Question

In each​ part, find the area under the standard normal curve that lies between the specified​ z-score, sketch a standard normal​ curve, and shade the area of interest.

a. minus1 and 1
b. minus2 and 2
c. minus3 and 3

Answer 1

a)
`P(-1<Z<1)= P(Z<1) -P(Z<-1)= 0.841-0.159= 0.682`

b)
`P(-2<Z<2)= P(Z<2) -P(Z<-2)= 0.977-0.0228= 0.954`

c)
`P(-3<Z<3)= P(Z<3) -P(Z<-3)= 0.999-0.0013= 0.998`

The results are on the fogure attached.

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

For this case we want to find this probability:

`P(-1<Z<1)`

And we can find this probability with this difference:

`P(-1<Z<1)= P(Z<1) -P(Z<-1)`

And if we find the probability using the normal standard distribution or excel we got:

`P(-1<Z<1)= P(Z<1) -P(Z<-1)= 0.841-0.159= 0.682`

Part b

For this case we want to find this probability:

`P(-2<Z<2)`

And we can find this probability with this difference:

`P(-2<Z<2)= P(Z<2) -P(Z<-2)`

And if we find the probability using the normal standard distribution or excel we got:

`P(-2<Z<2)= P(Z<2) -P(Z<-2)= 0.977-0.0228= 0.954`

Part c

For this case we want to find this probability:

`P(-3<Z<3)`

And we can find this probability with this difference:

`P(-3<Z<3)= P(Z<3) -P(Z<-3)`

And if we find the probability using the normal standard distribution or excel we got:

`P(-3<Z<3)= P(Z<3) -P(Z<-3)= 0.999-0.0013= 0.998`