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I have a pretty solid understanding of what a normal variable is but I am having difficulty understanding the graph. For example, what will the normal distribution probability be for these:

z = P(1<X<2)
x = P(-3<X<4)
y = P(X<=-2)
w = P(X>=3) 

My attempt:

I got z = .136 , x = .496 , y = 0.023 , and w = .001. I know I am wrong, can you clarify what I did wrong?

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is the mean 0 and the variance is 1? – binn Sep 26 '12 at 2:07
@binn Yes sir it is .. – Q.matin Sep 26 '12 at 2:11
See also, this answer to a related question. – Dilip Sarwate Sep 26 '12 at 4:20
up vote 1 down vote accepted

Judging by your answers, you’re using a table that shows the area to the left of a given value of $z$. Your first answer is right: using a $4$-place table, I make it


You seem to have misplaced the decimal point for $y$: I get $0.023$ to three decimal places, or $0.02275$ to four significant figures.

For $w$ you seem to have taken the area to the left of the cutoff 3 instead of to the right: you want $w=1-0.999=0.001$ (actually about $0.001350$). Remember, the table is giving you everything to the left of a given cutoff; if you’re interested in $X\ge \text{ some cutoff}$, you need to subtract the table value from $1$ to get the area to the right of the cutoff.

For $x$ you want the table value at $4$ minus the table value at $-3$; I don’t know how many places your table gives, but that’s about


it appears that you have about half the right value, so you may be using a one-sided table and not compensating properly.

Most people find it helpful to make a rough sketch of the normal curve and mark the cutoff(s) of interest. For $x$, for instance, they would be $-3$ and $4$. Then figure out how to get the area to the left of $4$ and the area to the left of $-3$, and then subtract the latter from the former.

The applet here may be helpful: you can set it to show areas to the left of a cutoff (Up to Z), to the right of a cutoff (Z onwards), or from the centre at $0$ up to some positive cutoff (0 to Z).

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Thank you a lot !!! This is very helpful ! – Q.matin Sep 26 '12 at 2:29
@Q.matin: I’m glad to hear it; you’re welcome. – Brian M. Scott Sep 26 '12 at 2:31

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