Range of Data Values That Includes 95% of my Data












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For a math assignment, I am required to find the range of data values that would include 95% of my data.



My data is the wait time for a specific bus. I have gathered 20 pieces of data for this, where each number represents how late(+)/early(-)/on time(0) the bus is.



This is the organized data in ascending order: 0, 1, -4, 0, -3, 0, -4, -2, 0, 1, -1, 2, 0, -4, 5, -2, 0, 3, -3, 0.



The link attached down below is my normal distribution graph where I have listed the mean (-0.55), standard deviation(2.4), and the graph information.



Normal Distribution Graph



What I think the range would be is (-5) - 4. I came to this conclusion by looking at the normal distribution graph above. Although I`m concerned about my answer.






normal-distribution standard-deviation means







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    0












    $begingroup$


    For a math assignment, I am required to find the range of data values that would include 95% of my data.



    My data is the wait time for a specific bus. I have gathered 20 pieces of data for this, where each number represents how late(+)/early(-)/on time(0) the bus is.



    This is the organized data in ascending order: 0, 1, -4, 0, -3, 0, -4, -2, 0, 1, -1, 2, 0, -4, 5, -2, 0, 3, -3, 0.



    The link attached down below is my normal distribution graph where I have listed the mean (-0.55), standard deviation(2.4), and the graph information.



    Normal Distribution Graph



    What I think the range would be is (-5) - 4. I came to this conclusion by looking at the normal distribution graph above. Although I`m concerned about my answer.






    normal-distribution standard-deviation means







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      For a math assignment, I am required to find the range of data values that would include 95% of my data.



      My data is the wait time for a specific bus. I have gathered 20 pieces of data for this, where each number represents how late(+)/early(-)/on time(0) the bus is.



      This is the organized data in ascending order: 0, 1, -4, 0, -3, 0, -4, -2, 0, 1, -1, 2, 0, -4, 5, -2, 0, 3, -3, 0.



      The link attached down below is my normal distribution graph where I have listed the mean (-0.55), standard deviation(2.4), and the graph information.



      Normal Distribution Graph



      What I think the range would be is (-5) - 4. I came to this conclusion by looking at the normal distribution graph above. Although I`m concerned about my answer.






      normal-distribution standard-deviation means







      share|cite|improve this question









      $endgroup$




      For a math assignment, I am required to find the range of data values that would include 95% of my data.



      My data is the wait time for a specific bus. I have gathered 20 pieces of data for this, where each number represents how late(+)/early(-)/on time(0) the bus is.



      This is the organized data in ascending order: 0, 1, -4, 0, -3, 0, -4, -2, 0, 1, -1, 2, 0, -4, 5, -2, 0, 3, -3, 0.



      The link attached down below is my normal distribution graph where I have listed the mean (-0.55), standard deviation(2.4), and the graph information.



      Normal Distribution Graph



      What I think the range would be is (-5) - 4. I came to this conclusion by looking at the normal distribution graph above. Although I`m concerned about my answer.






      normal-distribution standard-deviation means




      normal-distribution standard-deviation means






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 4 at 22:36









      Yashvi ShahYashvi Shah

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          $begingroup$

          If this data is normally distributed with $mu=-0.55$ and $sigma=2.4$ we can say that X, the wait time for a specific bus, is distributed as follows: $$Xsim N(-0.55,(2.4)^2)$$
          For 95% of the data to be included we want $P(mu-a<X<mu+a)=0.95$. As the normal distribution is symmetrical we can simplify this to finding: $P(X>mu+a)=frac{1-0.95}{2}=0.025$ or similarly $P(X<mu-a)=0.025$. Using the Inverse Normal function (a website can be found here to do this) we find that $mu-a=-5.2539$, such that: $$a=mu+5.2539=-0.55+5.2539=4.7039$$
          The range of data required is then $2cdot a =2cdot(4.7039)= 9.407$8 as the data ranges from $mu-a$ to $mu+a$.






