If $Z$ is a standard normal variate, find the value of $c$ for the following:
- $P \left(- c < Z < c\right) = 0.40$
- $P \left(Z > c\right) = 0.85$
- $\because \;$ $Z = - c$ and $Z = + c$ lie at equal distance from $Z = 0$,
we have $P \left(0 < Z < c\right) = \dfrac{0.40}{2} = 0.20$
$Z$ value for the area $0.20$ from the Normal Distribution table is $= 0.52$
$\therefore \;$ $- c = - 0.52$ $\;$ and $\;$ $+c = + 0.52$
- $P \left(Z > c\right) = 0.85$ $\implies$ $P \left(c < Z < \infty\right) = 0.85$
i.e. $\;$ $c$ lies to the right of $Z = 0$
Area to the right of $Z = 0$ is $= 0.5$
$\therefore \;$ We have, $P \left(0 < Z < \infty\right) - P \left(0 < Z < c\right) = 0.85$
i.e. $\;$ $0.5 - P \left(0 < Z < c\right) = 0.85$
i.e. $P \left(0 < Z < c\right) = 0.5 - 0.85 = - 0.35$
From the Normal Distribution table, $Z$ value for $- 0.35$ is $= - 1.04$
$\therefore \;$ $c = -1.04$