Normal Random Variables
If X is normal with a mean of 8.48 and standard deviation of 1.60, what are the two values of X which include the central 80% of the probability?
Thanks
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March 29, 2009 5:45 p.m. PHeller HalfDork
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March 29, 2009 5:57 p.m. Toyman01 Reader
Uh...Just a guess...Miata?
Makes me glad I'm not in school any more.
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March 29, 2009 6:51 p.m. mtn Dork
Crap... Let me go find my book, its buried somewhere... I should know this
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March 29, 2009 7:04 p.m. Per Schroeder Technical Editor/Advertising Director
6.432, 10.528
I thought the upper and lower deciles were 1.28 * sd in a normal distribution.
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March 29, 2009 7:06 p.m. mtn Dork
^^ He beat me to it.
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March 29, 2009 7:14 p.m. PHeller HalfDork
But how do you figure that out?
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March 29, 2009 7:24 p.m. Per Schroeder Technical Editor/Advertising Director
Z score table for .4000 which would have the 40% above and 40% below. I remembered it somewhat from teaching statistics in graduate school—enough so that a quick google search got me the exact number.
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March 29, 2009 7:27 p.m. Rangeball Reader
Z = (X - u)/(sigma)
Rearrange to get P(-1.28 < (X - u)/(sigma) < 1.28)
P(u - 1.28sigma < X < u + 1.28sigma) = .80
P(8.48 - 1.28*1.6 < X < 8.48 + 1.28*1.6) = P(6.432 < X < 10.528)
To get 1.28 I went to the Z table and found .10 (because [1 - .80] 20% = alpha/2 = .1). Then its just plugging into the formula. u is your mean and sigma is your standard deviation.
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March 29, 2009 7:49 p.m. PHeller HalfDork
I'll probably have more for you guys.
I need a Stat tutor because this crap just blows my mind. I think it's all the different terminology that gets me. Constantly switching things up.
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March 29, 2009 7:56 p.m. Rangeball Reader
I am always happy to help. I intend to post my statistic project once I get it finished. Its a neat little experiment in regressions, stock prices, and Google's search volume index.
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March 29, 2009 8:39 p.m. 11110000 New Reader
The Excel way:
=NORMINV(0.1,8.48,1.6)
=NORMINV(0.9,8.48,1.6)
You may have to go to 'Tools'-->'Add-ins' and add the 'Data Analysis' toolpack.
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March 29, 2009 8:53 p.m. PHeller HalfDork
µ = 629 σ = 60 Engineering school only accepts those in the top 15%, what is the minimum score in order to be accepted?
I can't figure how to go about doing these...I guess I just don't know the "steps" to approaching problems.
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March 29, 2009 9:40 p.m. Tommy Suddard SonDork
78.9% of statistics are made up, including this one.
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March 29, 2009 9:55 p.m. Tighe New Reader
I like Unibrogue.
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March 29, 2009 10:00 p.m. AngryCorvair Dork
PHeller wrote:
µ = 629 σ = 60 Engineering school only accepts those in the top 15%, what is the minimum score in order to be accepted?
I can't figure how to go about doing these...I guess I just don't know the "steps" to approaching problems.
from the answers others have posted, sounds like there's a table which relates decimal sigmas to percentages of populations. having almost no stats experience i will take them at their words. but i've got years and years of problem-solving experience, and i think what i've outlined below (leaning on what others have posted above) would be a reasonable way to approach the problem:
draw a normal distribution bell curve.
draw a vertical line through the center of the curve, that's your µ = 629.
σ is defined as the distance (in this case, score) you have to travel from the mean in order to capture some percentage of the population of scores.
in this case, you need to know how many sigmas away from the mean is the 85th percentile, which i believe you'll get from the table others have mentioned.
then multiply the value from the table by the value of sigma, and add that to the mean, and that's the score required.
remember the words of charles kettering: "A problem well-stated is a problem half-solved."
anyway, following Per's example i googled "Z table statistics" and found that the z value corresponding to a 15% population above is about 1.04. so 629 plus 1.04*60 is the test score required to get into engineering school.
answer = 691.4, unless i suck at teaching something i think i just learned.
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March 29, 2009 10:47 p.m. alex Reader
Tighe wrote:
I like Unibrogue.
Heh. Is that the accent in which one speaks after one too many Unibroue beers?
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March 30, 2009 9:07 a.m. Rangeball Reader
PHeller, are you allowed to use Excel on your test? If so, I highly recommend just learning how to use that. Saves time and valuable brain space. The Data Analysis Plus will have about 80% of what you need. I can send you an Excel spreadsheet that will help with pretty much everything your are trying to do.
Lastly, the table referenced is the Z table. It has a Z value around the boarder (+ is the inverse of - and vice versa) with p-values (percentage) in the middle. You can convert p-values to Z values or vice versa.
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March 30, 2009 8:20 p.m. Tighe New Reader
alex wrote:
Heh. Is that the accent in which one speaks after one too many Unibroue beers?
When I posted that I... wasn't performing at my peak, to say the least. Diageo products can be blamed this time however.
