Question

Determine the area under the standard normal curve that lies to the right of ​(a) Upper Z equals negative 0.77 comma ​(b) Upper Z equals negative 0.07 comma ​(c) Upper Z equals negative 0.92 comma and ​(d) Upper Z equals 0.39 .

Answer 1

Answer:

a) $P(Z> 0.07) = 1-P(Z$

And we can use the following excel code:

"=1-NORM.DIST(0.07,0,1,TRUE)"

And we got:

$P(Z>0.07) = 0.472$

b) $P(Z> 0.07) = 1-P(Z$

And we can use the following excel code:

"=1-NORM.DIST(0.07,0,1,TRUE)"

And we got:

$P(Z>0.07) = 0.472$

c) $P(Z> -0.92) = 1-P(Z$

And we can use the following excel code:

"=1-NORM.DIST(-0.92,0,1,TRUE)"

And we got:

$P(Z>-0.92) = 0.821$

d) $P(Z> 0.39) = 1-P(Z$

And we can use the following excel code:

"=1-NORM.DIST(0.39,0,1,TRUE)"

And we got:

$P(Z>0.39) = 0.348$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Part a

We want this probability:

$P(Z> -0.77)$

And using the complement rule and we got:

$P(Z> -0.77) = 1-P(Z$

And we can use the following excel code:

"=1-NORM.DIST(-0.77,0,1,TRUE)"

And we got:

$P(Z>-0.77) = 0.779$

Part b

We want this probability:

$P(Z> 0.07)$

And using the complement rule and we got:

$P(Z> 0.07) = 1-P(Z$

And we can use the following excel code:

"=1-NORM.DIST(0.07,0,1,TRUE)"

And we got:

$P(Z>0.07) = 0.472$

Part c

We want this probability:

$P(Z> -0.92)$

And using the complement rule and we got:

$P(Z> -0.92) = 1-P(Z$

And we can use the following excel code:

"=1-NORM.DIST(-0.92,0,1,TRUE)"

And we got:

$P(Z>-0.92) = 0.821$

Part d

We want this probability:

$P(Z> 0.39)$

And using the complement rule and we got:

$P(Z> 0.39) = 1-P(Z$

And we can use the following excel code:

"=1-NORM.DIST(0.39,0,1,TRUE)"

And we got:

$P(Z>0.39) = 0.348$