Elleanor Mckenzie

2021-08-21

Find the following probabilities for the standard normal variable z:
(a) $P\left(-1.96\le z\le 1.96\right)$
(b) $P\left(z>1.96\right)$

(c) $P\left(z<-1.96\right)$

Jayden-James Duffy

(a)
$P\left(-1.96\le z\le 1.96\right)$
Determine the corresponding probability using the normal probability table in the appendix.
$P\left(z<1.96\right)$is given in the row starting with 1.9 and in the column starting with 0.06 of the standard normal probability table in the appendix.
$P\left(z<1.96\right)=0.9750$
$P\left(z<-1.96\right)$ is given in the row starting with -1.9 and in the column starting with 0.06 of the standard normal probability $P\left(z<-1.96\right)=0.0250$
The probability between two boundaries is the difference between the probabilities to the left of both boundaries.
$P\left(-1.96\right)$ (b)
$P\left(z>1.96\right)$
Determine the corresponding probability using the normal probability table in the appendix. $P\left(z>1.96\right)$ is given in the row starting with 1.9 and in the column starting with 0.06 of the standard normal probability table in the appendix.
$P\left(z>1.96\right)=0.9750$
Use the Complement rule: $P\left({A}^{c}\right)=P\left(\mathrm{¬}A\right)=1-P\left(A\right)$
$P\left(z>1.96\right)=1-P\left(z<1.96\right)=1-0.9750=0.0250$
(c)
$P\left(z<-1.96\right)$
Determine the corresponding probability using the normal probability table in the appendix.
$P\left(z>1.96\right)$ is given in the row starting with 1.9 and in the column starting with 0.06 of the standard normal probability table in the appendix.
$P\left(z<-1.96\right)=0.0250$
Result:(a)0.9500.(b)0.0250.(c)0.0250.

Do you have a similar question?