Respuesta :
Answer:
[tex]P(0.02<X<5.66)=P(\frac{0.02-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{5.66-\mu}{\sigma})=P(\frac{0.02-2.84}{1.41}<Z<\frac{5.66-2.84}{1.41})=P(-2<z<2)[/tex]
And we can find this probability with the following difference:
[tex]P(-2<z<2)=P(z<2)-P(z<-2)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(-2<z<2)=P(z<2)-P(z<-2)=0.97725-0.02275=0.9545[/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 variable of interest of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(2.84,1.41)[/tex]
Where [tex]\mu=2.84[/tex] and [tex]\sigma=1.41[/tex]
We are interested on this probability
[tex]P(2.84-2*1.41<X<2.84+ 2*1.41)[/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(0.02<X<5.66)=P(\frac{0.02-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{5.66-\mu}{\sigma})=P(\frac{0.02-2.84}{1.41}<Z<\frac{5.66-2.84}{1.41})=P(-2<z<2)[/tex]
And we can find this probability with the following difference:
[tex]P(-2<z<2)=P(z<2)-P(z<-2)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(-2<z<2)=P(z<2)-P(z<-2)=0.97725-0.02275=0.9545[/tex]
