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    Can anyone help me out with this? I've looked it up on Solutionbank but I don't quite get how they make it to the answer. Thanks.

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    (Original post by ChrisWolff)
    Can anyone help me out with this? I've looked it up on Solutionbank but I don't quite get how they make it to the answer. Thanks.

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    What have you done so far?

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    (Original post by ChrisWolff)
    Can anyone help me out with this? I've looked it up on Solutionbank but I don't quite get how they make it to the answer. Thanks.

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    Knowing that the normal distribution is symmetrical about the mean, if p(-4 < x < 4) = 0.6, what can you say about p(0 geq x < 4) and why does this help?
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    (Original post by Matrix123)
    What have you done so far?

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    This is all I have so far.
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    (Original post by SeanFM)
    Knowing that the normal distribution is symmetrical about the mean, if p(-4 < x < 4) = 0.6, what can you say about p(0 geq x < 4) and why does this help?
    So I was assuming that P(X<4) - P(X<-4) = 0.6, so it's 0.3 on each side that's shaded in, and 0.2 on each side that isn't shaded in. Not quite sure how to move on from there.
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    (Original post by ChrisWolff)
    So I was assuming that P(X<4) - P(X<-4) = 0.6, so it's 0.3 on each side that's shaded in, and 0.2 on each side that isn't shaded in. Not quite sure how to move on from there.
    You therefore know P(X<4) don't you? You can then calculate this using the normal distribution table.

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    (Original post by ChrisWolff)
    So I was assuming that P(X<4) - P(X<-4) = 0.6, so it's 0.3 on each side that's shaded in, and 0.2 on each side that isn't shaded in. Not quite sure how to move on from there.
    Remember that the x values can be converted to z values, but at the moment you can't do that yet as you don't have anything of the form P(X<x) on its own (you have two in the equation in your post, which cant be used now but it is going in the right direction)

    You are 100% correct in that it is 0.3 on each side, so what is P(X<4) using that fact and one fact that you know about normal distribution?

    And why is knowing what P(X<4) is useful? I have dropped a hint as to why it is already.
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    (Original post by Matrix123)
    You therefore know P(X<4) don't you? You can then calculate this using the normal distribution table.

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    I might have sorted it - I'm not sure whether I've fluffed my way to an answer or if this genuinely works.
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    (Original post by SeanFM)
    Remember that the x values can be converted to z values, but at the moment you can't do that yet as you don't have anything of the form P(X<x) on its own (you have two in the equation in your post, which cant be used now but it is going in the right direction)

    You are 100% correct in that it is 0.3 on each side, so what is P(X<4) using that fact and one fact that you know about normal distribution?

    And why is knowing what P(X<4) is useful? I have dropped a hint as to why it is already.
    So, P(X<4) is 0.2, and therefore P(X<-4) is 0.2 due to symmetry. I feel really silly but I'm not connecting the dots here. I've posted another attempt at a solution above but I'm not sure whether I've just made it by luck or not.
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    (Original post by ChrisWolff)
    So, P(X<4) is 0.2, and therefore P(X<-4) is 0.2 due to symmetry. I feel really silly but I'm not connecting the dots here. I've posted another attempt at a solution above but I'm not sure whether I've just made it by luck or not.
    When picturing symmetry use the bell shaped sketch.

    P(X<4) is actually 0.8 which I think you knew, because there is 0.3 between 4 and 0 and you know that because it's symmetrical, 0.5 is greater than the mean and 0.5 less than the mean, which is this case = 0, and so P(X<4) = P(0<x<4) + P(0 geq 4) (see how this makes sense from the diagram) = 0.5 + 0.3 = 0,8 so you can now find what the standard deviation is.
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    (Original post by Bean sprout)
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    Thanks for the help mate - I'm a little confused about the Φ symbols, what I'm inferring is that's in place of a P(X) bracket? Your solution looks sort of what I've had an attempt at, do you reckon what I've done is okay? Name:  20170115_175551.jpg
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    (Original post by ChrisWolff)
    Thanks for the help mate - I'm a little confused about the Φ symbols, what I'm inferring is that's in place of a P(X) bracket? Your solution looks sort of what I've had an attempt at, do you reckon what I've done is okay? Name:  20170115_175551.jpg
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    Yes, you did it right. We have a slightly diferent answer purely because my data booklet only goes to two decimal places. As for the phi symbol, it represents the cumulative distribution function to the left of that point. e.g. phi(0)=0.5

    What exam board do you use?
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    (Original post by Bean sprout)
    Yes, you did it right. We have a slightly diferent answer purely because my data booklet only goes to two decimal places. As for the phi symbol, it represents the cumulative distribution function to the left of that point. e.g. phi(0)=0.5

    What exam board do you use?
    Edexcel, there might be a bit of a difference between them.
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    (Original post by ChrisWolff)
    Edexcel, there might be a bit of a difference between them.
    I'm on edexcell aswell, in the formula booklet they write it using Phi too.
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    (Original post by Bean sprout)
    I'm on edexcell aswell, in the formula booklet they write it using Phi too.
    Yeah, you're right, weirdly I've never used phi in an answer at all.
 
 
 
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