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Normal distrubution

  1. Nov 22, 2011 #1
    1. The problem statement, all variables and given/known data
    The door frames used by a builder are of one standard height, 1.83m. The heights of men are normally distributed with mean 1.73m and standard deviation 0.064m.
    The door frames are to be used in a department store. It is known that women outnumber men in the ratio 19:1 in the store and the proportion of women taller than the door frame is 0.00069.
    Find the proportion of people for whom a frame height of 1.83m would be too low.


    2. Relevant equations



    3. The attempt at a solution
    I thought it should be 19/20*(0.99931)+1/20*(0.9409)
    But the answer is 19/20(0.00069)+1/20(0.0591)
    But 0.00069 and 0.0591 are the probabilities for men and women taller than the door frames.
    So i don't understand.
     
  2. jcsd
  3. Nov 22, 2011 #2

    Matterwave

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    Gold Member

    The question is to find the proportion of people for whom the frame is "too low". This means that the door is too short for these people.
     
  4. Nov 22, 2011 #3
    oh.... thanks
     
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