Mixing
Normal distribtion (a little confused?)
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On a particular day, 50% of the employees arrive at work by 8.30 am, and 10% had not arrived by 8.55 am.
Assuming a normal model, find the standard deviation of the arrival times, in minutes?
What I tried: From what I understand, 40% of the people showed up in the gap of 25 minutes. Therefore, taking the difference between 0.90 - 0.50 = 0.40.
My query is how will I form an equation that I can solve because I don't have a x-bar, mean and standard deviation (3 unknowns?)
probability self-study normal-distribution
edited 3 hours ago
Kodiologist
15.8k22951
asked 3 hours ago
user585380
316
add a comment |Â
up vote
1
down vote
favorite
On a particular day, 50% of the employees arrive at work by 8.30 am, and 10% had not arrived by 8.55 am.
Assuming a normal model, find the standard deviation of the arrival times, in minutes?
What I tried: From what I understand, 40% of the people showed up in the gap of 25 minutes. Therefore, taking the difference between 0.90 - 0.50 = 0.40.
My query is how will I form an equation that I can solve because I don't have a x-bar, mean and standard deviation (3 unknowns?)
probability self-study normal-distribution
edited 3 hours ago
Kodiologist
15.8k22951
asked 3 hours ago
user585380
316
Thanks for specifying what you have already tried. This sounds like a homework or some type of self-directed study; if so, please add theself-studytag to the question. Such questions are encouraged on this site, but please read the info that explains how we try to provide help.
– EdM
3 hours ago
@EdM tag is added. Not a homework question - its a self-study one.
– user585380
3 hours ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
On a particular day, 50% of the employees arrive at work by 8.30 am, and 10% had not arrived by 8.55 am.
Assuming a normal model, find the standard deviation of the arrival times, in minutes?
What I tried: From what I understand, 40% of the people showed up in the gap of 25 minutes. Therefore, taking the difference between 0.90 - 0.50 = 0.40.
My query is how will I form an equation that I can solve because I don't have a x-bar, mean and standard deviation (3 unknowns?)
probability self-study normal-distribution
edited 3 hours ago
Kodiologist
15.8k22951
asked 3 hours ago
user585380
316
On a particular day, 50% of the employees arrive at work by 8.30 am, and 10% had not arrived by 8.55 am.
Assuming a normal model, find the standard deviation of the arrival times, in minutes?
What I tried: From what I understand, 40% of the people showed up in the gap of 25 minutes. Therefore, taking the difference between 0.90 - 0.50 = 0.40.
My query is how will I form an equation that I can solve because I don't have a x-bar, mean and standard deviation (3 unknowns?)
probability self-study normal-distribution
probability self-study normal-distribution
edited 3 hours ago
Kodiologist
15.8k22951
asked 3 hours ago
user585380
316
edited 3 hours ago
Kodiologist
15.8k22951
asked 3 hours ago
user585380
316
edited 3 hours ago
Kodiologist
15.8k22951
edited 3 hours ago
Kodiologist
15.8k22951
edited 3 hours ago
Kodiologist
15.8k22951
15.8k22951
asked 3 hours ago
user585380
316
asked 3 hours ago
user585380
316
asked 3 hours ago
user585380
316
316
Thanks for specifying what you have already tried. This sounds like a homework or some type of self-directed study; if so, please add theself-studytag to the question. Such questions are encouraged on this site, but please read the info that explains how we try to provide help.
– EdM
3 hours ago
@EdM tag is added. Not a homework question - its a self-study one.
– user585380
3 hours ago
add a comment |Â
Thanks for specifying what you have already tried. This sounds like a homework or some type of self-directed study; if so, please add theself-studytag to the question. Such questions are encouraged on this site, but please read the info that explains how we try to provide help.
– EdM
3 hours ago
@EdM tag is added. Not a homework question - its a self-study one.
– user585380
3 hours ago
Thanks for specifying what you have already tried. This sounds like a homework or some type of self-directed study; if so, please add the
self-study tag to the question. Such questions are encouraged on this site, but please read the info that explains how we try to provide help.– EdM
3 hours ago
Thanks for specifying what you have already tried. This sounds like a homework or some type of self-directed study; if so, please add the
self-study tag to the question. Such questions are encouraged on this site, but please read the info that explains how we try to provide help.– EdM
3 hours ago
@EdM tag is added. Not a homework question - its a self-study one.
– user585380
3 hours ago
@EdM tag is added. Not a homework question - its a self-study one.
