Given a standard normal distribution, find the area under the curve t

Step 1
a)
Let's find the area under the curve that is to the left of the ${\displaystyle z=-1.39}$. So, we need to find ${\displaystyle P(Z<-1.39)}$, where Z represent Standard Normal random variable.
By consulting a Normal Probability Table, we can readily obtain:
${\displaystyle P(Z<-1.39)=0.0823}$
Step 2
b)
Let's find the area under the curve that is to the right of the ${\displaystyle z=1.96}$. So, we need to find ${\displaystyle P\left(Z<1.96\right)}$, where Z represent Standard Normal random variable.
By consulting a Normal Probability Table, we can readily obtain:
${\displaystyle P\left(Z>1.96\right)=1-P\left(Z<1.96\right)}$
${\displaystyle =1-0.9750}$
${\displaystyle =0.025}$
Step 3
c)
Let's now find the area under the curve that is ${\displaystyle z=-2.16\text{and}z=-0.65}$.
So, we need to find ${\displaystyle P(-2.16<Z<-0.65)}$, where Z represent Standard Normal random variable.
By consulting a Normal Probability Table, we can readily obtain:
${\displaystyle P(-2.16<Z<0.65)=P(Z<-0.65)-P(Z<-2.16)}$
${\displaystyle =0.2578-0.0154}$
${\displaystyle =0.2424}$
Step 4
d)
Let's find the area under the curve that lies to the left of ${\displaystyle z=1.43}$. So, we need to fidn ${\displaystyle P\left(Z<1.43\right)}$, where Z represent Standard Normal random variable.
By consulting a Normal Probability Table, we can readily obtain:
${\displaystyle P\left(Z<1.43\right)=0.9236}$
Step 5
e)
Let's find the area under the curve that lies to the right of ${\displaystyle z=-0.89}$. Som we need to find ${\displaystyle P(Z\succ 0.89)}$, where Z represent Standard Normal random variable.
By consulting a Normal Probability Table, we can readily obtain:
${\displaystyle P(Z\succ 0.896)=1-P(Z<-0.89)}$
${\displaystyle =1-0.1867}$
${\displaystyle =0.8133}$
Step 6
f)
Let's now find the area under the curve that lies between ${\displaystyle z=-0.48\text{and}z=1.74}$.
So, we need to find ${\displaystyle P(-0.48<Z<1.74)}$, where Z represent Standard Normal random variable.
By consulting a Normal Probability Table, we can readily obtain:
${\displaystyle P(-0.48<Z<1.74)=P\left(Z<1.74\right)-P(Z<-0.48)}$
${\displaystyle =0.9591-0.3156}$
${\displaystyle =0.6435}$.

Answer: 
a) ${\displaystyle P(Z<-1.39)=0.082264}$ 
b) ${\displaystyle P\left(Z>1.96\right)=0.025}$ 
c) ${\displaystyle P(-2.16<Z<-0.65)=0.2457}$ 
d) ${\displaystyle P\left(Z<1.43\right)=0.92364}$ 
e) ${\displaystyle P(Z\succ 0.48)=0.68439}$ 
f) ${\displaystyle P(-0.48<Z<1.74)=0.6435}$ 
Step-by-step explanation: 
Considering a 
From the question we are told that 
The z-score is ${\displaystyle z=-1.39}$. 
Generally from the z table the area under the curve that lies to the left of ${\displaystyle z=-1.39}$ is ${\displaystyle P(Z<-1.39)=0.082264}$ 
Considering b 
From the question we are told that 
The z-score is ${\displaystyle z=1.96}$ 
Generally from the z table the area under the curve that lies to the right of ${\displaystyle z=1.96}$ is ${\displaystyle P\left(Z>1.96\right)=0.025}$ 
Considering c 
From the question we are told that 
The z-score is ${\displaystyle z=-2.16\text{and}z=-0.65}$ 
Generally from the z table the area under the curve that lies between ${\displaystyle z=-2.16\text{and}z=-0.65}$; 
${\displaystyle P(-2.16<Z<-0.65)=P(Z<-0.64)-P(Z<-2.16)}$ 
From the z table the area under ${\displaystyle (Z<-0.64)\text{and}(Z<-2.16)}$ is ${\displaystyle P(Z<-0.64)=0.26109}$ 
and ${\displaystyle P(Z<-2.16)=0.015386}$ 
So ${\displaystyle P(-2.16<Z<-0.65)=0.26109-0.015386}$ 
${\displaystyle \Rightarrow P(-2.16<Z<-0.65)=0.2457}$ 
Considering d 
From the question we are told that 
The z-score is ${\displaystyle z=1.45}$ 
Generally from the z table the area under the curve that lies to the left of ${\displaystyle z=1.43}$ is ${\displaystyle P\left(Z<1.43\right)=0.92364}$ 
Considering e 
From the question we are told that 
The z-score is ${\displaystyle z=-0.48}$ 
Generally from the z table the area under the curve that lies to the right of ${\displaystyle z=0.48}$ is ${\displaystyle P(Z\succ 0.48)=0.68439}$ 
Considering f 
From the question we are told that 
The z-score is ${\displaystyle z=-0.48\text{and}z=1.74}$ 
Generally from the z table the area under the curve that lies between ${\displaystyle z=-2.16\text{and}z=-0.65}$; 
${\displaystyle P(-0.48<Z<1.74)=P\left(Z<1.74\right)-P(Z<-0.48)}$ 
From the z table the area under ${\displaystyle \left(Z<1.74\right)\text{and}(Z<-0.48)}$ is ${\displaystyle P\left(Z<1.74\right)=0.95907}$ 
and ${\displaystyle P(Z<-0.48)=0.31561}$