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.
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
asked Jan 4 at 22:36
Yashvi ShahYashvi Shah
166
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add a comment |
$begingroup$
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.
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
asked Jan 4 at 22:36
Yashvi ShahYashvi Shah
166
$endgroup$
add a comment |
$begingroup$
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.
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
asked Jan 4 at 22:36
Yashvi ShahYashvi Shah
166
$endgroup$
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.
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
asked Jan 4 at 22:36
Yashvi ShahYashvi Shah
166
asked Jan 4 at 22:36
Yashvi ShahYashvi Shah
166
asked Jan 4 at 22:36
Yashvi ShahYashvi Shah
166
asked Jan 4 at 22:36
Yashvi ShahYashvi Shah
166
asked Jan 4 at 22:36
Yashvi ShahYashvi Shah
166
166
add a comment |
add a comment |
1 Answer
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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 $mu+a$.
answered Jan 4 at 23:01
Peter ForemanPeter Foreman
5,9091317
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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...
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– Yashvi Shah
Jan 4 at 23:49
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Yes your initial estimate was quite accurate
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– Peter Foreman
Jan 5 at 0:24
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Wait, how come did you do 1 - 0.95 / 2 and 2(a)? Why the /2/ for both of these?
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– Yashvi Shah
Jan 5 at 1:14
add a comment |
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1 Answer
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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 $mu+a$.
answered Jan 4 at 23:01
Peter ForemanPeter Foreman
5,9091317
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $mu+a$.
answered Jan 4 at 23:01
Peter ForemanPeter Foreman
5,9091317
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $mu+a$.
answered Jan 4 at 23:01
Peter ForemanPeter Foreman
5,9091317
$endgroup$
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 $mu+a$.
answered Jan 4 at 23:01
Peter ForemanPeter Foreman
5,9091317
answered Jan 4 at 23:01
Peter ForemanPeter Foreman
5,9091317
answered Jan 4 at 23:01
Peter ForemanPeter Foreman
5,9091317
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
add a comment |
$begingroup$
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
add a comment |
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