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Quick Normal Distribution Question [+rep] Watch

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    Hi,

    I've got normal distribution, mean=1, variance=16 and trying to find P(X>-1.04). I've looked up what it should be by using online normal distribution calculators but keep getting the wrong answers. Can anyone help? +rep for help

    Thank you!
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    Have you double checked you're using the right table?
    Sounds daft, but a lot of the time when I got it wrong it's because the table was for values above instead of below etc.
    Also - are you just using the variance value? Standard deviation is the square root of variance.
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    (Original post by sweetfloss)
    Have you double checked you're using the right table?
    Sounds daft, but a lot of the time when I got it wrong it's because the table was for values above instead of below etc.
    Also - are you just using the variance value? Standard deviation is the square root of variance.
    Thanks, I've been manipulating it to suit the particular table I'm using, and I am using the sd rather than the variance. Since you're 4th year maths, do I need to standardise before applying rules of symmetry or can I manipulate first and then standardise? My impression is it doesn't make a difference but I was only taught it 24 hours ago..
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    (Original post by GingerApple)
    Thanks, I've been manipulating it to suit the particular table I'm using, and I am using the sd rather than the variance. Since you're 4th year maths, do I need to standardise before applying rules of symmetry or can I manipulate first and then standardise? My impression is it doesn't make a difference but I was only taught it 24 hours ago..
    Shouldn't matter.
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    (Original post by sweetfloss)
    Shouldn't matter.
    If you say P(x>-1.04)=1-P(x<-1.04)=P(x<1.04)=P(z<0.01) or P(x>-1.04)=P(z>-0.51)=1-P(z<-0.51)=P(z<0.51) it does make a difference, doesn't it?
 
 
 
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