The packaging process in a breakfast cereal company has been adjusted so that an average of 13.0 oz of cereal is placed in each package. Of course, not all packages have precisely 13.0 oz because of random sources of variability. The standard deviation of the actual net weight is 0.1 oz, and the distribution of weights is known to follow the normal probability distribution. Determine the probability that a randomly chosen package will contain between 13.0 and 13.2oz of cereal and illustrate the proportion of area under the normal curve which is associated with this probability value.

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 weights of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(13,0.1)[/tex]

Where [tex]\mu=13[/tex] and [tex]\sigma=0.1[/tex]

We are interested on this probability

[tex]P(13<X<13.2)[/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(13<X<13.2)=P(\frac{13-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{13.2-\mu}{\sigma})=P(\frac{13-13}{0.1}<Z<\frac{13.2-13}{0.1})=P(0<z<2)[/tex]

And we can find this probability with this difference:

[tex]P(0<z<2)=P(z<2)-P(z<0)= 0.97725 -0.5=0.47725 [/tex]

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

[tex]P(-0.50<z<0.65)=P(z<0.65)-P(-0.5)=0.742-0.309=0.434[/tex]

The figure illustrate the result obtained.