Question

Use Cumulative Standardized Normal Probability table to find the value z* for which:

a) P( Z ≤ z*) = 0.95

b) P( Z ≤ z*) = 0.2

c) P( Z ≤ z*) = 0.25

d) P( Z ≥ z*) = 0.9

e) P( 0 ≤ Z ≤ z*) = 0.41

f) P( -z* ≤ Z ≤ z*) = 0.88

Answer 1

$\text{Using the table of the values of Laplace function }\\ \Phi(z)=\frac{1}{\sqrt{2\pi}}\int_0^z e^{-\frac{t^2}{2}}dt\text{ we get:}\\ a)\ P\{Z\leq z^*\}=0.95\\ \Phi(z^*)=0.95/2=0.475\\ z^*\approx 1.96\\ b)\ P\{Z\leq z^*\}=0.2\\ \Phi(z^*)=0.2/2=0.1\\ z^*\approx 0.25\\ c)\ P\{Z\leq z^*\}=0.25\\ \Phi(z^*)=0.25/2=0.125\\ z^*\approx 0.32\\ d)\ P\{Z\geq z^*\}=0.9\\ P\{Z\leq z^*\}=0.9-0.5\\ P\{Z\leq z^*\}=0.4\\ \Phi(z^*)=0.4\\ z^*\approx 1.28\\ \text{The answer is } \approx -1.28.\\ e)\ P\{0\leq Z\leq z^*\}=0.41\\ \Phi(z^*)=0.41\\ z^*\approx 1.34\\ f)\ P\{-z^*\leq Z\leq z^*\}=0.88\\ \Phi(z^*)=0.44\\ z^*\approx 1.55$