          share|cite|improve this answer









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          • $begingroup$
            So, 95% of my data falls between μ−a and μ+a, which is basically in between -5.2539 and 4.1539? Then this just proves that what I said in my question about the range of values being -5 and 4 is partly right...
            $endgroup$
            – Yashvi Shah
            Jan 4 at 23:49










          • $begingroup$
            Yes your initial estimate was quite accurate
            $endgroup$
            – Peter Foreman
            Jan 5 at 0:24










          • $begingroup$
            Wait, how come did you do 1 - 0.95 / 2 and 2(a)? Why the /2/ for both of these?
            $endgroup$
            – Yashvi Shah
            Jan 5 at 1:14












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          1 Answer
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          $begingroup$

          If this data is normally distributed with $mu=-0.55$ and $sigma=2.4$ we can say that X, the wait time for a specific bus, is distributed as follows: $$Xsim N(-0.55,(2.4)^2)$$
          For 95% of the data to be included we want $P(mu-a<X<mu+a)=0.95$. As the normal distribution is symmetrical we can simplify this to finding: $P(X>mu+a)=frac{1-0.95}{2}=0.025$ or similarly $P(X<mu-a)=0.025$. Using the Inverse Normal function (a website can be found here to do this) we find that $mu-a=-5.2539$, such that: $$a=mu+5.2539=-0.55+5.2539=4.7039$$
          The range of data required is then $2cdot a =2cdot(4.7039)= 9.407$8 as the data ranges from $mu-a$ to $mu+a$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So, 95% of my data falls between μ−a and μ+a, which is basically in between -5.2539 and 4.1539? Then this just proves that what I said in my question about the range of values being -5 and 4 is partly right...
            $endgroup$
            – Yashvi Shah
            Jan 4 at 23:49










          • $begingroup$
            Yes your initial estimate was quite accurate
            $endgroup$
            – Peter Foreman
            Jan 5 at 0:24










          • $begingroup$
            Wait, how come did you do 1 - 0.95 / 2 and 2(a)? Why the /2/ for both of these?
            $endgroup$
            – Yashvi Shah
            Jan 5 at 1:14
















          0












          $begingroup$

          If this data is normally distributed with $mu=-0.55$ and $sigma=2.4$ we can say that X, the wait time for a specific bus, is distributed as follows: $$Xsim N(-0.55,(2.4)^2)$$
          For 95% of the data to be included we want $P(mu-a<X<mu+a)=0.95$. As the normal distribution is symmetrical we can simplify this to finding: $P(X>mu+a)=frac{1-0.95}{2}=0.025$ or similarly $P(X<mu-a)=0.025$. Using the Inverse Normal function (a website can be found here to do this) we find that $mu-a=-5.2539$, such that: $$a=mu+5.2539=-0.55+5.2539=4.7039$$
          The range of data required is then $2cdot a =2cdot(4.7039)= 9.407$8 as the data ranges from $mu-a$ to $mu+a$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So, 95% of my data falls between μ−a and μ+a, which is basically in between -5.2539 and 4.1539? Then this just proves that what I said in my question about the range of values being -5 and 4 is partly right...
            $endgroup$
            – Yashvi Shah
            Jan 4 at 23:49










          • $begingroup$
            Yes your initial estimate was quite accurate
            $endgroup$
            – Peter Foreman
            Jan 5 at 0:24










          • $begingroup$
            Wait, how come did you do 1 - 0.95 / 2 and 2(a)? Why the /2/ for both of these?
            $endgroup$
            – Yashvi Shah
            Jan 5 at 1:14














