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dee
10-02-2005, 10:07 PM
how would one find quartiles in a normal distrubtion?
im just confused as to what im supposed to look for
i know the mean/median/q2 in this case would be @ 0 if im not mistaken...
do i use the probabilities table and find the number which represents the propertions for 25% and 75% of the values for q1 and q3 respectively?

quark
10-02-2005, 11:03 PM
Hi dee,

Welcome to TS.

I assume you want to find q1 and q3 for the standard normal distribution. For q3, since 75% of the scores are below this value and since the area under the curve is 1, the area below the score representing the 75th percentile is 0.75. To use the table, you would look for a score corresponding to 0.75. Looking at the body of the table, find the value closest to 0.75, and you can get the z-score from the column and row that corresponds to 0.75.

To find q1, keep in mind the normal curve is symmetric about zero...

Let me know if you need more help. :)

dee
10-03-2005, 10:19 AM
Considering all normal distrubtions are the same....
q1=-.68 and q3=0.68........approx.....correct?

quark
10-03-2005, 10:27 AM
Yes, those are for standard normal (mean=0, sd=1) distributions. If it is a normal distributions with different mean or sd, the quartiles will be different.

james__L
05-30-2007, 08:27 AM
i have found that

if fi (x) = 0.25 (LQ)
and fi (x) = 0.75 (UQ)

assuming X ~ N (0,1), x will equal -0.67448 (LQ) and +0.67448 (UQ) respectively

the difference between these two points (UQ and LQ) is the interquartile range and this is 0.67448-(-0.67448) = 1.34896

and therefore it is possible to deduce that:

interquartile range = 1.34896 x standard deviation

hope this is of help