Respuesta :
Answer:
[tex]P(334<\bar X<406)[/tex]
And we can use the following z score formula:
[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And replacing we got:
[tex] z = \frac{406 -370}{\frac{100}{\sqrt{25}}}= 1.8[/tex]
[tex] z = \frac{334 -370}{\frac{100}{\sqrt{25}}}= 1.8[/tex]
And we want thi probability:
[tex] P(-1.8 < Z<1.8)= P(Z<1.8) -P(Z<-1.8) = 0.964-0.036= 0.928[/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 expenditures of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(370,100)[/tex]
Where [tex]\mu=370[/tex] and [tex]\sigma=100[/tex]
We are interested on this probability
[tex]P(334<\bar X<406)[/tex]
And we can use the following z score formula:
[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And replacing we got:
[tex] z = \frac{406 -370}{\frac{100}{\sqrt{25}}}= 1.8[/tex]
[tex] z = \frac{334 -370}{\frac{100}{\sqrt{25}}}= 1.8[/tex]
And we want thi probability:
[tex] P(-1.8 < Z<1.8)= P(Z<1.8) -P(Z<-1.8) = 0.964-0.036= 0.928[/tex]
