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    Hey, I have a question. Say if X is distributed normally X~N(6600, 5500) where 6600 is the average time in minutes and 5500 is the variance.
    Then when the question says find P(X>180) where the 180 is in hours. So i decided to divide the mean and variance by 60 to get the time in hours but the answer comes out wrong. Can't I do this? Thanks!
    EDIT: When i convert 180 into minutes, it works just fine but I dont understand why...
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      Convert 180 hours to minutes from the beginning.
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      (Original post by Glutamic Acid)
      Convert 180 hours to minutes from the beginning.
      Why doesn't it work the other way? I just don't understand it.
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        (Original post by JBKProductions)
        Why doesn't it work the other way? I just don't understand it.
        Ah, ok. Let X be the time in minutes, and Z the time in hours.

        So Z = X/60.

        Thus E[Z] = E[X/60] = E[X]/60, which is ok.
        But Var[Z] = Var[X/60] = Var[X]/3600, which is presumably what went wrong.

        {I'm using Var[aX] = a^2Var[X].}
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        The variance is squared; you can't just divide it by 60.
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        Oh I see. Thanks!
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        The distances travelled to work D km, by the employees at a large company are normally distributed with D ˜ N (30, 8 squared)

        (a) Find the upper quartile, Q3, of D.

        (b) Write down the lower quartile, Q1, of D.

        (c) An outlier is defined as any value of D such that D < 8.4 or D > 51.6
        An employee is selected at random.
        Find the probability that the distance travelled to work by this employee is an outlier.
        ________________________________ ________________________________ ______________

        For (a) you do P (D < Q3) and make it equal to something that’s what someone told me! I was wondering why the sign was < and not > to find the P (D < Q3)?

        Can you please show your working. Thank you for your help!
       
       
       
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