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anonymous
 one year ago
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. since the height requirements have changed to a percentage of 0.986% What would then be the new height requirements?
anonymous
 one year ago
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. since the height requirements have changed to a percentage of 0.986% What would then be the new height requirements?

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amistre64
 one year ago
Best ResponseYou've already chosen the best response.1it 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%"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh well maybe I needed to make that decimal into a percentage...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1but what does the information mean? i cant make sense of what it is asking of us

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1does it mean that we want to find an 98.6% interval?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A 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?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1that one reads better

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1we are given percentages (areas), and need to find the z score related to them what function takes an area, and gives us a zscore?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.12nd vars is a button combination, which takes us to a distribution menu.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1press it and see what we get for function options ...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0id say the 3rd for invnorm

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1invnorm takes an area, and gives a zvalue. good what is the zvalue for 1%? invnorm(.01) what is the zvalue for 2%? invnorm(.02)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So Id then put in my new percentage and input (0.1) and (0.2)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1work it 2 times ... use .01 and .02

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1.10 is 10%; .01 is 1%

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So invnorm(98.6, .01,.02)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1no, its exactly like a wrote above invnorm(.01)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1and invnorm(.02) is?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Would I then take out the negative? Also, would I make them a percentage or leave as a decimal?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok, leave as decimal. I just read that

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1now 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)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So the top should be rounded to 2.33?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1no, i mentioned nothing about rounding.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh ok so leave as is

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1what is: 63.5 2.326(2.5) and what is: 63.5 +2.053(2.5)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1those are z values, we use them to define the interval with ..

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1of 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)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0.9999 for the first

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1when you get around to answering my question, let me know ...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and 1.4321 for the second

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1no, we are not using any functions now, we are doing elementary school math ... addition and multiplication.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I know. I was using the normalcdf. I thought that is what I needed to use..

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1no, we simply need to use this 63.5 2.326(2.5)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.068.6325 for the second

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry about the confusion

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1i 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you again for your help. I really do appreciate it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Math is just my absolute worst subject! Any form of medical term I can understand but this stuff is horrible!
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