Question

Thickness measurement ancient prehistoric Native American Pot Shards discovered in Hopi Village are approximately normally distributed with the mean of 5.1 millimeters and standard deviation of 0.9 millimeters. For a randomly found shard, What is the probability that the thickness is: a) Less than 3.0 millimeters b) More than 7.0 millimeters Present your answer in three decimal places. Present your answer: answer for "a",answer for "b" Group of answer choices

Answer 1

a)
`P(X<3)=P((X-\mu)/(\sigma)<(3-\mu)/(\sigma))=P(Z<(3-5.1)/(0.9))=P(z<-2.33)`

And we can find this probability using the normal standard table and we got:

`P(z<-2.33)=0.010`

b)
`P(X>7)=P((X-\mu)/(\sigma)>(7-\mu)/(\sigma))=P(Z>(7-5.1)/(0.9))=P(z>2.11)`

And we can find this probability using the complement rule and the normal standard table and we got:

`P(z>2.11)=1-P(Z<2.11) = 1-0.983 = 0.017`

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

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

`X \sim N(5.1,0.9)`

Where
`\mu=5.1` and
`\sigma=0.9`

Part a

We are interested on this probability

`P(X<3)`

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)=P((X-\mu)/(\sigma)<(3-\mu)/(\sigma))=P(Z<(3-5.1)/(0.9))=P(z<-2.33)`

And we can find this probability using the normal standard table and we got:

`P(z<-2.33)=0.010`

Part b

We are interested on this probability

`P(X>7)`

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)=P((X-\mu)/(\sigma)>(7-\mu)/(\sigma))=P(Z>(7-5.1)/(0.9))=P(z>2.11)`

And we can find this probability using the complement rule and the normal standard table and we got:

`P(z>2.11)=1-P(Z<2.11) = 1-0.983 = 0.017`