    Jan 8 at 21:25














    $begingroup$
    @GeorgeHarrison Let $overline X sim mathcal N(mu, sigma^2)$, then $Z=frac{overline X-mu}{sigma}sim N(0,1)$. In your exercise $mu=0$ and $sigma=sqrt{frac1{16}}=frac14$.
    $endgroup$
    – callculus
    Jan 8 at 21:31






    $begingroup$
    @GeorgeHarrison Let $overline X sim mathcal N(mu, sigma^2)$, then $Z=frac{overline X-mu}{sigma}sim N(0,1)$. In your exercise $mu=0$ and $sigma=sqrt{frac1{16}}=frac14$.
    $endgroup$
    – callculus
    Jan 8 at 21:31














    $begingroup$
    I am not getting where you are getting the mean is 0 and the standard deviation is 0.25.
    $endgroup$
    – George Harrison
    Jan 8 at 21:36




    $begingroup$
    I am not getting where you are getting the mean is 0 and the standard deviation is 0.25.
    $endgroup$
    – George Harrison
    Jan 8 at 21:36












    $begingroup$
    In your example $P(overline Xleq c)$ is the same as $P(Zleq 4c)$ due standardizing.
    $endgroup$
    – callculus
    Jan 8 at 21:37




    $begingroup$
    In your example $P(overline Xleq c)$ is the same as $P(Zleq 4c)$ due standardizing.
    $endgroup$
    – callculus
    Jan 8 at 21:37












    $begingroup$
    @GeorgeHarrison The expected value of the mean of n variables is the mean of the expected values. And for $n$ i.i.d random variables we have $Varleft (frac1n sumlimits_{i=1}^n X_i right)=frac1{n^2}cdot Varleft (sumlimits_{i=1}^n X_i right)=Varleft (sumlimits_{i=1}^n X_i right)$ $=frac1{n^2}cdotleft(Var(X_1)+Var(X_2)+ldots+Var(X_n)right)$. Since the $X_i$´s has the same variance we get: $=frac1{n^2}cdot ncdot Var(X_1)=frac1ncdot sigma^2$
    $endgroup$
    – callculus
    Jan 8 at 21:48






    $begingroup$
    @GeorgeHarrison The expected value of the mean of n variables is the mean of the expected values. And for $n$ i.i.d random variables we have $Varleft (frac1n sumlimits_{i=1}^n X_i right)=frac1{n^2}cdot Varleft (sumlimits_{i=1}^n X_i right)=Varleft (sumlimits_{i=1}^n X_i right)$ $=frac1{n^2}cdotleft(Var(X_1)+Var(X_2)+ldots+Var(X_n)right)$. Since the $X_i$´s has the same variance we get: $=frac1{n^2}cdot ncdot Var(X_1)=frac1ncdot sigma^2$
    $endgroup$
    – callculus
    Jan 8 at 21:48













    0












    $begingroup$

    HINT- The sum of independent normal variables also follows the normal distribution with



    $$N(sum_{i=1}^n {mu_i},sum_{i=1}^n {sigma_i^2})$$



    Also, if a random variable $X$ has mean $mu$ and variance $sigma^2$, then the random variable $Y=kX$ (where $k$ is a constant) has mean $kmu$ and variance $k^2sigma^2$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      HINT- The sum of independent normal variables also follows the normal distribution with



      $$N(sum_{i=1}^n {mu_i},sum_{i=1}^n {sigma_i^2})$$



      Also, if a random variable $X$ has mean $mu$ and variance $sigma^2$, then the random variable $Y=kX$ (where $k$ is a constant) has mean $kmu$ and variance $k^2sigma^2$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        HINT- The sum of independent normal variables also follows the normal distribution with



        $$N(sum_{i=1}^n {mu_i},sum_{i=1}^n {sigma_i^2})$$



        Also, if a random variable $X$ has mean $mu$ and variance $sigma^2$, then the random variable $Y=kX$ (where $k$ is a constant) has mean $kmu$ and variance $k^2sigma^2$






        share|cite|improve this answer









        $endgroup$



        HINT- The sum of independent normal variables also follows the normal distribution with



        $$N(sum_{i=1}^n {mu_i},sum_{i=1}^n {sigma_i^2})$$



        Also, if a random variable $X$ has mean $mu$ and variance $sigma^2$, then the random variable $Y=kX$ (where $k$ is a constant) has mean $kmu$ and variance $k^2sigma^2$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 17:40









        Sauhard SharmaSauhard Sharma

        953318




        953318






























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