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

Cynthia Bell 2021-12-20 Answered

Given a standard normal distribution, find the area under the curve that lies
(a) to the left of $z=-1.39$
(b) to the right of $z=1.96$
(c) between
(d) to the left of $z=1.43$
(e) to the right of $z=-0.89$
(f) between .

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turtletalk75

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 $P\left(Z<-1.39\right)$, where Z represent Standard Normal random variable.
Using Normal Probability Table, we easily obtain:
$P\left(Z<-1.39\right)=0.0823$
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 .
So, we need to find $P\left(-2.16, where Z represent Standard Normal random variable.
Using Normal Probability Table, we obtain:
$P\left(-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 $P\left(Z\succ 0.89\right)$, where Z represent Standard Normal random variable.
Using Normal Probability Table, we obtain:
$P\left(Z\succ 0.896\right)=1-P\left(Z<-0.89\right)$
$=1-0.1867$
$=0.8133$
Step 6
f)
Let's now find the area under the curve that lies between .
So, we need to find $P\left(-0.48, where Z represent Standard Normal random variable.
Using Normal Probability Table, we obtain:
$P\left(-0.48
$=0.9591-0.3156$
$=0.6435$.

###### Not exactly what you’re looking for?
Bubich13

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