Probability Distribution Table of Bus Arrival












0














For an assignment in my high school Data Management class, I am required to make a probability distribution table for my data, which is "Wait Time for a Bus." It is not just any bus. It is a specific bus that arrives at a bus stop near my school, and I know someone who takes this bus, so I ask her to record the data down for me.



This bus is supposed to arrive at 12:15 pm every day, but sometimes its a little late or a little early, and sometimes its on time.



I have gathered the data and timings for the past 20 days.



Now, what I believe I should do is that I should arrange my data in ascending order and group them in ranges. To find the probability, my classmate said to do "Success over total." I think she means the total number of outcomes. This is what I`ve been told by a classmate, at least, this is what she is doing, and to be honest, what she told me sounds a little confusing because she did not elaborate on it.



Here is my organized data set in ascending order: -4, -4, -4, -3, -3, -2, -2, -1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 3, 5.



I am quite confused on how to do this, and I`ve asked around as well, but I am not getting quick responses. Am I on the right track? Is what my classmate told me to do right?



THIS PART HAS BEEN EDITED



This is what I did for the probability distribution table (click the Probability Distribution Table to view the table if you cannot view it on here):



Probability Distribution Table



NEWLY EDITED



I have done the continuous probability distribution graph (the y-axis is the probability, and the x-axis is the wait time for the bus in minutes). Although, to me,This graph does not look right. Im pretty sure this is how the continuous probability distribution with a normal distribution curve is supposed to look like, considering Ive searched up and seen some things similar to what I have done.










share|cite|improve this question









New contributor




Yashvi Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

























    0














    For an assignment in my high school Data Management class, I am required to make a probability distribution table for my data, which is "Wait Time for a Bus." It is not just any bus. It is a specific bus that arrives at a bus stop near my school, and I know someone who takes this bus, so I ask her to record the data down for me.



    This bus is supposed to arrive at 12:15 pm every day, but sometimes its a little late or a little early, and sometimes its on time.



    I have gathered the data and timings for the past 20 days.



    Now, what I believe I should do is that I should arrange my data in ascending order and group them in ranges. To find the probability, my classmate said to do "Success over total." I think she means the total number of outcomes. This is what I`ve been told by a classmate, at least, this is what she is doing, and to be honest, what she told me sounds a little confusing because she did not elaborate on it.



    Here is my organized data set in ascending order: -4, -4, -4, -3, -3, -2, -2, -1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 3, 5.



    I am quite confused on how to do this, and I`ve asked around as well, but I am not getting quick responses. Am I on the right track? Is what my classmate told me to do right?



    THIS PART HAS BEEN EDITED



    This is what I did for the probability distribution table (click the Probability Distribution Table to view the table if you cannot view it on here):



    Probability Distribution Table



    NEWLY EDITED



    I have done the continuous probability distribution graph (the y-axis is the probability, and the x-axis is the wait time for the bus in minutes). Although, to me,This graph does not look right. Im pretty sure this is how the continuous probability distribution with a normal distribution curve is supposed to look like, considering Ive searched up and seen some things similar to what I have done.










    share|cite|improve this question









    New contributor




    Yashvi Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.























      0












      0








      0







      For an assignment in my high school Data Management class, I am required to make a probability distribution table for my data, which is "Wait Time for a Bus." It is not just any bus. It is a specific bus that arrives at a bus stop near my school, and I know someone who takes this bus, so I ask her to record the data down for me.



      This bus is supposed to arrive at 12:15 pm every day, but sometimes its a little late or a little early, and sometimes its on time.



      I have gathered the data and timings for the past 20 days.



      Now, what I believe I should do is that I should arrange my data in ascending order and group them in ranges. To find the probability, my classmate said to do "Success over total." I think she means the total number of outcomes. This is what I`ve been told by a classmate, at least, this is what she is doing, and to be honest, what she told me sounds a little confusing because she did not elaborate on it.



      Here is my organized data set in ascending order: -4, -4, -4, -3, -3, -2, -2, -1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 3, 5.



      I am quite confused on how to do this, and I`ve asked around as well, but I am not getting quick responses. Am I on the right track? Is what my classmate told me to do right?



