Calculating standard deviation without a data set.












1












$begingroup$


I know how to calculate SD when given data points by using: $ displaystyle mathrm{SD} = sqrt{sum(x^2 - text{mean}^2) / n} $.



I have been given just the sum of $x$ and sum of $x^2$. How do I calculate SD from this?!



An example question I am stuck on:
Sum of $x = 1303$



Sum of $x^2 = 123557.$



There are 14 years for which the data is given - I would assume this is n...






statistics standard-deviation means







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$endgroup$












  • $begingroup$
    would you like to describe how do you compute SD given data points? which formula do you use?
    $endgroup$
    – Siong Thye Goh
    Jan 24 '18 at 18:44






  • 2




    $begingroup$
    Hint: The variance is $s^2 = frac{1}{n-1}(Q - T^2/n),$ where $Q = sum_i X_i^2$ and $T = sum_i X_i.$ This is proved by performing the square in $frac{1}{n-2}sum_i(X_i - bar X)^2,$ using the distributive law, and using the definition of $bar X.$
    $endgroup$
    – BruceET
    Jan 24 '18 at 19:12


















1












$begingroup$


I know how to calculate SD when given data points by using: $ displaystyle mathrm{SD} = sqrt{sum(x^2 - text{mean}^2) / n} $.



I have been given just the sum of $x$ and sum of $x^2$. How do I calculate SD from this?!



An example question I am stuck on:
Sum of $x = 1303$



Sum of $x^2 = 123557.$



There are 14 years for which the data is given - I would assume this is n...






statistics standard-deviation means







share|cite|improve this question











$endgroup$












  • $begingroup$
    would you like to describe how do you compute SD given data points? which formula do you use?
    $endgroup$
    – Siong Thye Goh
    Jan 24 '18 at 18:44






  • 2




    $begingroup$
    Hint: The variance is $s^2 = frac{1}{n-1}(Q - T^2/n),$ where $Q = sum_i X_i^2$ and $T = sum_i X_i.$ This is proved by performing the square in $frac{1}{n-2}sum_i(X_i - bar X)^2,$ using the distributive law, and using the definition of $bar X.$
    $endgroup$
    – BruceET
    Jan 24 '18 at 19:12
















1












1








1





$begingroup$


I know how to calculate SD when given data points by using: $ displaystyle mathrm{SD} = sqrt{sum(x^2 - text{mean}^2) / n} $.



I have been given just the sum of $x$ and sum of $x^2$. How do I calculate SD from this?!



An example question I am stuck on:
Sum of $x = 1303$



Sum of $x^2 = 123557.$



There are 14 years for which the data is given - I would assume this is n...






statistics standard-deviation means







share|cite|improve this question











$endgroup$




I know how to calculate SD when given data points by using: $ displaystyle mathrm{SD} = sqrt{sum(x^2 - text{mean}^2) / n} $.



I have been given just the sum of $x$ and sum of $x^2$. How do I calculate SD from this?!



An example question I am stuck on:
Sum of $x = 1303$



Sum of $x^2 = 123557.$



There are 14 years for which the data is given - I would assume this is n...






statistics standard-deviation means




statistics standard-deviation means






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Sep 14 '18 at 16:51









Michael Hardy

1




1










asked Jan 24 '18 at 18:38









Ben HughesBen Hughes

17819




17819












  • $begingroup$
    would you like to describe how do you compute SD given data points? which formula do you use?
    $endgroup$
    – Siong Thye Goh
    Jan 24 '18 at 18:44






  • 2




    $begingroup$
    Hint: The variance is $s^2 = frac{1}{n-1}(Q - T^2/n),$ where $Q = sum_i X_i^2$ and $T = sum_i X_i.$ This is proved by performing the square in $frac{1}{n-2}sum_i(X_i - bar X)^2,$ using the distributive law, and using the definition of $bar X.$
    $endgroup$
    – BruceET
    Jan 24 '18 at 19:12




















