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My teacher really can't teach this class! This class is so bad and I regret taking it..here are some questions that I don't get at all
6. The duration of Alzheimer’s disease, from the onset of symptoms until death, ranges from 3 to 20 years, with a mean of 8 years and a standard deviation of 4 years. The administrator of a large medical center randomly selects the medical records of 30 deceased Alzheimer’s patients and records the duration of the disease for each. Find the probability that the average duration of the 30 durations will lie within 1 year of the overall mean of 8 years.
a) 99
b) 0.4147
c) 0.1706
d) 0.1191
e) 0.8291
What the fuck?? What formula do I use? I know what the mean and standard deviation are...now what? Do I use a z-curve? lol i'm so bad at this FUCK ME
7. The duration of Alzheimer’s disease, from the onset of symptoms until death, ranges from 3 to 20 years, with a mean of 8 years and a standard deviation of 4 years. The administrator of a large medical center randomly selects the medical records of 30 deceased Alzheimer’s patients and records the duration of the disease for each. Find the value L such that there is a probability of .99 that the average duration of the 30 patients lies less than L years above the overall mean of 8 years.
a) 0.73
b) 0.94
c) 1.48
d) 1.70
e) 2.33
Same as #6, how do I do it?
I'm surprised I somehow maintain a B in this class -_-
 
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trust me, nobody can teach statistics. The only people who understand it are the really smart people, who don't become teachers anyway
anyway
ugh I don't remember this stuff, you don't use the normal Z tests, you do something with your calculator... I'll get back to this if I remember...
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Alzheimer patient - 8 year mean, 4 years standard deviation.
The sum of thirty Alzheimer's patients is 240 years with 4*sqrt(30) standard deviation - remember, the variance of the sum is the sum of the variances.
Divide by thirty to get the average - the average distribution can be expected to have 8 year mean and 4*sqrt(30)/30 standard deviation.
You want the chance that the average will lie within one year of eight years, but since you know the standard deviation and can express the range in terms of standard deviations you can do a two-sided Z-test/whatever off your table to get the chances.
Seven is done similarly. Same mean and standard deviation - use Z-test again, but one-sided this time - you want to look up .99 in the table and check how many standard deviations it corresponds to, then add that to the mean.
You do have a Z-table, right?
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Yeah! I have a z-table. Meanwhile I will see what you did =D
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On February 14 2008 14:50 Sunyveil wrote:
trust me, nobody can teach statistics. The only people who understand it are the really smart people, who don't become teachers anyway
anyway
ugh I don't remember this stuff, you don't use the normal Z tests, you do something with your calculator... I'll get back to this if I remember...
i had a rocket scientist for a teacher. she taught statistics pretty fucking well if you ask me.
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Is it possible to plug this in the calculator without doing all the z-table stuff? :X
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On February 14 2008 15:00 EmeraldSparks wrote:
Alzheimer patient - 8 year mean, 4 years standard deviation.
The sum of thirty Alzheimer's patients is 240 years with 4*sqrt(30) standard deviation - remember, the variance of the sum is the sum of the variances.
Divide by thirty to get the average - the average distribution can be expected to have 8 year mean and 4*sqrt(30)/30 standard deviation.
You want the chance that the average will lie within one year of eight years, but since you know the standard deviation and can express the range in terms of standard deviations you can do a two-sided Z-test/whatever off your table to get the chances.
Seven is done similarly. Same mean and standard deviation - use Z-test again, but one-sided this time - you want to look up .99 in the table and check how many standard deviations it corresponds to, then add that to the mean.
You do have a Z-table, right?
What worries me is about the limits of 3 and 20 years for the duration. Firstly I'm guessing each duration is modelled by a normal distribution, even though the question doesn't state that. In that case, it is not totally correct to say that the distribution of the mean of 30 patients is a normal distribution with mean 8 and standard deviation 4*sqrt(30)/30) because since the ends are "cut off", it's no longer a normal distrubtion. Since it's impossible to now have durations lower than 3 or higher than 20, the mean also can't be lower than 3 or higher than 20, which increases the probably that the mean is within 1 year of 8 years.
In fact I have no idea how to do this because of that complication, cause you can't use your normal distribution tables for this.
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Have you realized that ALL AP Stat questions seem to deal with death or disease?
Why can't it be happy things T_T
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On February 14 2008 22:07 LiLu wrote:
Have you realized that ALL AP Stat questions seem to deal with death or disease?
Why can't it be happy things T_T
lol I guess.
This test was ridiculously easy..two wrong max. But what the FUCK!
I don't get this class and this course lol.
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In fact I have no idea how to do this because of that complication, cause you can't use your normal distribution tables for this.
In AP Stats it's generally accepted that you ignore all such complications. (The only possible reason you might need to know them is if they specifically ask what could be wrong, and even then you don't need to do calculations for it. Further, it's general practice to report mean and standard deviation to describe a distribution only if its approximately normal.
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Man im glad I just finished my statistics exams. Hope I passed and dont have to see this for a while.
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EPT