– user585380
3 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
As median = mean for the normal distribution you know that: $$/mu=8.30$$. Furthermore you know that $$/Phi_8.30,/sigma ^2(8.55)=0.9$$ (denoting with $/Phi_/mu,/sigma ^2$ the distribution function of a $N(/mu, /sigma ^2)$
This is equivalent to
$$/Phi_0,1(/frac8.55-8.30/sigma)=0.9$$
Now you only have to apply the inverse oh the distribution function of the $N(0,1)$ and solve for $/sigma$
answered 1 hour ago
0rangetree
456212
So basically mu = 8.30?
– user585380
1 hour ago
Yes because 50% arrive before 8.30 and therefore 50% After 8.30. This means that the Median is 8.30 and as Median = mean for normal distributions you know that mean = 8.30
– 0rangetree
39 mins ago
My sigma value comes out to be 0.194. Their value is 19.50. Why is an *100 difference here? Do they expect to take 830 and not 8.30?
– user585380
29 mins ago
add a comment |Â
up vote
1
down vote
The normal probability distribution has several nice properties that can help resolve your confusion. Consider in particular its symmetry: half of the cases have values lower than the mean of the distribution (the median equals the mean). So you do have information about the mean of the distribution of arrival times. You also have information about the 90th percentile of arrival times, when 10% hadn't yet arrived. That's enough information to estimate the standard deviation based on normal probability tables.
answered 3 hours ago
EdM
19.9k23388
Yes, I do know about this property. But there is only a 'timing' given. How do I convert that into a mean? Will mean be 30?
– user585380
2 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
As median = mean for the normal distribution you know that: $$/mu=8.30$$. Furthermore you know that $$/Phi_8.30,/sigma ^2(8.55)=0.9$$ (denoting with $/Phi_/mu,/sigma ^2$ the distribution function of a $N(/mu, /sigma ^2)$
This is equivalent to
$$/Phi_0,1(/frac8.55-8.30/sigma)=0.9$$
Now you only have to apply the inverse oh the distribution function of the $N(0,1)$ and solve for $/sigma$
answered 1 hour ago
0rangetree
456212
So basically mu = 8.30?
– user585380
1 hour ago
Yes because 50% arrive before 8.30 and therefore 50% After 8.30. This means that the Median is 8.30 and as Median = mean for normal distributions you know that mean = 8.30
– 0rangetree
39 mins ago
My sigma value comes out to be 0.194. Their value is 19.50. Why is an *100 difference here? Do they expect to take 830 and not 8.30?
– user585380
29 mins ago
add a comment |Â
up vote
1
down vote
accepted
As median = mean for the normal distribution you know that: $$/mu=8.30$$. Furthermore you know that $$/Phi_8.30,/sigma ^2(8.55)=0.9$$ (denoting with $/Phi_/mu,/sigma ^2$ the distribution function of a $N(/mu, /sigma ^2)$
This is equivalent to
$$/Phi_0,1(/frac8.55-8.30/sigma)=0.9$$
Now you only have to apply the inverse oh the distribution function of the $N(0,1)$ and solve for $/sigma$
answered 1 hour ago
0rangetree
456212
So basically mu = 8.30?
– user585380
1 hour ago
Yes because 50% arrive before 8.30 and therefore 50% After 8.30. This means that the Median is 8.30 and as Median = mean for normal distributions you know that mean = 8.30
– 0rangetree
39 mins ago
My sigma value comes out to be 0.194. Their value is 19.50. Why is an *100 difference here? Do they expect to take 830 and not 8.30?
– user585380
29 mins ago
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
As median = mean for the normal distribution you know that: $$/mu=8.30$$. Furthermore you know that $$/Phi_8.30,/sigma ^2(8.55)=0.9$$ (denoting with $/Phi_/mu,/sigma ^2$ the distribution function of a $N(/mu, /sigma ^2)$
This is equivalent to
$$/Phi_0,1(/frac8.55-8.30/sigma)=0.9$$
Now you only have to apply the inverse oh the distribution function of the $N(0,1)$ and solve for $/sigma$
answered 1 hour ago
0rangetree
456212
As median = mean for the normal distribution you know that: $$/mu=8.30$$. Furthermore you know that $$/Phi_8.30,/sigma ^2(8.55)=0.9$$ (denoting with $/Phi_/mu,/sigma ^2$ the distribution function of a $N(/mu, /sigma ^2)$
This is equivalent to
$$/Phi_0,1(/frac8.55-8.30/sigma)=0.9$$
Now you only have to apply the inverse oh the distribution function of the $N(0,1)$ and solve for $/sigma$
answered 1 hour ago
0rangetree
456212
answered 1 hour ago
0rangetree
456212
answered 1 hour ago
0rangetree
456212
answered 1 hour ago
0rangetree
456212
456212
So basically mu = 8.30?
