I too am doing this exact same question right now, and I have no idea how to do it. If anyone can help out, it would be greatly appreciated.
So far, all I've been able to figure out is the IQR in percentage is 50%. This is the problem.
The percent of the observations that are classified as outliers by the 1.5x IQR rule is the same in any normal distribution. What is this percent? Show your method clearly.
I really don't have a clue how to begin to solve this. If someone can get me started, I would much appreciate it.
I too am doing this exact same question right now, and I have no idea how to do it. If anyone can help out, it would be greatly appreciated.
I'm sorry fellas but there doesn't seem to be anyone answering in this forum, only people doing questions. I'm new to statistics forums, so maybe this is the case in a such specilized context as statistics...
Think of a normal density plot. What is the range of x values which would contain 50% of the values centered on the mean? Now, how many values would be contained in +/- 1.5 IQR and what is the x range? How many values are left over?
Hints: think z-scores; boxplots; the meaning of 1.5 IQR
Sorry if this seems cryptic but, really, it doesn't get much simpler than this.
Last edited by DAV; 10-04-2008 at 03:29 PM. Reason: Slight clarification
The interquartile range is
Q_3 - Q_1
Try to write both quartiles in terms of the mean and standard deviation of the distribution, and that will give you the spread about the mean in language the normal distribution is designed to use
look up Q1 and Q3 in the Stand.Norm.Table they are +/- 0.675. Then take Q3 -Q1 and you get 1.35. Then use Tukey's outlier rule and multiply by 1.5 to get 2.025. Add to the Q3 and get cutoffs of z = 2.70. Then look up the area to the right 0.0034 and double, multiply by 100 and get 0.69%
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