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Mathematics 17 Online
OpenStudy (anonymous):

the quarterly returns for a group of 79 mutual funds with a mean of 1.1% and a standard deviation of 4.5% can be modeled by a normal model. based on the model N(0.011,0.045), what are the cutoff values for the a) highest 40% of these funds? b) lowest 20%? c) middle 80%? d) highest 80%?

OpenStudy (anonymous):

Do you have inverse z-tables readily available? If so, find the z-numbers from those probabilities, then set \[\mu = 1.1, \ \sigma = 4.5\] and determine the x-values at which those z-numbers are satisfied. If not, you'll need to use the original integral formulation of the normal pdf, which is...annoying..

OpenStudy (anonymous):

it keeps telling me to use technology to determine the cutoff z-value(s) for the given portion of the standard normal distribution.

OpenStudy (anonymous):

i have a z table. will that work?

OpenStudy (anonymous):

You don't have an inverse z-table? That's the one you need to get from probability to z-number, a z-table gives you the opposite. Try this link, and tell me if it gets you anywhere. It is essentially an instruction manual to using your calculator to do these types of calculations. http://www.tc3.edu/instruct/sbrown/ti83/normcalc.htm

OpenStudy (anonymous):

no i don't have the inverse z table, but i have a midterm coming up on monday, and i'm not allowed to use a graphing calculator for it, i don't know why the hw is telling me to use it but if you can, do you think you could tell me how to solve everything using a z table?

OpenStudy (anonymous):

the inverse of the z table isn't called the t table is it? cause i have that table as well

OpenStudy (kropot72):

I used the standard normal distributopn table to find the value for z to solve a). Simply find the value of probability .6000 and read off the value for z = 0.253. Then substitute the values for z, mu and sigma to find the answer.

OpenStudy (kropot72):

\[z=0.253=\frac{X-0.011}{0.045}\] X = 0.011 + 0.011385 = 0.22385 = 2.2385%

OpenStudy (anonymous):

thank you so much! i was just having trouble figuring out how to find the z score since i didn't want to use the graphing calculator. so whenever i'm given a percentage, i subtract that from 100% and then i find that difference in the z table, to find the value of z?

OpenStudy (kropot72):

To find the z value for b) look up the z value for probability .8000 and assign the z value a negative sign. Can you do this an post back your result?

OpenStudy (anonymous):

yes of course! thank you :)

OpenStudy (anonymous):

i got -.84 for the z value

OpenStudy (anonymous):

so would it be -.84 = (x-.011)/(.045)

OpenStudy (kropot72):

Yes. That's what I get. But I am now confused about the question. The standard deviation appears to be about 4 times the mean. If so my calculations are not valid, the reason being that the question now does not make sense :(

OpenStudy (anonymous):

hmm, may i attach a file, i took a print screen shot of a similar question, just different #'s, i tried following it the way they did it, but the answer was still incorrect

OpenStudy (kropot72):

Please attach the file.

OpenStudy (anonymous):

i attached the file, twice, i don't know if it's uploaded onto here yet

OpenStudy (anonymous):

oh! i think i got it

OpenStudy (anonymous):

i think for this particular question, we find the z value for 20% or .2, which would be -.84

OpenStudy (kropot72):

No sign of the file. Is it possible that the standard deviation of 4.5% means 4.5% of the mean? 1.1 * 0.045 = 0.0495% If this were true, the total range of interest rates would be from about 0.95% to 1.25% , these values being minus 3 standard deviations from the mean to plus 3 standard deviations from the mean.

OpenStudy (anonymous):

yup i actually just solved it now looking over carefully the steps from the similar problem. sorry for all the trouble! and thank you once again :)

OpenStudy (kropot72):

You're welcome :)

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