Respuesta :
Answer:
If we assume that the deviation is [tex]\sigma=3[/tex] then the solution is:
[tex]P(2.55<X<64.95)=P(-9.15<z<11.65)=P(z<11.65)-P(z<-9.15)[/tex]
f) None of the above
If we assume that the deviation is [tex]\sigma=15[/tex] then the solution is:
[tex]P(2.55<X<64.95)=P(-1.83<z<2.33)=0.956[/tex]
b) 0.956
Step-by-step explanation:
1) 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".
2) Solution to the problem
Let X the random variable that represent the variable of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(30,3)[/tex]
Where [tex]\mu=30[/tex] and [tex]\sigma=3[/tex]
We are interested on this probability
[tex]P(2.55<X<64.95)[/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(2.55<X<64.95)=P(\frac{2.55-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{64.95-\mu}{\sigma})=P(\frac{2.55-30}{3}<Z<\frac{64.95-30}{3})=P(-9.15<Z<11.65)[/tex]
And we can find this probability on this way:
[tex]P(-9.15<z<11.65)=P(z<11.65)-P(z<-9.15)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(-9.15<z<11.65)=0.99999[/tex]
If we assume that the deviation is [tex]\sigma=15[/tex] then the solution is:
[tex]P(2.55<X<64.95)=P(\frac{2.55-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{64.95-\mu}{\sigma})=P(\frac{2.55-30}{15}<Z<\frac{64.95-30}{15})=P(-1.83<Z<2.33)[/tex]
And we can find this probability on this way:
[tex]P(-1.83<z<2.33)=P(z<2.33)-P(z<-1.83)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(-1.83<z<2.33)=0.956[/tex]
