Normal Distribution Curve

summertigger

New Member
I just want to make sure I am getting this right.
0.0<z<2.85
.5000-.4978
=.0022
so I would shade in to the right of the curve.

-1.38<z<0
.4162 + .5000
= .9162
I would shade in to the right of the curve labeling .5000 and to the left labeling .4162

-2.85<z<1.38
.4978 + .5000
=.9978
I would make to graphs
On the first shade the right labeling .5000
and the left labeling .4978
On the second shade the left .5000 and the right .9162
.5000+.4162
=.9162

JohnM

TS Contributor
On each of these it looks like you need to find the probability of getting a z score in between two values (the proportion of the curve between those values, in other words).

First, find the proportion between negative infinity and the larger z score. Then find the proportion between negative infinity and the smaller z score. Then, lastly, subtract the second proportion from the first, and you'll be left with the proportion in the "middle" (in between the scores).

In the first problem, the proportion between neg infinity and z=2.85 is 0.9978. The proportion between neg infinity and z=0 is 0.50, so we then subtract the second proportion from the first to get .9978 - .50 = .4978.

Follow this process for all three problems.

There is a post in our Examples section called "The Vaunted Normal Distribution" that covers this.

summertigger

New Member
Normal Curve

So then I would make the curve graph, shading the right side and labeling it
.4978

JohnM

TS Contributor
You would shade in almost all of the right half of the curve.

summertigger

New Member
Normal Distribution

So I re-did them and this is what I got

0.0 < z< 2.85
=P(z<2.85) - P(z< 0.0)
=.9978 - .50
= .4978

-1.38 < z < 0
=P(z<0) - [1-P(z<-1.38)]
=.50 - [1- .9162]
=.4162

-2.85 < z < 1.38
=P(z< 1.38) - P(z<-2.85)
=P(z<1.38) - [1-P(z<-2.85)]
=.9162 - [1-.9978]
=.9140

1.38 < z < 2.06
=P(z < 2.06) - P(z<1.38)
=P(z<2.06) - [1-P(z<1.38)]
=.9803 - [1- .9664]
=.9467

z < 2.06
=1-P(z<=2.060
=1- .9803
=.0197

z> -1.38
=1-P(z <= -1.38)
=1- .9162
=.0838

summertigger

New Member
On the second one I would graph it almost all the right half and label it .4162

3rd
I would shade in the whole right half labeling it .500 and shade in almost all the left side labeling it .9140

4th
I would shade in the whole left half labeling it .500 and shade in almost all the right side labeling it .9467

5th
I would shade in a small portion of the right side labeling it .0197

6th
I would shade in a small portion of the left side labeling it .0838

feldner

New Member
I am confused. I derived the same answer for the first three problems, but these last three I don't understand where the (1) is coming from. Is it because the problems do not have the equal sign? Is that necessary with these type problems? Or is it understood? My examples are below original examples. Could you please explain?

1.38 < z < 2.06
=P(z < 2.06) - P(z<1.38)
=P(z<2.06) - [1-P(z<1.38)]
=.9803 - [1- .9664]
=.9467

1.38 < z < 2.06
=P(0 < z < 2.06) - P(0 < z<1.38)
=.4803 - .4162

z < 2.06
=1-P(z<=2.060)
=1- .9803
=.0197
=1-P(z<=2.060)
=1- .9803
=.0197

z < 2.06
= .5 + P(z < 0 < 2.06)
= 0.5 + .4803

z> -1.38
=1-P(z <= -1.38)
=1- .9162
=.0838

z> -1.38
-1.38 < z = P(-1.38 < z < 0)
= P(z < 0) + P(0 < z < 1.38)
=0.5 + .4162