Question

Thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally distributed, with a mean of 4.5 millimeters (mm) and a standard deviation of 1.7 mm. For a randomly found shard, find the following probabilities. (Round your answers to four decimal places.)

(a) the thickness is less than 3.0 mm.

(b) the thickness is more than 7.0 mm.

(c) the thickness is between 3.0 mm and 7.0 mm.

Answer 1

a)
`P(X<3.0)=P((X-\mu)/(\sigma)<(3-\mu)/(\sigma))=P(Z<(3-4.5)/(1.7))=P(z<-0.882)`

`P(z<-0.882)=0.189`

b)
`P(X>7.0)=P((X-\mu)/(\sigma)>(7-\mu)/(\sigma))=P(Z<(7-4.5)/(1.7))=P(z>1.47)`

`P(z>1.47)=1-P(z<1.47) = 1-0.929=0.071`

c)
`P(3<X<7)=P((3-\mu)/(\sigma)<(X-\mu)/(\sigma)<(7-\mu)/(\sigma))=P((3-4.5)/(1.7)<Z<(7-4.5)/(1.7))=P(-0.882<z<1.47)`

`P(-0.882<z<1.47)=P(z<1.47)-P(z<-0.882)=0.929-0.189=0.740`

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".

Part a

Let X the random variable that represent the thickness of a population, and for this case we know the distribution for X is given by:

`X \sim N(4.5,1.7)`

Where
`\mu=4.5` and
`\sigma=1.7`

We are interested on this probability

`P(X<3.0)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

If we apply this formula to our probability we got this:

`P(X<3.0)=P((X-\mu)/(\sigma)<(3-\mu)/(\sigma))=P(Z<(3-4.5)/(1.7))=P(z<-0.882)`

And we can find this probability using excel or the normal standard table:

`P(z<-0.882)=0.189`

Part b

We are interested on this probability

`P(X>3.0)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

If we apply this formula to our probability we got this:

`P(X>7.0)=P((X-\mu)/(\sigma)>(7-\mu)/(\sigma))=P(Z<(7-4.5)/(1.7))=P(z>1.47)`

And we can find this probability using excel or the normal standard table:

`P(z>1.47)=1-P(z<1.47) = 1-0.929=0.071`

Part c

`P(3<X<7)=P((3-\mu)/(\sigma)<(X-\mu)/(\sigma)<(7-\mu)/(\sigma))=P((3-4.5)/(1.7)<Z<(7-4.5)/(1.7))=P(-0.882<z<1.47)`

And we can find this probability using excel or the normal standard table liek this:

`P(-0.882<z<1.47)=P(z<1.47)-P(z<-0.882)=0.929-0.189=0.740`