Mixing
Probability Distribution Table of Bus Arrival
$begingroup$
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):
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,
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
asked Jan 2 at 21:53
Yashvi ShahYashvi Shah
144
$endgroup$
add a comment |
$begingroup$
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):
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, 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
asked Jan 2 at 21:53
Yashvi ShahYashvi Shah
144
$endgroup$
add a comment |
$begingroup$
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):
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, 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
asked Jan 2 at 21:53
Yashvi ShahYashvi Shah
144
$endgroup$
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):
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, 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
probability probability-distributions uniform-continuity
asked Jan 2 at 21:53
Yashvi ShahYashvi Shah
144
asked Jan 2 at 21:53
Yashvi ShahYashvi Shah
144
edited Jan 6 at 3:38
Yashvi Shah
asked Jan 2 at 21:53
Yashvi ShahYashvi Shah
144
asked Jan 2 at 21:53
Yashvi ShahYashvi Shah
144
asked Jan 2 at 21:53
Yashvi ShahYashvi Shah
144
144
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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
$$
text{Probability of something happening} = frac{text{Number of favorable outcomes}}{text{Total number of outcomes}}
$$
answered Jan 5 at 5:17
silent_monksilent_monk
16
$endgroup$
$begingroup$
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)
$endgroup$
– Yashvi Shah
Jan 5 at 5:59
$begingroup$
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!
$endgroup$
– silent_monk
Jan 5 at 6:16
$begingroup$
Haha well isn’t that amazing! Thank you for clarifying for me though. Very appreciated (:
$endgroup$
– Yashvi Shah
Jan 5 at 6:20
$begingroup$
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?
$endgroup$
– Yashvi Shah
Jan 5 at 16:56
$begingroup$
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) + ..).
$endgroup$
– silent_monk
Jan 5 at 20:01
|
show 8 more comments
$begingroup$
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.
answered Jan 4 at 0:35
Aditya DuaAditya Dua
94918
$endgroup$
$begingroup$
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.
$endgroup$
– Yashvi Shah
Jan 5 at 4:35
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
$$
text{Probability of something happening} = frac{text{Number of favorable outcomes}}{text{Total number of outcomes}}
$$
answered Jan 5 at 5:17
silent_monksilent_monk
16
$endgroup$
$begingroup$
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)
$endgroup$
– Yashvi Shah
Jan 5 at 5:59
$begingroup$
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!
$endgroup$
– silent_monk
Jan 5 at 6:16
$begingroup$
Haha well isn’t that amazing! Thank you for clarifying for me though. Very appreciated (:
$endgroup$
– Yashvi Shah
Jan 5 at 6:20
$begingroup$
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?
$endgroup$
– Yashvi Shah
Jan 5 at 16:56
$begingroup$
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) + ..).
$endgroup$
– silent_monk
Jan 5 at 20:01
|
show 8 more comments
$begingroup$
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
$$
text{Probability of something happening} = frac{text{Number of favorable outcomes}}{text{Total number of outcomes}}
$$
answered Jan 5 at 5:17
silent_monksilent_monk
16
$endgroup$
$begingroup$
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)
$endgroup$
– Yashvi Shah
Jan 5 at 5:59
$begingroup$
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!
$endgroup$
– silent_monk
Jan 5 at 6:16
$begingroup$
Haha well isn’t that amazing! Thank you for clarifying for me though. Very appreciated (:
$endgroup$
– Yashvi Shah
Jan 5 at 6:20
$begingroup$
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?
$endgroup$
– Yashvi Shah
Jan 5 at 16:56
$begingroup$
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) + ..).
$endgroup$
– silent_monk
Jan 5 at 20:01
|
show 8 more comments
$begingroup$
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
$$
text{Probability of something happening} = frac{text{Number of favorable outcomes}}{text{Total number of outcomes}}
$$
answered Jan 5 at 5:17
silent_monksilent_monk
16
$endgroup$
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
$$
text{Probability of something happening} = frac{text{Number of favorable outcomes}}{text{Total number of outcomes}}
$$
answered Jan 5 at 5:17
silent_monksilent_monk
16
answered Jan 5 at 5:17
silent_monksilent_monk
16
answered Jan 5 at 5:17
silent_monksilent_monk
16
answered Jan 5 at 5:17
silent_monksilent_monk
16
16
$begingroup$
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)
$endgroup$
– Yashvi Shah
Jan 5 at 5:59
$begingroup$
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!
