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If Z is a random variable with a standard normal distribution, what is $P(Z^2 \lt 3.841)$?

Can I just square root $3.841$ so that it becomes $P(Z \lt \sqrt{3.841})$ and use the normal distribution table to obtain the probability?

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Hint: The square root of $3.841$ can be positive or negative, so don't forget to take this into account. Also, if this is a homework (or self-study) problem, please add the homework or self-study tag. –  Dilip Sarwate Jan 19 '14 at 5:25
What is the distribution of the square of a standard normal random variable? –  JohnK Jan 19 '14 at 12:44

2 Answers 2

Your method would obviously not work if the bound was very small: for example $\Pr(Z^2 \lt 0.0001)$ is small but your suggested method would give an answer over $0.5$: more precisely the true probability would be about $0.00008$ while your method would give about $0.50004$.

If $\Phi(x)$ is the cumulative distribution function for a standard normal random variable $X$, i.e. $\Pr(X \le x) = \Phi(x)$ then for positive $z$,

$$\Pr(Z^2 \lt z) = \Pr(-\sqrt{z} \lt Z \lt \sqrt{z}) = \Phi(\sqrt{z} ) - \Phi(-\sqrt{z} ) = 2 \Phi(\sqrt{z} ) - 1.$$

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Note that : If $\displaystyle Z\sim N(0,1)$ then $\displaystyle Z^2\sim \chi^2_1$. It may help you.

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Only if you have a table for $\chi^2_1$ –  Henry Jan 19 '14 at 12:48
@Henry :Yes of course. –  Argha Jan 19 '14 at 12:50

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