How to get probability from a normal distribution?

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To calculate the probability of Z=Y^3 where Y is a standard normal distribution, one can approximate P(Z ≤ 1) by first determining P(Y ≤ 1), which is equivalent to P(X^3 ≤ 1) for X ~ N(0,1). The value for P(Y ≤ 1) is found to be approximately 0.84134 from standard normal distribution tables. For odd powers like Y^3, the appropriate root of the constant can be used for calculations. Additionally, to visualize the probability density function (pdf) for Y^3, one can plot the pdf of the normal distribution and adjust the X coordinates to Y^(1/3) for a linear representation in Y. The discussion emphasizes the relationship between transformations of normal variables and their probabilities.
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If I had Z=Y^3 where Y is a standard normal distribution. How would I approx. calculate the probability of Z<=1 ?, I would understand it if it was Z=Y^2 which is chi-square...
 
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P(Z ≤ 1)=P(Y ≤ 1). In general for odd powers, you just need to take the appropriate root of the constant of interest.
 
well i have to do P(Y<=1) = P(X^3<=1) = P(X<=1), and since X ~N(0,1), so I can just look in the book for the standard normal distribution values
which is just 0.84134, is that right?
By the way out of curiosity how would the pdf graph look for Y^3 ?
 
I assume you meant X^3 = Y. Plot pdf for normal distribution. Change X coordinates to Y^(1/3) and replot to be linear in Y.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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