Thread: find area under non-standard normal curve (i.e. units not percentages)

Hello,

I would like some advice on how to calculate the area under a non-standard normal curve, in its own units, not relative to a standard-normal curve. Example distribution characteristics:

Mean ：500
Standard Deviation: 100
Point of interest: 550

By finding the relevant Z-score ((550-500)/100), I know that 31% of the area under the curve lies above 550. How do I convert that 31% back into units/terms of the non-standard normal curve?

Through simulation, I believe the answer is roughly 570, but I don’t know how to truly calculate it.

Thank you.

It would just depend on the size of the population. How did you arrive at 570, if not by finding 31% of roughly 1775, the latter being the population?

Also, what do you mean by a non-standard normal curve: the probability density function of the normal distribution whose mean and variance are different from 0 and 1? But the area under every probability density function is 1...

I am confused by your question. You say "point of interest 550" so I would think you were looking for the probability "y< 550" or "500< y< 550" which is what you seem to be doing when you calculate the "standard z"- but then you turn around and say "I think it is roughly 550". Given what you said, because "550" gave you the 0.31, 0.31 must correspond to 550.
I am assuming that by "non-standard normal distribution" you simply mean that the mean is not 0 and the standard deviation is not 1.

Hmmm- that can't be what you mean- z= (550- 500)/100= .5 and, according to my table of the standard normal distribution, the probability that z is less 0 and 0.5 is 0.191, not 0.31.
0.31 corresponds to z= 0.88. That would be equivalent to the value x= 544, not 550.

Originally Posted by HallsofIvy

Hmmm- that can't be what you mean- z= (550- 500)/100= .5 and, according to my table of the standard normal distribution, the probability that z is less 0 and 0.5 is 0.191, not 0.31.
0.31 corresponds to z= 0.88. That would be equivalent to the value x= 544, not 550.

From this table, P(z < 0.5) is 0.6915, so P(z > 0.5) is indeed about 31%. OP writes, "31% of the area under the curve lies above 550."