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?

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dfbustos dfbustos · Answer 1 · 2020-07-17T02:28:07+02:00

Answer:

[tex]P(X<0.4375)=P(\frac{X-\mu}{\sigma}<\frac{0.4375-\mu}{\sigma})=P(Z<\frac{0.4375-0.635}{0.082})=P(z<-2.41)[/tex]

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

[tex]P(z<-2.41)=0.0080[/tex]

Step-by-step 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:

[tex]X \sim N(0.635,0.082)[/tex]

Where [tex]\mu=0.635[/tex] and [tex]\sigma=0.032[/tex]

We are interested on this probability

[tex]P(X<0.4375)[/tex]

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

[tex]z=\frac{x-\mu}{\sigma}[/tex]

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

[tex]P(X<0.4375)=P(\frac{X-\mu}{\sigma}<\frac{0.4375-\mu}{\sigma})=P(Z<\frac{0.4375-0.635}{0.082})=P(z<-2.41)[/tex]

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

[tex]P(z<-2.41)=0.0080[/tex]