Nicole missed the lesson on normal distribution and needs to do her homework. Explain to Nicole how to use the mean and standard deviation of a normal distribution to determine the top 5% of the population.
@SarahEZZMcK Do you know anyone who can help with this?
sounds like an emiprical rule approximation
I'm learning about Conditional Probablity
have you covered what a zscore is? (might be called a standard score)
No.
@aegidious @stonewoods @kag
hmm, then just knowing a mean and standard distribution might not be all that useful in determining the top 5% the empirical rule is 64, 95, 99 giving us at best 2.5 % of the top 100 - 95 -------- = 2.5% 2 100-64 ------- = 50-32 = 18% which is a far cry from 5% 2
so the thing is, the mean and sd can be used to determine a zscore, that can then be tabled up to find a probability the table can also be used to find an appropriate zscore to work it backwards to find the value as well
\[z=\frac{x-mean}{sd}\] if the z of .0500 is known than \[mean+z(sd)=x\]defines the cutoff point for the top 5%
ok. So in order to find the answer we have to do everything backwards?
if you have a table, yes. if you have a ti83, no :)
I have a TI-84
Silver Edition. Will that work?
then to find the zscore, we would use the invnorm function under the distribution menu
2nd, VARS, invnorm: enter .95, since its a left tail measure. invnorm(.05) would produce a negative z value, but due to symmetry of the curve, we would just ignore the sign. we would still need to use that z value in the setup: mean + z(sd) to determine the cutoff point tho
Ummmm.... HUH?? Can you say that again. Except this time dumb it down a little.... or a lot. A lot is good too.
lets start with the basics, do you recall what a normal distibution curve looks like? and what the sd (standard deviation) measures on it?
Yes. A normal distribution curve looks a lot like a hill. and the (sd) measures the spread of normal distribution.
The spread of a normal distribution is controlled by the standard deviation, The smaller the standard deviation the more concentrated the data
good, the zscore tells us how many (sd) fit within a specific interval from the mean. let say we have a mean of 'u' and have some value 'x', we want to know the distance that x is from the mean. |dw:1400270051502:dw| would you agree that the distance between them is simply (x - u)?
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