Respuesta :
Answer:
[tex]P(16.5<\bar X <18.5)=P(\frac{16.5-17.5}{\frac{4}{\sqrt{50}}}<Z<\frac{18.5-17.5}{\frac{4}{\sqrt{50}}})[/tex]
And using the normal standard table or excel we find the probability:
[tex] P(-1.768< Z< 1.768) = P(Z<1.768)-P(Z<-1.768)=0.961- 0.039=0.9229[/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 avergae number of weeks an individual is unemployed of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(17.5,4)[/tex]
Where [tex]\mu=17.5[/tex] and [tex]\sigma=4[/tex]
Since the distribution for X is normal then, the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu= 17.5, \frac{\sigma}{\sqrt{n}}= \frac{4}{\sqrt{50}}=0.566)[/tex]
We select a sample of n =50 people. And we want to find the following probability
[tex]P(16.5<\bar X <18.5)=P(\frac{16.5-17.5}{\frac{4}{\sqrt{50}}}<Z<\frac{18.5-17.5}{\frac{4}{\sqrt{50}}})[/tex]
And using the normal standard table or excel we find the probability:
[tex] P(-1.768< Z< 1.768) = P(Z<1.768)-P(Z<-1.768)=0.961- 0.039=0.9229[/tex]
