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it sounds like its asking for an interval but this part is hard to understand: "since the height requirements have changed to a percentage of 0.986%"
Oh well maybe I needed to make that decimal into a percentage...
but what does the information mean? i cant make sense of what it is asking of us
does it mean that we want to find an 98.6% interval?
A survey found that women’s heights are normally distributed with mean 63.5 in and standard deviation 2.5 in. A branch of the military requires women’s heights to be between 58in and 80 in. b. If this branch of the military changes the height requirements so that all women are eligible except the shortest 1% and the tallest 2%, what are the new height requirements?
that one reads better
we are given percentages (areas), and need to find the z score related to them what function takes an area, and gives us a zscore?
2nd vars is a button combination, which takes us to a distribution menu.
press it and see what we get for function options ...
id say the 3rd for invnorm
invnorm takes an area, and gives a zvalue. good what is the zvalue for 1%? invnorm(.01) what is the zvalue for 2%? invnorm(.02)
So Id then put in my new percentage and input (0.1) and (0.2)
work it 2 times ... use .01 and .02
.10 is 10%; .01 is 1%
So invnorm(98.6, .01,.02)
no, its exactly like a wrote above invnorm(.01)
and invnorm(.02) is?
Would I then take out the negative? Also, would I make them a percentage or leave as a decimal?
Ok, leave as decimal. I just read that
now comes some mental gymnastics ... this gave us the bottom 1% and bottom 2% bottom 1% is fine, its what we want TOP 2% is what we desire, the good news is, the top is just the opposite of the bottom TOP 2% = 2.053 -------------------- our interval is therefore the calculations of: 63.5 -2.326(2.5) 63.5 +2.053(2.5)
So the top should be rounded to 2.33?
for the 1%
no, i mentioned nothing about rounding.
Oh ok so leave as is
what is: 63.5 -2.326(2.5) and what is: 63.5 +2.053(2.5)
2.326 and 2.053
those are z values, we use them to define the interval with ..
of course not ... you have to work out the actual calculations now, and im only going to post this this last time ... what is: 63.5 -2.326(2.5) and what is: 63.5 +2.053(2.5)
-.9999 for the first
when you get around to answering my question, let me know ...
and -1.4321 for the second
no, we are not using any functions now, we are doing elementary school math ... addition and multiplication.
I know. I was using the normalcdf. I thought that is what I needed to use..
no, we simply need to use this 63.5 -2.326(2.5)
for the first
68.6325 for the second
Sorry about the confusion
i never made any mention of, nor gave reference to, any other function other than the invnorm ... and that was to determine some z values which we can then use to define our interval with 57.6850 is good for the low end 68.6325 is good for the high end
Thank you again for your help. I really do appreciate it.
Math is just my absolute worst subject! Any form of medical term I can understand but this stuff is horrible!