      THIS PART HAS BEEN EDITED



      This is what I did for the probability distribution table (click the Probability Distribution Table to view the table if you cannot view it on here):



      Probability Distribution Table



      NEWLY EDITED



      I have done the continuous probability distribution graph (the y-axis is the probability, and the x-axis is the wait time for the bus in minutes). Although, to me,This graph does not look right. Im pretty sure this is how the continuous probability distribution with a normal distribution curve is supposed to look like, considering Ive searched up and seen some things similar to what I have done.










      share|cite|improve this question









      New contributor




      Yashvi Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      For an assignment in my high school Data Management class, I am required to make a probability distribution table for my data, which is "Wait Time for a Bus." It is not just any bus. It is a specific bus that arrives at a bus stop near my school, and I know someone who takes this bus, so I ask her to record the data down for me.



      This bus is supposed to arrive at 12:15 pm every day, but sometimes its a little late or a little early, and sometimes its on time.



      I have gathered the data and timings for the past 20 days.



      Now, what I believe I should do is that I should arrange my data in ascending order and group them in ranges. To find the probability, my classmate said to do "Success over total." I think she means the total number of outcomes. This is what I`ve been told by a classmate, at least, this is what she is doing, and to be honest, what she told me sounds a little confusing because she did not elaborate on it.



      Here is my organized data set in ascending order: -4, -4, -4, -3, -3, -2, -2, -1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 3, 5.



      I am quite confused on how to do this, and I`ve asked around as well, but I am not getting quick responses. Am I on the right track? Is what my classmate told me to do right?



      THIS PART HAS BEEN EDITED



      This is what I did for the probability distribution table (click the Probability Distribution Table to view the table if you cannot view it on here):



      Probability Distribution Table



      NEWLY EDITED



      I have done the continuous probability distribution graph (the y-axis is the probability, and the x-axis is the wait time for the bus in minutes). Although, to me,This graph does not look right. Im pretty sure this is how the continuous probability distribution with a normal distribution curve is supposed to look like, considering Ive searched up and seen some things similar to what I have done.







      probability probability-distributions uniform-continuity






      share|cite|improve this question









      New contributor




      Yashvi Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      Yashvi Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question








      edited 22 hours ago





















      New contributor




      Yashvi Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked Jan 2 at 21:53









      Yashvi Shah

      124




      124




      New contributor




      Yashvi Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      Yashvi Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Yashvi Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















          2 Answers
          2






          active

          oldest

          votes


















          0














          You populated the distribution table correctly so yes, you are doing it right. You can understand "success over total" as the probability. A more formal way to understand it would be as below
          $$
          text{Probability of something happening} = frac{text{Number of favorable outcomes}}{text{Total number of outcomes}}
          $$






          share|cite|improve this answer








          New contributor




          silent_monk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.


















          • For the probability distribution graph, would I do that vertical line graph? My data is continuous, so I’m thinking not the vertical line graph. Maybe a continuous probability distribution graph (I searched it up and I’m getting the bell curve for normal distribution)
            – Yashvi Shah
            yesterday










          • That would again depend on how the data set is defined. If the data is continuous and not discrete then your probability distribution is also continuous as you mentioned. Here is what I think about your case, the bus could be late by 1 or 2 or 3 or N minutes and it could take values in between as well like 1 minute and 15 seconds (equal to 1.25 minutes). So it is continuous and you got that right as well!
            – silent_monk
            yesterday












          • Haha well isn’t that amazing! Thank you for clarifying for me though. Very appreciated (:
            – Yashvi Shah
            yesterday










          • I have a question: I added all the probabilities up together and Im getting 3.7. Shouldnt a probability distribution table`s probabilities all add up to exactly 1?
            – Yashvi Shah
            yesterday










          • The reason you get P(total) > 1 is because you added up the probabilities corresponding to every occurrence of each event. We have considered the number of occurrences of each event (late by 1 minute, late by -2 minutes etc) already, while calculating the probability of each event (last column in the table). Now while adding them up, you just need to sum the probability corresponding to each event (P(late by 1 minute) + P(late by -2 minutes) + ..).
            – silent_monk
            yesterday



















          0














          Now that you have binned the data, all you need to do is to divide the size of each bin by the total number of observations. In your example, there are 20 observations. There are 3 entries in the -4 bin, so the probability of the wait time being -4 = 3/20. And so on.



          The choice of bins is somewhat arbitrary. For example, you could have created coarser bins of the type -4:-2, -1:1, etc. Or finer bins with higher resolution, though in your example that wouldn't be meaningful because your data resolution is 1 minute.



          To summarize, just think of the probability distribution as the histogram normalized by the total number of observations.






          share|cite|improve this answer





















          • I have included the probability distribution table in the edited part of my question. I`ve done it by taking what you told me into consideration here.
            – Yashvi Shah
            yesterday











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });






          Yashvi Shah is a new contributor. Be nice, and check out our Code of Conduct.










