OpenStudy (anonymous):

.

3 years ago
OpenStudy (amistre64):

oy vey ... it may be best to redirect this to the answer given yesterday ... let me find the link in my profile

3 years ago
OpenStudy (anonymous):

Ty.:-)

3 years ago
OpenStudy (amistre64):

good luck with it, if you have any questions about what was gone thru, feel free to ask ...

3 years ago
OpenStudy (anonymous):

what is a ti83? Is it that using the z-formula and curve chart can find the sd and mean?

3 years ago
OpenStudy (amistre64):

a ti83 is a calculator that makes looking at tables obsolete

3 years ago
OpenStudy (amistre64):

if given the mean and (sd), those are simply not sufficient to determine a zscore of a given percentage

3 years ago
OpenStudy (amistre64):

only by knowing the zscore, the mean, and the (sd) can we accurately calculate the point that is the 5% cutoff

3 years ago
OpenStudy (amistre64):

.... you dont have to do this with integration do you?

3 years ago
OpenStudy (anonymous):

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?

3 years ago
OpenStudy (amistre64):

if memory serves we could integrate it by something that looks like this \[\frac{1}{\sqrt{2\pi}}\int_{x_0}^{x_1}\exp(-\left(\frac{x-\mu}{2\sigma}\right)^2)~dx\]

3 years ago
OpenStudy (amistre64):

a zscore is just a change of variable: such that mu = 0 and sigma=1

3 years ago
OpenStudy (anonymous):

Do i have to use that for the 5%? I don't see how I can find mean and sd with just that number. ?

3 years ago
OpenStudy (amistre64):

so when... \[z=\frac{1}{\sqrt{2\pi}}\int^{b}_{-\infty}\exp(-\left(\frac{x}{2}\right)^2)~dx=.9500\] or \[z=\frac{1}{\sqrt{2\pi}}\int_{a}^{\infty}\exp(-\left(\frac{x}{2}\right)^2)~dx=.0500\]

3 years ago
OpenStudy (amistre64):

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

3 years ago
OpenStudy (amistre64):

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

3 years ago
OpenStudy (amistre64):

\[e^u\approx\sum_{n=0}^{k}\frac{u^n}{n!}\] \[e^{(-(x/2)^2)}\approx\sum_{n=0}^{k}\frac{(-x^2/4)^n}{n!}\] \[e^{(-(x/2)^2)}\approx\sum_{n=0}^{k}\frac{(-1)^n}{4^nn!}x^{2n}\] 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

3 years ago
OpenStudy (amistre64):

im sure none of this is in your material tho ... which is why the question appears decevingly simple at first

3 years ago
OpenStudy (anonymous):

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.

3 years ago
OpenStudy (amistre64):

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.

3 years ago
OpenStudy (amistre64):

well, b=z, if we want to see this in the right light :)

3 years ago
OpenStudy (amistre64):

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

3 years ago
OpenStudy (anonymous):

lol.:) o.k. .4500 x 6.28 and this is like the z-score?

3 years ago
OpenStudy (amistre64):

no, thats would be the value to assess b at: like saying x^2 + 6x = -9, what is x?

3 years ago
OpenStudy (amistre64):

x does not equal -9 in that instance

3 years ago
OpenStudy (amistre64):

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. \[\large mean+z_{.95}(sd)=x\]

3 years ago
OpenStudy (amistre64):

if they want you to calculate the zscore, theyll have to give you more information than whats provided.

3 years ago
OpenStudy (anonymous):

Oh, so this is basically what I would use and I can't actually determine the mean and sd?

3 years ago
OpenStudy (anonymous):

I need to solve for x?

3 years ago
OpenStudy (amistre64):

correct

3 years ago
OpenStudy (amistre64):

\[mean+1.645(sd)=x\] amounts to a top 95/5 split of the data at x

3 years ago
OpenStudy (anonymous):

cool! and I need to use the .95 into the formula?

3 years ago
OpenStudy (amistre64):

your missing what a zscore tells us about .95

3 years ago
OpenStudy (anonymous):

That is in the empirical rule, no?

3 years ago
OpenStudy (amistre64):

|dw:1400350260799:dw|

3 years ago
Similar Questions: