There are two machines available for cutting corks intended for use in wine bottles. The first produces corks with diameters that are normally distributed with mean 3 cm and standard deviation 0.10 cm. The second machine produces corks with diameters that have a normal distribution with mean 3.04 cm and standard deviation 0.04 cm. Acceptable corks have diameters between 2.9 cm and 3.1 cm.

a. What is the probability that the first machine produces an acceptable cork? (Round your answer to four decimal places.)
b. What is the probability that the second machine produces an acceptable cork? (Round your answer to four decimal places.)

b) [tex]P(2.9<X<3.1)=P(\frac{2.9-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{3.1-\mu}{\sigma})=P(\frac{2.9-3.04}{0.04}<Z<\frac{3.1-3.04}{0.04})=P(-3.5<z<1.5)[/tex]

And we can find this probability with the following difference:

[tex]P(-3.5<z<1.5)=P(z<1.5)-P(z<-3.5)= 0.9332-0.0002=0.9330[/tex]

Step-by-step 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 diameters of a population 1, and for this case we know the distribution for X is given by:

[tex]X \sim N(3,0.1)[/tex]

Where [tex]\mu=2[/tex] and [tex]\sigma=0.1[/tex]

We are interested on this probability

[tex]P(2.9<X<3.1)[/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(2.9<X<3.1)=P(\frac{2.9-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{3.1-\mu}{\sigma})=P(\frac{2.9-3}{0.1}<Z<\frac{3.1-3}{0.1})=P(-1<z<1)[/tex]

And we can find this probability with the following difference:

[tex]P(-1<z<1)=P(z<1)-P(z<-1)= 0.8413-0.1587=0.6827[/tex]

Part b

[tex]X \sim N(3.04,0.04)[/tex]

[tex]P(2.9<X<3.1)=P(\frac{2.9-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{3.1-\mu}{\sigma})=P(\frac{2.9-3.04}{0.04}<Z<\frac{3.1-3.04}{0.04})=P(-3.5<z<1.5)[/tex]

And we can find this probability with the following difference:

[tex]P(-3.5<z<1.5)=P(z<1.5)-P(z<-3.5)= 0.9332-0.0002=0.9330[/tex]