          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060019%2fprobability-distribution-table-of-bus-arrival%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0














          You populated the distribution table correctly so yes, you are doing it right. You can understand "success over total" as the probability. A more formal way to understand it would be as below
          $$
          text{Probability of something happening} = frac{text{Number of favorable outcomes}}{text{Total number of outcomes}}
          $$






          share|cite|improve this answer








          New contributor




          silent_monk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.


















          • For the probability distribution graph, would I do that vertical line graph? My data is continuous, so I’m thinking not the vertical line graph. Maybe a continuous probability distribution graph (I searched it up and I’m getting the bell curve for normal distribution)
            – Yashvi Shah
            yesterday










          • That would again depend on how the data set is defined. If the data is continuous and not discrete then your probability distribution is also continuous as you mentioned. Here is what I think about your case, the bus could be late by 1 or 2 or 3 or N minutes and it could take values in between as well like 1 minute and 15 seconds (equal to 1.25 minutes). So it is continuous and you got that right as well!
            – silent_monk
            yesterday












          • Haha well isn’t that amazing! Thank you for clarifying for me though. Very appreciated (:
            – Yashvi Shah
            yesterday










          • I have a question: I added all the probabilities up together and Im getting 3.7. Shouldnt a probability distribution table`s probabilities all add up to exactly 1?
            – Yashvi Shah
            yesterday










          • The reason you get P(total) > 1 is because you added up the probabilities corresponding to every occurrence of each event. We have considered the number of occurrences of each event (late by 1 minute, late by -2 minutes etc) already, while calculating the probability of each event (last column in the table). Now while adding them up, you just need to sum the probability corresponding to each event (P(late by 1 minute) + P(late by -2 minutes) + ..).
            – silent_monk
            yesterday
















          0














          You populated the distribution table correctly so yes, you are doing it right. You can understand "success over total" as the probability. A more formal way to understand it would be as below
          $$
          text{Probability of something happening} = frac{text{Number of favorable outcomes}}{text{Total number of outcomes}}
          $$






          share|cite|improve this answer








          New contributor




          silent_monk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.


















          • For the probability distribution graph, would I do that vertical line graph? My data is continuous, so I’m thinking not the vertical line graph. Maybe a continuous probability distribution graph (I searched it up and I’m getting the bell curve for normal distribution)
            – Yashvi Shah
            yesterday










          • That would again depend on how the data set is defined. If the data is continuous and not discrete then your probability distribution is also continuous as you mentioned. Here is what I think about your case, the bus could be late by 1 or 2 or 3 or N minutes and it could take values in between as well like 1 minute and 15 seconds (equal to 1.25 minutes). So it is continuous and you got that right as well!
            – silent_monk
            yesterday












          • Haha well isn’t that amazing! Thank you for clarifying for me though. Very appreciated (:
            – Yashvi Shah
            yesterday










          • I have a question: I added all the probabilities up together and Im getting 3.7. Shouldnt a probability distribution table`s probabilities all add up to exactly 1?
            – Yashvi Shah
            yesterday










          • The reason you get P(total) > 1 is because you added up the probabilities corresponding to every occurrence of each event. We have considered the number of occurrences of each event (late by 1 minute, late by -2 minutes etc) already, while calculating the probability of each event (last column in the table). Now while adding them up, you just need to sum the probability corresponding to each event (P(late by 1 minute) + P(late by -2 minutes) + ..).
            – silent_monk
            yesterday














          0












          0








          0






          You populated the distribution table correctly so yes, you are doing it right. You can understand "success over total" as the probability. A more formal way to understand it would be as below
          $$
          text{Probability of something happening} = frac{text{Number of favorable outcomes}}{text{Total number of outcomes}}
          $$






          share|cite|improve this answer








          New contributor




          silent_monk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          You populated the distribution table correctly so yes, you are doing it right. You can understand "success over total" as the probability. A more formal way to understand it would be as below
          $$
          text{Probability of something happening} = frac{text{Number of favorable outcomes}}{text{Total number of outcomes}}
          $$







          share|cite|improve this answer








          New contributor




          silent_monk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          share|cite|improve this answer



          share|cite|improve this answer






          New contributor




          silent_monk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          answered yesterday









          silent_monk

          16




          16




          New contributor




          silent_monk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.





          New contributor





          silent_monk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          silent_monk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.












