You may need to use the appropriate appendix table or technology to answer this question.

For the standard normal random variable $z,$ compute the following probabilities. $($Round your answers to four decimal places.$)$

$($a$)P(0z0\mathrm{.}81)$

$($b$)P(-1\mathrm{.}57z0)$

$($c$)P(z>0\mathrm{.}46)$

$($d$)P(z-0\mathrm{.}24)$

$($e$)P(z<mn\; id="EditorElement\_385442">1</mn><mn\; id="EditorElement\_385443">.</mn><mn\; id="EditorElement\_385444">6</mn><mn\; id="EditorElement\_385445">0</mn><mi\; id="EditorElement\_385447">)</mi>$

$($f$)P(z-0\mathrm{.}76)$