Question

g Steel used for water pipelines is often coated on the inside with cement mortar to prevent corrosion. In a study of the mortar coatings of the pipeline used in a water transmission project in California, researchers noted that the mortar thickness was specified to be 7/16 inch. A very large sample of thickness measurements produced a mean equal to 0.635 inch and astandard deviation equal to 0.082 inch. If the thickness measurements were normally distributed, approximately what proportion were less than 7/16 inch?

Answer 1

`P(X<0.4375)=P((X-\mu)/(\sigma)<(0.4375-\mu)/(\sigma))=P(Z<(0.4375-0.635)/(0.082))=P(z<-2.41)`

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

`P(z<-2.41)=0.0080`

Explanation:

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(0.635,0.082)`

Where
`\mu=0.635` and
`\sigma=0.032`

We are interested on this probability

`P(X<0.4375)`

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<0.4375)=P((X-\mu)/(\sigma)<(0.4375-\mu)/(\sigma))=P(Z<(0.4375-0.635)/(0.082))=P(z<-2.41)`

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

`P(z<-2.41)=0.0080`