          • For the probability distribution graph, would I do that vertical line graph? My data is continuous, so I’m thinking not the vertical line graph. Maybe a continuous probability distribution graph (I searched it up and I’m getting the bell curve for normal distribution)
            – Yashvi Shah
            yesterday










          • That would again depend on how the data set is defined. If the data is continuous and not discrete then your probability distribution is also continuous as you mentioned. Here is what I think about your case, the bus could be late by 1 or 2 or 3 or N minutes and it could take values in between as well like 1 minute and 15 seconds (equal to 1.25 minutes). So it is continuous and you got that right as well!
            – silent_monk
            yesterday












          • Haha well isn’t that amazing! Thank you for clarifying for me though. Very appreciated (:
            – Yashvi Shah
            yesterday










          • I have a question: I added all the probabilities up together and Im getting 3.7. Shouldnt a probability distribution table`s probabilities all add up to exactly 1?
            – Yashvi Shah
            yesterday










          • The reason you get P(total) > 1 is because you added up the probabilities corresponding to every occurrence of each event. We have considered the number of occurrences of each event (late by 1 minute, late by -2 minutes etc) already, while calculating the probability of each event (last column in the table). Now while adding them up, you just need to sum the probability corresponding to each event (P(late by 1 minute) + P(late by -2 minutes) + ..).
            – silent_monk
            yesterday


















          • For the probability distribution graph, would I do that vertical line graph? My data is continuous, so I’m thinking not the vertical line graph. Maybe a continuous probability distribution graph (I searched it up and I’m getting the bell curve for normal distribution)
            – Yashvi Shah
            yesterday










          • That would again depend on how the data set is defined. If the data is continuous and not discrete then your probability distribution is also continuous as you mentioned. Here is what I think about your case, the bus could be late by 1 or 2 or 3 or N minutes and it could take values in between as well like 1 minute and 15 seconds (equal to 1.25 minutes). So it is continuous and you got that right as well!
            – silent_monk
            yesterday












          • Haha well isn’t that amazing! Thank you for clarifying for me though. Very appreciated (:
            – Yashvi Shah
            yesterday










          • I have a question: I added all the probabilities up together and Im getting 3.7. Shouldnt a probability distribution table`s probabilities all add up to exactly 1?
            – Yashvi Shah
            yesterday










          • The reason you get P(total) > 1 is because you added up the probabilities corresponding to every occurrence of each event. We have considered the number of occurrences of each event (late by 1 minute, late by -2 minutes etc) already, while calculating the probability of each event (last column in the table). Now while adding them up, you just need to sum the probability corresponding to each event (P(late by 1 minute) + P(late by -2 minutes) + ..).
            – silent_monk
            yesterday
















          For the probability distribution graph, would I do that vertical line graph? My data is continuous, so I’m thinking not the vertical line graph. Maybe a continuous probability distribution graph (I searched it up and I’m getting the bell curve for normal distribution)
          – Yashvi Shah
          yesterday




          For the probability distribution graph, would I do that vertical line graph? My data is continuous, so I’m thinking not the vertical line graph. Maybe a continuous probability distribution graph (I searched it up and I’m getting the bell curve for normal distribution)
          – Yashvi Shah
          yesterday












          That would again depend on how the data set is defined. If the data is continuous and not discrete then your probability distribution is also continuous as you mentioned. Here is what I think about your case, the bus could be late by 1 or 2 or 3 or N minutes and it could take values in between as well like 1 minute and 15 seconds (equal to 1.25 minutes). So it is continuous and you got that right as well!
          – silent_monk
          yesterday






          That would again depend on how the data set is defined. If the data is continuous and not discrete then your probability distribution is also continuous as you mentioned. Here is what I think about your case, the bus could be late by 1 or 2 or 3 or N minutes and it could take values in between as well like 1 minute and 15 seconds (equal to 1.25 minutes). So it is continuous and you got that right as well!
          – silent_monk
          yesterday














          Haha well isn’t that amazing! Thank you for clarifying for me though. Very appreciated (:
          – Yashvi Shah
          yesterday




          Haha well isn’t that amazing! Thank you for clarifying for me though. Very appreciated (:
          – Yashvi Shah
          yesterday












          I have a question: I added all the probabilities up together and Im getting 3.7. Shouldnt a probability distribution table`s probabilities all add up to exactly 1?
          – Yashvi Shah
          yesterday




          I have a question: I added all the probabilities up together and Im getting 3.7. Shouldnt a probability distribution table`s probabilities all add up to exactly 1?
          – Yashvi Shah
          yesterday












