OpenStudy (anonymous):

The results of a race are normally distributed with a mean of 28 minutes and a standard deviation of 4 minutes. The fastest 20% go to districts. Using standard normal probability table, find the 2 probability values which bound the exact probability of finishing in the top 20%

OpenStudy (amistre64):

so, i assume you are trying to find the exact minutes?

OpenStudy (amistre64):

you are going to have to compare a zscore from a field table ... do you have a table for it?

OpenStudy (anonymous):

Yes

OpenStudy (anonymous):

Sorry was having trouble responding

OpenStudy (amistre64):

do you know how to read the table?

OpenStudy (anonymous):

Yes but what values do I look for and if there is an equation how do I set it up?

OpenStudy (amistre64):

the table is not going to be exact, but it will give us a field value lower and higher than the exact value ... these define the 'bounds' that they are looking for

OpenStudy (amistre64):

there are different tables out there so id have to know what format your table takes. is it a left tail, a right tail, or a shade from the mean?

OpenStudy (anonymous):

.0793 = x - 28/4 ?

OpenStudy (anonymous):

It's set up like a multiplication table

OpenStudy (amistre64):

IF we knew the exact value, \(\chi\) to begin with, we would use the formula:\[z=\frac{\chi-\mu}{\sigma}\] this would most likely produce a z score that has more than 2 decimal places to play with; the table most likely only gives us a error rate of 2 decimal places

OpenStudy (amistre64):

\[\chi=z(\sigma)+\mu\] but we cant find an exact z; but can find something a little lower, zo, and a little higher, z1, such that:\[[z_0(\sigma)+\mu]\le \chi\le [z_1(\sigma)+\mu]\]

OpenStudy (amistre64):

they are all setup like a multiplication table .... they just have different values depending on how they actually measure the area under the normal distribution curve.

OpenStudy (amistre64):

can you show us a picture of your table?

OpenStudy (anonymous):

It goes up two decimals on the top and the values have 4 decimal places

OpenStudy (amistre64):

sigh ... they all do that. your simply not understanding what im saying about the format of the tables. most likely because youve only seen one format

OpenStudy (amistre64):

left tailed format http://4.bp.blogspot.com/_5u1UHojRiJk/TEh9BHxxPUI/AAAAAAAAAIQ/DafeQNMYFoE/s1600/ztable.gif right tailed format http://www.rochester.edu/college/psc/clarke/201/ztable2.jpg from the mean fromat http://1.bp.blogspot.com/-X8R15VpvgBI/TuWryLCWR_I/AAAAAAAABZs/1i5yuVpW5oQ/s1600/z-table.jpeg

OpenStudy (anonymous):

OpenStudy (amistre64):

but regardless of the format, we can know the zscore for 20% left tail of a normal distribution http://www.wolframalpha.com/input/?i=invnormal%28.2000%29 z = -0.842 ; so your table will most likely have boxed between -0.84 and -0.85

OpenStudy (anonymous):

I also need to find the z score in which the most runners will be qualified

OpenStudy (amistre64):

nice pic :) yeah, IF we only use that part, then we want to find 1-.2000 = .8000 in the field ... then negate the results since we want the other side of the mean

OpenStudy (amistre64):

notice that we can find: .7995 and .8023 in the field; .8000 fits right in between them

OpenStudy (amistre64):

so, using the formulas to entrap \(\chi\) we have: xo = -.84(sd) + mean x1 = -.85(sd) + mean

OpenStudy (amistre64):

as for the second part, im not real sure what "find the z score in which the most runners will be qualified" means for us

OpenStudy (amistre64):

i see one typo so far ... since -85 is smaller than -84, i got my xo and x1 mixed up

OpenStudy (anonymous):

Thanks is there a special notation needed to write it?