Respuesta :
Answer:
[tex] Lower = \mu -3 \frac{\sigma}{\sqrt{n}}= 16- 3* \frac{0.2}{\sqrt{49}}= 15.91[/tex]
[tex] Upper = \mu +3 \frac{\sigma}{\sqrt{n}}= 16+ 3* \frac{0.2}{\sqrt{49}}= 16.09[/tex]
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 weigths for the Boces of Honey Nut meal of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(16,0.2)[/tex]
Where [tex]\mu=16[/tex] and [tex]\sigma=0.2[/tex]
Since the distribution for X is normal, we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
And using the 3 sigma rule we can find the interval like this:
[tex] Lower = \mu -3 \frac{\sigma}{\sqrt{n}}= 16- 3* \frac{0.2}{\sqrt{49}}= 15.91[/tex]
[tex] Upper = \mu +3 \frac{\sigma}{\sqrt{n}}= 16+ 3* \frac{0.2}{\sqrt{49}}= 16.09[/tex]
