Toms of Maine sells toothpaste in a tube advertised to contain 8 ounces. In their continuous production process, the amount of toothpaste put in a tube is normally distributed with a mean of 8.21 ounces and a standard deviation of 0.09 ounces. If the actual capacity of the tubes used is 8.45 ounces, what proportion of the tubes will be filled beyond capacity?

Respuesta :

Answer:

[tex]P(X>8.45)=P(\frac{X-\mu}{\sigma}>\frac{8.45-\mu}{\sigma})=P(Z>\frac{8.45-8.21}{0.09})=P(Z>2.67)[/tex]

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

[tex]P(Z>2.67)=1-P(Z<2.67)=1-0.996=0.004 [/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".  

Solution to the problem

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

[tex]X \sim N(8.21,0.09)[/tex]  

Where [tex]\mu=8.21[/tex] and [tex]\sigma=0.09[/tex]

The capacity is 8.45 ounces and we are interested on this probability

[tex]P(X>8.45)[/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>8.45)=P(\frac{X-\mu}{\sigma}>\frac{8.45-\mu}{\sigma})=P(Z>\frac{8.45-8.21}{0.09})=P(Z>2.67)[/tex]

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

[tex]P(Z>2.67)=1-P(Z<2.67)=1-0.996=0.004 [/tex]