– user585380
1 hour ago
Yes because 50% arrive before 8.30 and therefore 50% After 8.30. This means that the Median is 8.30 and as Median = mean for normal distributions you know that mean = 8.30
– 0rangetree
39 mins ago
My sigma value comes out to be 0.194. Their value is 19.50. Why is an *100 difference here? Do they expect to take 830 and not 8.30?
– user585380
29 mins ago
add a comment |Â
So basically mu = 8.30?
– user585380
1 hour ago
Yes because 50% arrive before 8.30 and therefore 50% After 8.30. This means that the Median is 8.30 and as Median = mean for normal distributions you know that mean = 8.30
– 0rangetree
39 mins ago
My sigma value comes out to be 0.194. Their value is 19.50. Why is an *100 difference here? Do they expect to take 830 and not 8.30?
– user585380
29 mins ago
So basically mu = 8.30?
– user585380
1 hour ago
So basically mu = 8.30?
– user585380
1 hour ago
Yes because 50% arrive before 8.30 and therefore 50% After 8.30. This means that the Median is 8.30 and as Median = mean for normal distributions you know that mean = 8.30
– 0rangetree
39 mins ago
Yes because 50% arrive before 8.30 and therefore 50% After 8.30. This means that the Median is 8.30 and as Median = mean for normal distributions you know that mean = 8.30
– 0rangetree
39 mins ago
My sigma value comes out to be 0.194. Their value is 19.50. Why is an *100 difference here? Do they expect to take 830 and not 8.30?
– user585380
29 mins ago
My sigma value comes out to be 0.194. Their value is 19.50. Why is an *100 difference here? Do they expect to take 830 and not 8.30?
– user585380
29 mins ago
add a comment |Â
up vote
1
down vote
The normal probability distribution has several nice properties that can help resolve your confusion. Consider in particular its symmetry: half of the cases have values lower than the mean of the distribution (the median equals the mean). So you do have information about the mean of the distribution of arrival times. You also have information about the 90th percentile of arrival times, when 10% hadn't yet arrived. That's enough information to estimate the standard deviation based on normal probability tables.
answered 3 hours ago
EdM
19.9k23388
Yes, I do know about this property. But there is only a 'timing' given. How do I convert that into a mean? Will mean be 30?
– user585380
2 hours ago
add a comment |Â
up vote
1
down vote
The normal probability distribution has several nice properties that can help resolve your confusion. Consider in particular its symmetry: half of the cases have values lower than the mean of the distribution (the median equals the mean). So you do have information about the mean of the distribution of arrival times. You also have information about the 90th percentile of arrival times, when 10% hadn't yet arrived. That's enough information to estimate the standard deviation based on normal probability tables.
answered 3 hours ago
EdM
19.9k23388
Yes, I do know about this property. But there is only a 'timing' given. How do I convert that into a mean? Will mean be 30?
– user585380
2 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The normal probability distribution has several nice properties that can help resolve your confusion. Consider in particular its symmetry: half of the cases have values lower than the mean of the distribution (the median equals the mean). So you do have information about the mean of the distribution of arrival times. You also have information about the 90th percentile of arrival times, when 10% hadn't yet arrived. That's enough information to estimate the standard deviation based on normal probability tables.
answered 3 hours ago
EdM
19.9k23388
The normal probability distribution has several nice properties that can help resolve your confusion. Consider in particular its symmetry: half of the cases have values lower than the mean of the distribution (the median equals the mean). So you do have information about the mean of the distribution of arrival times. You also have information about the 90th percentile of arrival times, when 10% hadn't yet arrived. That's enough information to estimate the standard deviation based on normal probability tables.
answered 3 hours ago
EdM
19.9k23388
answered 3 hours ago
EdM
19.9k23388
answered 3 hours ago
EdM
19.9k23388
answered 3 hours ago
EdM
19.9k23388
19.9k23388
Yes, I do know about this property. But there is only a 'timing' given. How do I convert that into a mean? Will mean be 30?
– user585380
2 hours ago
add a comment |Â
Yes, I do know about this property. But there is only a 'timing' given. How do I convert that into a mean? Will mean be 30?
– user585380
2 hours ago
Yes, I do know about this property. But there is only a 'timing' given. How do I convert that into a mean? Will mean be 30?
– user585380
2 hours ago
Yes, I do know about this property. But there is only a 'timing' given. How do I convert that into a mean? Will mean be 30?
– user585380
2 hours ago
add a comment |Â
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Thanks for specifying what you have already tried. This sounds like a homework or some type of self-directed study; if so, please add the
self-studytag to the question. Such questions are encouraged on this site, but please read the info that explains how we try to provide help.– EdM
3 hours ago
@EdM tag is added. Not a homework question - its a self-study one.
– user585380
3 hours ago