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    Question:

    The random variable X ~ N(20,12)

    Find the value of 'a' such that: P(X < a) = 0.40


    The problem I have is this:

    the probability 0.40 is not even listed on the normal distribution function table at the back of my textbook. The table lists probabilities from 0.5 to 1, so anything below 0.5 is not there. This table is supposed to be the one we get in the exam, right?

    and I can't use the other table (percentage points of the normal distribution) because that requires P(X > a), whereas I have P(X < a)


    Help!
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    Well that doesn't matter too much, since \mathbb{P}(X \ge a) = 1 - \mathbb{P}(X&lt;a).
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    (Original post by nuodai)
    Well that doesn't matter too much, since \mathbb{P}(X \ge a) = 1 - \mathbb{P}(X&lt;a).
    I've tried that, but I get this:

    P(X > a) = 1 - 0.4

    P(X > a) = 0.6

    I can't use the 'percentage points of the normal distribution' table because the probability 0.6 is simply not listed there (it doesn't list any higher than 0.5000). I can't use the 'normal distribution function' table because I have P(X > a) rather than P(X < a)

    The solution page has this:

    p(X<a) = 0.40 Use P=0.4000

    which suggest that I shouldn't need to do any re-arranging and that I could just find 'a' with P = 0.4 in the table
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    Consider the symmetry of the Normal distribution with parameters N(0, 1)...
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    I think it's -0.2533
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    (Original post by perrytheplatypus)
    I think it's -0.2533
    yeah thats it, but how did you work it out?
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    (Original post by Rational Paradox)
    Consider the symmetry of the Normal distribution with parameters N(0, 1)...
    Can you elaborate on that please?
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      (Original post by perrytheplatypus)
      I think it's -0.2533
      :yep: That's what I got too.

      OP: it's a negative value as 0.4 < 0.5.

      EDIT:
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      (Original post by W.H.T)
      yeah thats it, but how did you work it out?
      Try drawing a sketch of it. It's symmetrical, so the distance from the middle is the same for more than 0.4 and more than 0.6

      which means the numbers will be the same, but one negative and one positive.
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      (Original post by im so academic)
      :yep: That's what I got too.

      OP: it's a negative value as 0.4 < 0.5.

      EDIT:
      You have the same book as me

      anyway, I understand that it must be a negative value. But how can you use that table (percentage points of the normal distribution) ?

      I thought you can only use it if you have P( Z > z), whereas I have P(Z < z)
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      (Original post by perrytheplatypus)
      Try drawing a sketch of it. It's symmetrical, so the distance from the middle is the same for more than 0.4 and more than 0.6

      which means the numbers will be the same, but one negative and one positive.
      sorry still not sure about this..


      if I were to draw this, what values would be on the horizontal axis?

      and what do you mean by 'more than 0.4 and more than 0.6'? (this? Z > 0.4?)
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        (Original post by W.H.T)
        You have the same book as me

        anyway, I understand that it must be a negative value. But how can you use that table (percentage points of the normal distribution) ?

        I thought you can only use it if you have P( Z &gt; z), whereas I have P(Z &lt; z)
        I see.

        Yeah, when it's P(Z < z), it's typically a negative value. But really it's just easier to work with using the graph method:


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        (Original post by W.H.T)
        sorry still not sure about this..


        if I were to draw this, what values would be on the horizontal axis?

        and what do you mean by 'more than 0.4 and more than 0.6'? (this? Z > 0.4?)
        no, those are the probabilities. The z from Z > z is what's on the axis.
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        I get it now!

        Thanks to everyone who helped: perrytheplatypus, im so academic, rational paradox and nuodai

        really appreciated it
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        My name's first

        ahem...I'll be leaving now...>_>
       
       
       
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