$endgroup$
– silent_monk
Jan 5 at 6:16
$begingroup$
Haha well isn’t that amazing! Thank you for clarifying for me though. Very appreciated (:
$endgroup$
– Yashvi Shah
Jan 5 at 6:20
$begingroup$
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?
$endgroup$
– Yashvi Shah
Jan 5 at 16:56
$begingroup$
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) + ..).
$endgroup$
– silent_monk
Jan 5 at 20:01
|
show 8 more comments
$begingroup$
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)
$endgroup$
– Yashvi Shah
Jan 5 at 5:59
$begingroup$
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!
$endgroup$
– silent_monk
Jan 5 at 6:16
$begingroup$
Haha well isn’t that amazing! Thank you for clarifying for me though. Very appreciated (:
$endgroup$
– Yashvi Shah
Jan 5 at 6:20
$begingroup$
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?
$endgroup$
– Yashvi Shah
Jan 5 at 16:56
$begingroup$
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) + ..).
$endgroup$
– silent_monk
Jan 5 at 20:01
$begingroup$
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)
$endgroup$
– Yashvi Shah
Jan 5 at 5:59
$begingroup$
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)
$endgroup$
– Yashvi Shah
Jan 5 at 5:59
$begingroup$
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!
$endgroup$
– silent_monk
Jan 5 at 6:16
$begingroup$
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!
$endgroup$
– silent_monk
Jan 5 at 6:16
$begingroup$
Haha well isn’t that amazing! Thank you for clarifying for me though. Very appreciated (:
$endgroup$
– Yashvi Shah
Jan 5 at 6:20
$begingroup$
Haha well isn’t that amazing! Thank you for clarifying for me though. Very appreciated (:
$endgroup$
– Yashvi Shah
Jan 5 at 6:20
$begingroup$
I have a question: I added all the probabilities up together and I
m getting 3.7. Shouldnt a probability distribution table`s probabilities all add up to exactly 1?$endgroup$
– Yashvi Shah
Jan 5 at 16:56
$begingroup$
I have a question: I added all the probabilities up together and I
m getting 3.7. Shouldnt a probability distribution table`s probabilities all add up to exactly 1?$endgroup$
– Yashvi Shah
Jan 5 at 16:56
$begingroup$
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) + ..).
$endgroup$
– silent_monk
Jan 5 at 20:01
$begingroup$
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) + ..).
$endgroup$
– silent_monk
Jan 5 at 20:01
|
show 8 more comments
$begingroup$
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.
answered Jan 4 at 0:35
Aditya DuaAditya Dua
94918
$endgroup$
$begingroup$
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.
$endgroup$
– Yashvi Shah
Jan 5 at 4:35
add a comment |
$begingroup$
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.
answered Jan 4 at 0:35
Aditya DuaAditya Dua
94918
$endgroup$
$begingroup$
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.
$endgroup$
– Yashvi Shah
Jan 5 at 4:35
add a comment |
$begingroup$
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.
answered Jan 4 at 0:35
Aditya DuaAditya Dua
94918
$endgroup$
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.
answered Jan 4 at 0:35
Aditya DuaAditya Dua
94918
answered Jan 4 at 0:35
Aditya DuaAditya Dua
94918
answered Jan 4 at 0:35
Aditya DuaAditya Dua
94918
answered Jan 4 at 0:35
Aditya DuaAditya Dua
94918
94918
$begingroup$
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.
$endgroup$
– Yashvi Shah
Jan 5 at 4:35
add a comment |
$begingroup$
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.
$endgroup$
– Yashvi Shah
Jan 5 at 4:35
$begingroup$
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.
$endgroup$
– Yashvi Shah
Jan 5 at 4:35
$begingroup$
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.
$endgroup$
– Yashvi Shah
Jan 5 at 4:35
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