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:

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

**Expert Answer**

**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:

\[P( Z< – 1.30 ) = 0.0968 \]