Step 1

a)

Let's find the area under the curve that lies to the left of $z=-1.39$. So, we need to find $P(Z<-1.39)$, where Z represent Standard Normal random variable.

Using Normal Probability Table, we easily obtain:

$P(Z<-1.39)=0.0823$

Step 2

b)

Let's now find the area the curve that lies to the right of $z=1.96$. So, we need to find $P\left(Z<1.96\right)$, where Z represent Standard Normal random variable.

Using Normal Probability Table, we obtain:

$P\left(Z>1.96\right)=1-P\left(Z<1.96\right)$

$=1-0.9750$

$=0.025$

Step 3

c)

Let's now find the area under the curve that lies between $z=-2.16\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}z=-0.65$.

So, we need to find $P(-2.16<Z<-0.65)$, where Z represent Standard Normal random variable.

Using Normal Probability Table, we obtain:

$P(-2.16<Z<0.65)=P(Z<-0.65)-P(Z<-2.16)$

$=0.2578-0.0154$

$=0.2424$

Step 4

d)

Let's find the area under the curve that lies to the left of $z=1.43$. So, we need to fidn $P\left(Z<1.43\right)$, where Z represent Standard Normal random variable.

Using Normal Probability Table, we obtain:

$P\left(Z<1.43\right)=0.9236$

Step 5

e)

Let's find the area under the curve that lies to the right of $z=-0.89$. Som we need to find $P(Z\succ 0.89)$, where Z represent Standard Normal random variable.

Using Normal Probability Table, we obtain:

$P(Z\succ 0.896)=1-P(Z<-0.89)$

$=1-0.1867$

$=0.8133$

Step 6

f)

Let's now find the area under the curve that lies between $z=-0.48\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}z=1.74$.

So, we need to find $P(-0.48<Z<1.74)$, where Z represent Standard Normal random variable.

Using Normal Probability Table, we obtain:

$P(-0.48<Z<1.74)=P\left(Z<1.74\right)-P(Z<-0.48)$

$=0.9591-0.3156$

$=0.6435$.