anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
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%"
anonymous
  • anonymous
Oh well maybe I needed to make that decimal into a percentage...
anonymous
  • anonymous
98.6% oops

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amistre64
  • amistre64
but what does the information mean? i cant make sense of what it is asking of us
amistre64
  • amistre64
does it mean that we want to find an 98.6% interval?
anonymous
  • anonymous
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?
amistre64
  • amistre64
that one reads better
anonymous
  • anonymous
Sorry :/
amistre64
  • amistre64
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?
anonymous
  • anonymous
2nd vars
anonymous
  • anonymous
or stats?
amistre64
  • amistre64
2nd vars is a button combination, which takes us to a distribution menu.
amistre64
  • amistre64
press it and see what we get for function options ...
anonymous
  • anonymous
id say the 3rd for invnorm
amistre64
  • amistre64
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)
anonymous
  • anonymous
So Id then put in my new percentage and input (0.1) and (0.2)
amistre64
  • amistre64
work it 2 times ... use .01 and .02
amistre64
  • amistre64
.10 is 10%; .01 is 1%
anonymous
  • anonymous
So invnorm(98.6, .01,.02)
anonymous
  • anonymous
or (98.6,1,2)
amistre64
  • amistre64
no, its exactly like a wrote above invnorm(.01)
anonymous
  • anonymous
-2.326
amistre64
  • amistre64
and invnorm(.02) is?
anonymous
  • anonymous
-2.053
anonymous
  • anonymous
Would I then take out the negative? Also, would I make them a percentage or leave as a decimal?
anonymous
  • anonymous
Ok, leave as decimal. I just read that
amistre64
  • amistre64
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)
anonymous
  • anonymous
So the top should be rounded to 2.33?
anonymous
  • anonymous
for the 1%
amistre64
  • amistre64
no, i mentioned nothing about rounding.
anonymous
  • anonymous
Oh ok so leave as is
amistre64
  • amistre64
what is: 63.5 -2.326(2.5) and what is: 63.5 +2.053(2.5)
anonymous
  • anonymous
2.326 and 2.053
amistre64
  • amistre64
those are z values, we use them to define the interval with ..
anonymous
  • anonymous
invnormcdf?
anonymous
  • anonymous
normalcdf***
amistre64
  • amistre64
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)
anonymous
  • anonymous
-.9999 for the first
amistre64
  • amistre64
when you get around to answering my question, let me know ...
anonymous
  • anonymous
and -1.4321 for the second
amistre64
  • amistre64
no, we are not using any functions now, we are doing elementary school math ... addition and multiplication.
anonymous
  • anonymous
I know. I was using the normalcdf. I thought that is what I needed to use..
amistre64
  • amistre64
no, we simply need to use this 63.5 -2.326(2.5)
anonymous
  • anonymous
57.685
anonymous
  • anonymous
for the first
anonymous
  • anonymous
68.6325 for the second
anonymous
  • anonymous
Sorry about the confusion
amistre64
  • amistre64
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
anonymous
  • anonymous
Thank you again for your help. I really do appreciate it.
amistre64
  • amistre64
your welcome
anonymous
  • anonymous
Math is just my absolute worst subject! Any form of medical term I can understand but this stuff is horrible!

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