  • $begingroup$
    would you like to describe how do you compute SD given data points? which formula do you use?
    $endgroup$
    – Siong Thye Goh
    Jan 24 '18 at 18:44






  • 2




    $begingroup$
    Hint: The variance is $s^2 = frac{1}{n-1}(Q - T^2/n),$ where $Q = sum_i X_i^2$ and $T = sum_i X_i.$ This is proved by performing the square in $frac{1}{n-2}sum_i(X_i - bar X)^2,$ using the distributive law, and using the definition of $bar X.$
    $endgroup$
    – BruceET
    Jan 24 '18 at 19:12


















$begingroup$
would you like to describe how do you compute SD given data points? which formula do you use?
$endgroup$
– Siong Thye Goh
Jan 24 '18 at 18:44




$begingroup$
would you like to describe how do you compute SD given data points? which formula do you use?
$endgroup$
– Siong Thye Goh
Jan 24 '18 at 18:44




2




2




$begingroup$
Hint: The variance is $s^2 = frac{1}{n-1}(Q - T^2/n),$ where $Q = sum_i X_i^2$ and $T = sum_i X_i.$ This is proved by performing the square in $frac{1}{n-2}sum_i(X_i - bar X)^2,$ using the distributive law, and using the definition of $bar X.$
$endgroup$
– BruceET
Jan 24 '18 at 19:12






$begingroup$
Hint: The variance is $s^2 = frac{1}{n-1}(Q - T^2/n),$ where $Q = sum_i X_i^2$ and $T = sum_i X_i.$ This is proved by performing the square in $frac{1}{n-2}sum_i(X_i - bar X)^2,$ using the distributive law, and using the definition of $bar X.$
$endgroup$
– BruceET
Jan 24 '18 at 19:12












2 Answers
2






active

oldest

votes


















0












$begingroup$

In a not confusing manner (hopefully) the way i would start is to work out the variance using:
Sxx = (Sum of)x^2 - n(Mean)^2



Then from there to find the standard deviation i would use:



srqroot(Sxx/n-1)



hopefully that has helped!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Your formula for SD seems slightly wrong. Assuming the mean $m$ is known then the variance $V=frac{sum (x_k-m)^2}{n}=frac{sum x_k^2}{n}-m^2$ so $SD=sqrt{frac{sum x_k^2}{n}-m^2}$. When the mean is being estimated by the average of the $x's$, then the division is by $n-1$ rather than $n$.
    $endgroup$
    – herb steinberg
    Jul 13 '18 at 3:02





















0












$begingroup$

Hint: You need a formula, where you can enter the sum of squares and the square of the sum. Let us first define these as follows:
$$
SSQ =sumlimits_{k=1}^{n} x_k^2 quadtext{and}quad SQS = sumlimits_{k=1}^{n} x_k ;.
$$



A famous formula of the (population)1 variance is
$$
mathrm{Var}(X)
= dfrac{1}{n} sumlimits_{k=1}^{n} x_k^2 - left( dfrac{1}{n} sumlimits_{k=1}^{n} x_k right)^2 = dfrac{SSQ}{n} - left( dfrac{SQS}{n} right)^2 ;.
$$
The (population) standard deviation, therefore, is
$$
boxed
{
mathrm{SD}(X)
= sqrt{dfrac{SSQ}{n} - left( dfrac{SQS}{n} right)^2}
} ;.
$$
You see that you also need $n$, the number of samples.




1 The population statistics can be corrected, when sample statistics are required.






share|cite|improve this answer











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    In a not confusing manner (hopefully) the way i would start is to work out the variance using:
    Sxx = (Sum of)x^2 - n(Mean)^2



    Then from there to find the standard deviation i would use:



    srqroot(Sxx/n-1)



    hopefully that has helped!






