# 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 z=-2.16 and z = -0.65; (d) to the left of z=1.43 ; (e) to the right of z=-0.89; (f) between z=-0.48 and z= 1.74.

This article aims to find the area under the curve for a standard normal distribution. A normal probability table is used to find the area under the curve. The formula for the probability density function is:

$f ( x ) = \dfrac{ 1 }{ \sigma \sqrt 2 \pi } e ^ {-\dfrac{ 1 }{ 2 } ( \dfrac { x -\mu}{\sigma}) ^ {2}}$

Part ( a )

Let’s find the area under the curve to the left of  $z = – 1.39$. So we need to see $P( Z< – 1.39 )$, where $Z$ represents a standard normal random variable.

Using a normal probability table, we easily obtain:

$P( Z< – 1.39 ) = 0.0823$

Part ( b )

Let’s find area under the curve that lies to right of  $z = 1.96$. So we need to determine $P( Z > 1.96 )$, where $Z$ represents a standard normal random variable.

Using a normal probability table, we easily obtain:

$P( Z > 1.96 ) = 1- P ( Z < 1.96)$

$= 1 – 0.9750$

$P ( Z > 1.96) = 0.025$

Part ( c )

Let’s find area under the curve that lies between $z = – 2.16$ and $z = -0.65$. So we need to find $P( -2.16 < Z< – 0.65 )$, where $Z$ represents a standard normal random variable.

Using a normal probability table, we easily obtain:

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

$=0.2578-0.0154$

$P(-2.16<Z<- 0.65 )=0.2424$

Part ( d )

Let’s find area under the curve that lies to the left of  $z=1.43$. So we need to find $P(Z<1.43 )$, where $Z$ represents a standard normal random variable.

Using a normal probability table, we easily obtain:

$P(Z<1.43 )=0.9236$

Part ( e )

Let’s find area under the curve that lies to the right of  $z=-0.89$. So we need to find $P(Z>-0.89 )$, where $Z$ represents a standard normal random variable.

Using a normal probability table, we easily obtain:

$P( Z>-0.89 ) = 1- P (Z<-0.89)$

$=1-0.1867$

$P( Z>-0.89 )=0.8133$

Part ( f )

Using a normal probability table, we easily find:

$P(-0.48 < Z < 1.74 ) = P(Z < 1.74) – P(Z<-0.48)$

$=0.9591-0.3156$

$P(-0.48 < Z < 1.74 )=0.6435$

## Numerical Result

(a) $P( Z< – 1.39 ) = 0.0823$

(b) $P(Z>1.96)= 0.025$

(c) $P(-2.16<Z<- 0.65 )=0.2424$

(d) $P(Z<1.43 )=0.9236$

(e) $P( Z>-0.89 )=0.8133$

(f) $P(-0.48<Z<1.74 )=0.6435$

## Example

Find area under the curve that lies for the standard normal distribution.

(1) to the left of  $z = -1.30$.

Solution

Let’s find the area under the curve to the left of  $z = – 1.30$. So we need to find $P( Z< – 1.30 )$, where $Z$ represents a standard normal random variable.

Using a normal probability table, we easily obtain:

$P( Z< – 1.30 ) = 0.0968$