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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?

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  1. amistre64
    • one year ago
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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%"

  2. anonymous
    • one year ago
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    Oh well maybe I needed to make that decimal into a percentage...

  3. anonymous
    • one year ago
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    98.6% oops

  4. amistre64
    • one year ago
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    but what does the information mean? i cant make sense of what it is asking of us

  5. amistre64
    • one year ago
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    does it mean that we want to find an 98.6% interval?

  6. anonymous
    • one year ago
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    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?

  7. amistre64
    • one year ago
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    that one reads better

  8. anonymous
    • one year ago
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    Sorry :/

  9. amistre64
    • one year ago
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    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?

  10. anonymous
    • one year ago
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    2nd vars

  11. anonymous
    • one year ago
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    or stats?

  12. amistre64
    • one year ago
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    2nd vars is a button combination, which takes us to a distribution menu.

  13. amistre64
    • one year ago
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    press it and see what we get for function options ...

  14. anonymous
    • one year ago
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    id say the 3rd for invnorm

  15. amistre64
    • one year ago
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    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)

  16. anonymous
    • one year ago
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    So Id then put in my new percentage and input (0.1) and (0.2)

  17. amistre64
    • one year ago
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    work it 2 times ... use .01 and .02

  18. amistre64
    • one year ago
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    .10 is 10%; .01 is 1%

  19. anonymous
    • one year ago
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    So invnorm(98.6, .01,.02)

  20. anonymous
    • one year ago
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    or (98.6,1,2)

  21. amistre64
    • one year ago
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    no, its exactly like a wrote above invnorm(.01)

  22. anonymous
    • one year ago
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    -2.326

  23. amistre64
    • one year ago
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    and invnorm(.02) is?

  24. anonymous
    • one year ago
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    -2.053

  25. anonymous
    • one year ago
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    Would I then take out the negative? Also, would I make them a percentage or leave as a decimal?

  26. anonymous
    • one year ago
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    Ok, leave as decimal. I just read that

  27. amistre64
    • one year ago
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    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)

  28. anonymous
    • one year ago
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    So the top should be rounded to 2.33?

  29. anonymous
    • one year ago
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    for the 1%

  30. amistre64
    • one year ago
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    no, i mentioned nothing about rounding.

  31. anonymous
    • one year ago
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    Oh ok so leave as is

  32. amistre64
    • one year ago
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    what is: 63.5 -2.326(2.5) and what is: 63.5 +2.053(2.5)

  33. anonymous
    • one year ago
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    2.326 and 2.053

  34. amistre64
    • one year ago
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    those are z values, we use them to define the interval with ..

  35. anonymous
    • one year ago
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    invnormcdf?

  36. anonymous
    • one year ago
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    normalcdf***

  37. amistre64
    • one year ago
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    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)

  38. anonymous
    • one year ago
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    -.9999 for the first

  39. amistre64
    • one year ago
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    when you get around to answering my question, let me know ...

  40. anonymous
    • one year ago
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    and -1.4321 for the second

  41. amistre64
    • one year ago
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    no, we are not using any functions now, we are doing elementary school math ... addition and multiplication.

  42. anonymous
    • one year ago
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    I know. I was using the normalcdf. I thought that is what I needed to use..

  43. amistre64
    • one year ago
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    no, we simply need to use this 63.5 -2.326(2.5)

  44. anonymous
    • one year ago
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    57.685

  45. anonymous
    • one year ago
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    for the first

  46. anonymous
    • one year ago
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    68.6325 for the second

  47. anonymous
    • one year ago
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    Sorry about the confusion

  48. amistre64
    • one year ago
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    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

  49. anonymous
    • one year ago
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    Thank you again for your help. I really do appreciate it.

  50. amistre64
    • one year ago
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    your welcome

  51. anonymous
    • one year ago
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    Math is just my absolute worst subject! Any form of medical term I can understand but this stuff is horrible!

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