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Your formula for SD seems slightly wrong. Assuming the mean $m$ is known then the variance $V=frac{sum (x_k-m)^2}{n}=frac{sum x_k^2}{n}-m^2$ so $SD=sqrt{frac{sum x_k^2}{n}-m^2}$. When the mean is being estimated by the average of the $x's$, then the division is by $n-1$ rather than $n$.
      $endgroup$
      – herb steinberg
      Jul 13 '18 at 3:02


















    0












    $begingroup$

    In a not confusing manner (hopefully) the way i would start is to work out the variance using:
    Sxx = (Sum of)x^2 - n(Mean)^2



    Then from there to find the standard deviation i would use:



    srqroot(Sxx/n-1)



    hopefully that has helped!






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Your formula for SD seems slightly wrong. Assuming the mean $m$ is known then the variance $V=frac{sum (x_k-m)^2}{n}=frac{sum x_k^2}{n}-m^2$ so $SD=sqrt{frac{sum x_k^2}{n}-m^2}$. When the mean is being estimated by the average of the $x's$, then the division is by $n-1$ rather than $n$.
      $endgroup$
      – herb steinberg
      Jul 13 '18 at 3:02
















    0












    0








    0





    $begingroup$

    In a not confusing manner (hopefully) the way i would start is to work out the variance using:
    Sxx = (Sum of)x^2 - n(Mean)^2



    Then from there to find the standard deviation i would use:



    srqroot(Sxx/n-1)



    hopefully that has helped!






    share|cite|improve this answer









    $endgroup$



    In a not confusing manner (hopefully) the way i would start is to work out the variance using:
    Sxx = (Sum of)x^2 - n(Mean)^2



    Then from there to find the standard deviation i would use:



    srqroot(Sxx/n-1)



    hopefully that has helped!







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 25 '18 at 15:57









    JoeJoe

    1




    1












    • $begingroup$
      Your formula for SD seems slightly wrong. Assuming the mean $m$ is known then the variance $V=frac{sum (x_k-m)^2}{n}=frac{sum x_k^2}{n}-m^2$ so $SD=sqrt{frac{sum x_k^2}{n}-m^2}$. When the mean is being estimated by the average of the $x's$, then the division is by $n-1$ rather than $n$.
      $endgroup$
      – herb steinberg
      Jul 13 '18 at 3:02




















    • $begingroup$
      Your formula for SD seems slightly wrong. Assuming the mean $m$ is known then the variance $V=frac{sum (x_k-m)^2}{n}=frac{sum x_k^2}{n}-m^2$ so $SD=sqrt{frac{sum x_k^2}{n}-m^2}$. When the mean is being estimated by the average of the $x's$, then the division is by $n-1$ rather than $n$.
      $endgroup$
      – herb steinberg
      Jul 13 '18 at 3:02


















    $begingroup$
    Your formula for SD seems slightly wrong. Assuming the mean $m$ is known then the variance $V=frac{sum (x_k-m)^2}{n}=frac{sum x_k^2}{n}-m^2$ so $SD=sqrt{frac{sum x_k^2}{n}-m^2}$. When the mean is being estimated by the average of the $x's$, then the division is by $n-1$ rather than $n$.
    $endgroup$
    – herb steinberg
    Jul 13 '18 at 3:02






    $begingroup$
    Your formula for SD seems slightly wrong. Assuming the mean $m$ is known then the variance $V=frac{sum (x_k-m)^2}{n}=frac{sum x_k^2}{n}-m^2$ so $SD=sqrt{frac{sum x_k^2}{n}-m^2}$. When the mean is being estimated by the average of the $x's$, then the division is by $n-1$ rather than $n$.
    $endgroup$
    – herb steinberg
    Jul 13 '18 at 3:02













    0












    $begingroup$

    Hint: You need a formula, where you can enter the sum of squares and the square of the sum. Let us first define these as follows:
    $$
    SSQ =sumlimits_{k=1}^{n} x_k^2 quadtext{and}quad SQS = sumlimits_{k=1}^{n} x_k ;.
    $$



