Needing help (percentiles): Can anybody solve for the 68th and 32nd percentile?
- Thread starter mspre89
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If the distribution is symmetric, then we have
p(x<32nd percentile) = a
p(x>68th percentile) = a
Notice the change of direction. The first is the left tail and the second is the right tail.
If it is true that that p(x < 32nd percentile) = p(x < 68th percentile), then it must also be the case that p(32nd percentile < x < 68th percentile) = 0. You've a very large hole in the middle of your distribution.
If there were some specifics, we wouldn't have to guess.
p(x<32nd percentile) = a
p(x>68th percentile) = a
Notice the change of direction. The first is the left tail and the second is the right tail.
If it is true that that p(x < 32nd percentile) = p(x < 68th percentile), then it must also be the case that p(32nd percentile < x < 68th percentile) = 0. You've a very large hole in the middle of your distribution.
If there were some specifics, we wouldn't have to guess.
Dr.Peterson
Elite Member
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Please state the entire problem you are solving, so we can see the context of your question, and have some idea what distribution you are talking about, and where you might be confused.
See the Read Before Posting announcement, which tells you this.