          The reason you get P(total) > 1 is because you added up the probabilities corresponding to every occurrence of each event. We have considered the number of occurrences of each event (late by 1 minute, late by -2 minutes etc) already, while calculating the probability of each event (last column in the table). Now while adding them up, you just need to sum the probability corresponding to each event (P(late by 1 minute) + P(late by -2 minutes) + ..).
          – silent_monk
          yesterday




          The reason you get P(total) > 1 is because you added up the probabilities corresponding to every occurrence of each event. We have considered the number of occurrences of each event (late by 1 minute, late by -2 minutes etc) already, while calculating the probability of each event (last column in the table). Now while adding them up, you just need to sum the probability corresponding to each event (P(late by 1 minute) + P(late by -2 minutes) + ..).
          – silent_monk
          yesterday











          0














          Now that you have binned the data, all you need to do is to divide the size of each bin by the total number of observations. In your example, there are 20 observations. There are 3 entries in the -4 bin, so the probability of the wait time being -4 = 3/20. And so on.



          The choice of bins is somewhat arbitrary. For example, you could have created coarser bins of the type -4:-2, -1:1, etc. Or finer bins with higher resolution, though in your example that wouldn't be meaningful because your data resolution is 1 minute.



          To summarize, just think of the probability distribution as the histogram normalized by the total number of observations.






          share|cite|improve this answer





















          • I have included the probability distribution table in the edited part of my question. I`ve done it by taking what you told me into consideration here.
            – Yashvi Shah
            yesterday
















          0














          Now that you have binned the data, all you need to do is to divide the size of each bin by the total number of observations. In your example, there are 20 observations. There are 3 entries in the -4 bin, so the probability of the wait time being -4 = 3/20. And so on.



          The choice of bins is somewhat arbitrary. For example, you could have created coarser bins of the type -4:-2, -1:1, etc. Or finer bins with higher resolution, though in your example that wouldn't be meaningful because your data resolution is 1 minute.



          To summarize, just think of the probability distribution as the histogram normalized by the total number of observations.






          share|cite|improve this answer





















          • I have included the probability distribution table in the edited part of my question. I`ve done it by taking what you told me into consideration here.
            – Yashvi Shah
            yesterday














          0












          0








          0






          Now that you have binned the data, all you need to do is to divide the size of each bin by the total number of observations. In your example, there are 20 observations. There are 3 entries in the -4 bin, so the probability of the wait time being -4 = 3/20. And so on.



          The choice of bins is somewhat arbitrary. For example, you could have created coarser bins of the type -4:-2, -1:1, etc. Or finer bins with higher resolution, though in your example that wouldn't be meaningful because your data resolution is 1 minute.



          To summarize, just think of the probability distribution as the histogram normalized by the total number of observations.






          share|cite|improve this answer












          Now that you have binned the data, all you need to do is to divide the size of each bin by the total number of observations. In your example, there are 20 observations. There are 3 entries in the -4 bin, so the probability of the wait time being -4 = 3/20. And so on.



          The choice of bins is somewhat arbitrary. For example, you could have created coarser bins of the type -4:-2, -1:1, etc. Or finer bins with higher resolution, though in your example that wouldn't be meaningful because your data resolution is 1 minute.



          To summarize, just think of the probability distribution as the histogram normalized by the total number of observations.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 0:35









          Aditya Dua

          86418




          86418












          • I have included the probability distribution table in the edited part of my question. I`ve done it by taking what you told me into consideration here.
            – Yashvi Shah
            yesterday


















          • I have included the probability distribution table in the edited part of my question. I`ve done it by taking what you told me into consideration here.
            – Yashvi Shah
            yesterday
















          I have included the probability distribution table in the edited part of my question. I`ve done it by taking what you told me into consideration here.
          – Yashvi Shah
          yesterday




          I have included the probability distribution table in the edited part of my question. I`ve done it by taking what you told me into consideration here.
          – Yashvi Shah
          yesterday










          Yashvi Shah is a new contributor. Be nice, and check out our Code of Conduct.










          draft saved

          draft discarded


















          Yashvi Shah is a new contributor. Be nice, and check out our Code of Conduct.













          Yashvi Shah is a new contributor. Be nice, and check out our Code of Conduct.












          Yashvi Shah is a new contributor. Be nice, and check out our Code of Conduct.
















          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • Please be sure to answer the question. Provide details and share your research!

          But avoid



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060019%2fprobability-distribution-table-of-bus-arrival%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          Mario Kart Wii

          What does “Dominus providebit” mean?

          Antonio Litta Visconti Arese