Question

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Answer 1

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

Answer 2

http://openstudy.com/users/amistre64#/updates/537669ace4b0e27c43145fa4

Answer 3

Ty.:-)

Answer 4

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

Answer 5

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

Answer 6

a ti83 is a calculator that makes looking at tables obsolete

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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?

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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.

Answer 20

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.

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

no, thats would be the value to assess b at: like saying

x^2 + 6x = -9, what is x?

Answer 25

x does not equal -9 in that instance

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

I need to solve for x?

Answer 30

correct

Answer 31

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

Answer 32

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

Answer 33

your missing what a zscore tells us about .95

Answer 34

That is in the empirical rule, no?

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