.
oy vey ... it may be best to redirect this to the answer given yesterday ... let me find the link in my profile
Ty.:-)
good luck with it, if you have any questions about what was gone thru, feel free to ask ...
what is a ti83? Is it that using the z-formula and curve chart can find the sd and mean?
a ti83 is a calculator that makes looking at tables obsolete
if given the mean and (sd), those are simply not sufficient to determine a zscore of a given percentage
only by knowing the zscore, the mean, and the (sd) can we accurately calculate the point that is the 5% cutoff
.... you dont have to do this with integration do you?
ah, o.k.,,i mean if I use those to find the mean and sd, even though the z-score can't be filled out. What is integration?
if memory serves we could integrate it by something that looks like this \[\frac{1}{\sqrt{2\pi}}\int_{x_0}^{x_1}\exp(-\left(\frac{x-\mu}{2\sigma}\right)^2)~dx\]
a zscore is just a change of variable: such that mu = 0 and sigma=1
Do i have to use that for the 5%? I don't see how I can find mean and sd with just that number. ?
so when... \[z=\frac{1}{\sqrt{2\pi}}\int^{b}_{-\infty}\exp(-\left(\frac{x}{2}\right)^2)~dx=.9500\] or \[z=\frac{1}{\sqrt{2\pi}}\int_{a}^{\infty}\exp(-\left(\frac{x}{2}\right)^2)~dx=.0500\]
this is how the z tables were created, so that you would simply open up a table instead of trying to do the integration every time
and since there is no function that this integrates into, approximation of a taylor series polynomial can be used, or reiman summs of tiny rectangles
\[e^u\approx\sum_{n=0}^{k}\frac{u^n}{n!}\] \[e^{(-(x/2)^2)}\approx\sum_{n=0}^{k}\frac{(-x^2/4)^n}{n!}\] \[e^{(-(x/2)^2)}\approx\sum_{n=0}^{k}\frac{(-1)^n}{4^nn!}x^{2n}\] and integrating the poly .. \[\int e^{(-(x/2)^2)}\approx\sum_{n=0}^{k}\frac{(-1)^n}{4^nn!~(2n+1)}x^{2n+1}\] or some such notion
im sure none of this is in your material tho ... which is why the question appears decevingly simple at first
Yea, this is new to me! I learned about the z-score and curve chart/empirical rule so i thought that's what they wanted from me.
just for giggles ... lets try this: \[\sum_{n=0}^{4}\frac{(-1)^n}{4^nn!}x^{2n}=1-\frac{1}{4^1~1!}x^{2}+\frac{1}{4^22!}x^{4}-\frac{1}{4^33!}x^{6}+\frac{1}{4^44!}x^{8}\] integrate this from 0 to b, to find the area from the mean to be .4500 sqrt(2pi) \[b-\frac{1}{4^1~1!~3}b^{3}+\frac{1}{4^22!~5}b^{5}-\frac{1}{4^33!~7}b^{7}+\frac{1}{4^44!~9}b^{9}\approx.4500\sqrt{2\pi}\] and solve for b.
well, b=z, if we want to see this in the right light :)
the wolf says z=1.645, so lets see how accurate this is http://www.wolframalpha.com/input/?i=b-%5Cfrac%7B1%7D%7B4*3%7Db%5E%7B3%7D%2B%5Cfrac%7B1%7D%7B4%5E2*10%7Db%5E%7B5%7D-%5Cfrac%7B1%7D%7B4%5E3*42%7Db%5E%7B7%7D%2B%5Cfrac%7B1%7D%7B4%5E4*24*9%7Db%5E%7B9%7D-.4500%5Csqrt%7B2%5Cpi%7D%2C+b%3D1.645 the wolf says that we are bout .21 off using that many terms, so more terms would be needed to shrink the error ... but thats the idea if we wanted to be technical about it lol
lol.:) o.k. .4500 x 6.28 and this is like the z-score?
no, thats would be the value to assess b at: like saying x^2 + 6x = -9, what is x?
x does not equal -9 in that instance
keeping it simple: we can use a known mean and sd to calculate the cutoff point for the top 5% by using the z formula:\[z=\frac{x-mean }{sd}\] solving for x we get:\[mean+z(sd)=x\] for some zscore related to 95% of the area in the left tail. \[\large mean+z_{.95}(sd)=x\]
if they want you to calculate the zscore, theyll have to give you more information than whats provided.
Oh, so this is basically what I would use and I can't actually determine the mean and sd?
I need to solve for x?
correct
\[mean+1.645(sd)=x\] amounts to a top 95/5 split of the data at x
cool! and I need to use the .95 into the formula?
your missing what a zscore tells us about .95
That is in the empirical rule, no?
|dw:1400350260799:dw|
Join our real-time social learning platform and learn together with your friends!