    A famous formula of the (population)1 variance is
    $$
    mathrm{Var}(X)
    = dfrac{1}{n} sumlimits_{k=1}^{n} x_k^2 - left( dfrac{1}{n} sumlimits_{k=1}^{n} x_k right)^2 = dfrac{SSQ}{n} - left( dfrac{SQS}{n} right)^2 ;.
    $$
    The (population) standard deviation, therefore, is
    $$
    boxed
    {
    mathrm{SD}(X)
    = sqrt{dfrac{SSQ}{n} - left( dfrac{SQS}{n} right)^2}
    } ;.
    $$
    You see that you also need $n$, the number of samples.




    1 The population statistics can be corrected, when sample statistics are required.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Hint: You need a formula, where you can enter the sum of squares and the square of the sum. Let us first define these as follows:
      $$
      SSQ =sumlimits_{k=1}^{n} x_k^2 quadtext{and}quad SQS = sumlimits_{k=1}^{n} x_k ;.
      $$



      A famous formula of the (population)1 variance is
      $$
      mathrm{Var}(X)
      = dfrac{1}{n} sumlimits_{k=1}^{n} x_k^2 - left( dfrac{1}{n} sumlimits_{k=1}^{n} x_k right)^2 = dfrac{SSQ}{n} - left( dfrac{SQS}{n} right)^2 ;.
      $$
      The (population) standard deviation, therefore, is
      $$
      boxed
      {
      mathrm{SD}(X)
      = sqrt{dfrac{SSQ}{n} - left( dfrac{SQS}{n} right)^2}
      } ;.
      $$
      You see that you also need $n$, the number of samples.




      1 The population statistics can be corrected, when sample statistics are required.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Hint: You need a formula, where you can enter the sum of squares and the square of the sum. Let us first define these as follows:
        $$
        SSQ =sumlimits_{k=1}^{n} x_k^2 quadtext{and}quad SQS = sumlimits_{k=1}^{n} x_k ;.
        $$



        A famous formula of the (population)1 variance is
        $$
        mathrm{Var}(X)
        = dfrac{1}{n} sumlimits_{k=1}^{n} x_k^2 - left( dfrac{1}{n} sumlimits_{k=1}^{n} x_k right)^2 = dfrac{SSQ}{n} - left( dfrac{SQS}{n} right)^2 ;.
        $$
        The (population) standard deviation, therefore, is
        $$
        boxed
        {
        mathrm{SD}(X)
        = sqrt{dfrac{SSQ}{n} - left( dfrac{SQS}{n} right)^2}
        } ;.
        $$
        You see that you also need $n$, the number of samples.




        1 The population statistics can be corrected, when sample statistics are required.






        share|cite|improve this answer











        $endgroup$



        Hint: You need a formula, where you can enter the sum of squares and the square of the sum. Let us first define these as follows:
        $$
        SSQ =sumlimits_{k=1}^{n} x_k^2 quadtext{and}quad SQS = sumlimits_{k=1}^{n} x_k ;.
        $$



        A famous formula of the (population)1 variance is
        $$
        mathrm{Var}(X)
        = dfrac{1}{n} sumlimits_{k=1}^{n} x_k^2 - left( dfrac{1}{n} sumlimits_{k=1}^{n} x_k right)^2 = dfrac{SSQ}{n} - left( dfrac{SQS}{n} right)^2 ;.
        $$
        The (population) standard deviation, therefore, is
        $$
        boxed
        {
        mathrm{SD}(X)
        = sqrt{dfrac{SSQ}{n} - left( dfrac{SQS}{n} right)^2}
        } ;.
        $$
        You see that you also need $n$, the number of samples.




        1 The population statistics can be corrected, when sample statistics are required.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 18 '18 at 12:53

























        answered Dec 18 '18 at 12:46









        Björn FriedrichBjörn Friedrich

        2,67461831




        2,67461831






























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