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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:

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

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:

\[P( Z< – 1.30 ) = 0.0968 \]

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