Respuesta :
Answer:
[tex]P(32<X<52)=P(\frac{42-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{52-\mu}{\sigma})=P(\frac{32-42}{5}<Z<\frac{52-42}{5})=P(-2<z<2)[/tex]
And we can find this probability with ths 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)=0.978-0.02275=0.9545[/tex]
So then we expect:
x = 0.9545*800= 763.6 or approximately 764 respondent within 2 deviations from the mean
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(42,5)[/tex]
Where [tex]\mu=42[/tex] and [tex]\sigma=5[/tex]
We are interested on this probability
[tex]P(42-2*5<X<42+2*5)[/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(32<X<52)=P(\frac{42-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{52-\mu}{\sigma})=P(\frac{32-42}{5}<Z<\frac{52-42}{5})=P(-2<z<2)[/tex]
And we can find this probability with ths 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)=0.978-0.02275=0.9545[/tex]
So then we expect:
x = 0.9545*800= 763.6 or approximately 764 respondent within 2 deviations from the mean