          0












          0








          0





          $begingroup$

          If this data is normally distributed with $mu=-0.55$ and $sigma=2.4$ we can say that X, the wait time for a specific bus, is distributed as follows: $$Xsim N(-0.55,(2.4)^2)$$
          For 95% of the data to be included we want $P(mu-a<X<mu+a)=0.95$. As the normal distribution is symmetrical we can simplify this to finding: $P(X>mu+a)=frac{1-0.95}{2}=0.025$ or similarly $P(X<mu-a)=0.025$. Using the Inverse Normal function (a website can be found here to do this) we find that $mu-a=-5.2539$, such that: $$a=mu+5.2539=-0.55+5.2539=4.7039$$
          The range of data required is then $2cdot a =2cdot(4.7039)= 9.407$8 as the data ranges from $mu-a$ to $mu+a$.






          share|cite|improve this answer









          $endgroup$



          If this data is normally distributed with $mu=-0.55$ and $sigma=2.4$ we can say that X, the wait time for a specific bus, is distributed as follows: $$Xsim N(-0.55,(2.4)^2)$$
          For 95% of the data to be included we want $P(mu-a<X<mu+a)=0.95$. As the normal distribution is symmetrical we can simplify this to finding: $P(X>mu+a)=frac{1-0.95}{2}=0.025$ or similarly $P(X<mu-a)=0.025$. Using the Inverse Normal function (a website can be found here to do this) we find that $mu-a=-5.2539$, such that: $$a=mu+5.2539=-0.55+5.2539=4.7039$$
          The range of data required is then $2cdot a =2cdot(4.7039)= 9.407$8 as the data ranges from $mu-a$ to $mu+a$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 23:01









          Peter ForemanPeter Foreman

          5,9091317




          5,9091317












          • $begingroup$
            So, 95% of my data falls between μ−a and μ+a, which is basically in between -5.2539 and 4.1539? Then this just proves that what I said in my question about the range of values being -5 and 4 is partly right...
            $endgroup$
            – Yashvi Shah
            Jan 4 at 23:49










          • $begingroup$
            Yes your initial estimate was quite accurate
            $endgroup$
            – Peter Foreman
            Jan 5 at 0:24










          • $begingroup$
            Wait, how come did you do 1 - 0.95 / 2 and 2(a)? Why the /2/ for both of these?
            $endgroup$
            – Yashvi Shah
            Jan 5 at 1:14


















          • $begingroup$
            So, 95% of my data falls between μ−a and μ+a, which is basically in between -5.2539 and 4.1539? Then this just proves that what I said in my question about the range of values being -5 and 4 is partly right...
            $endgroup$
            – Yashvi Shah
            Jan 4 at 23:49










          • $begingroup$
            Yes your initial estimate was quite accurate
            $endgroup$
            – Peter Foreman
            Jan 5 at 0:24










          • $begingroup$
            Wait, how come did you do 1 - 0.95 / 2 and 2(a)? Why the /2/ for both of these?
            $endgroup$
            – Yashvi Shah
            Jan 5 at 1:14
















          $begingroup$
          So, 95% of my data falls between μ−a and μ+a, which is basically in between -5.2539 and 4.1539? Then this just proves that what I said in my question about the range of values being -5 and 4 is partly right...
          $endgroup$
          – Yashvi Shah
          Jan 4 at 23:49




          $begingroup$
          So, 95% of my data falls between μ−a and μ+a, which is basically in between -5.2539 and 4.1539? Then this just proves that what I said in my question about the range of values being -5 and 4 is partly right...
          $endgroup$
          – Yashvi Shah
          Jan 4 at 23:49












          $begingroup$
          Yes your initial estimate was quite accurate
          $endgroup$
          – Peter Foreman
          Jan 5 at 0:24




          $begingroup$
          Yes your initial estimate was quite accurate
          $endgroup$
          – Peter Foreman
          Jan 5 at 0:24












          $begingroup$
          Wait, how come did you do 1 - 0.95 / 2 and 2(a)? Why the /2/ for both of these?
          $endgroup$
          – Yashvi Shah
          Jan 5 at 1:14




          $begingroup$
          Wait, how come did you do 1 - 0.95 / 2 and 2(a)? Why the /2/ for both of these?
          $endgroup$
          – Yashvi Shah
          Jan 5 at 1:14


















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