Answer :
Answer:
a) [tex]P(55<X<65)=P(\frac{55-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{65-\mu}{\sigma})=P(\frac{55-60}{5}<Z<\frac{65-60}{5})=P(-1<z<1)[/tex]
And we can find this probability with this difference:
[tex]P(-1<z<1)=P(z<1)-P(z<-1)[/tex]
And in order to find these probabilities we can using tables for the normal standard distribution, excel or a calculator.
[tex]P(-1<z<1)=P(z<1)-P(z<-1)=0.841-0.159=0.682[/tex]
And that represent 68.2%
b) [tex]P(50<X<70)=P(\frac{50-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{70-\mu}{\sigma})=P(\frac{50-60}{5}<Z<\frac{70-60}{5})=P(-2<z<2)[/tex]
And we can find this probability with this difference:
[tex]P(-2<z<2)=P(z<2)-P(z<-2)[/tex]
And in order to find these probabilities we can using tables for the normal standard distribution, excel or a calculator.
[tex]P(-2<z<2)=P(z<2)-P(z<-2)=0.977-0.0228=0.9542[/tex]
And that represent 95.42%
c) [tex]P(45<X<75)=P(\frac{45-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{75-\mu}{\sigma})=P(\frac{45-60}{5}<Z<\frac{75-60}{5})=P(-3<z<3)[/tex]
And we can find this probability with this difference:
[tex]P(-3<z<3)=P(z<3)-P(z<-3)[/tex]
And in order to find these probabilities we can using tables for the normal standard distribution, excel or a calculator.
[tex]P(-3<z<3)=P(z<3)-P(z<-3)=0.99865-0.00135=0.9973[/tex]
And that represent 97.73%
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".
Part a
Let X the random variable that represent the amount spent per month by employees of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(60,5)[/tex]
Where [tex]\mu=60[/tex] and [tex]\sigma=5[/tex]
We are interested on this probability
[tex]P(55<X<65)[/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(55<X<65)=P(\frac{55-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{65-\mu}{\sigma})=P(\frac{55-60}{5}<Z<\frac{65-60}{5})=P(-1<z<1)[/tex]
And we can find this probability with this difference:
[tex]P(-1<z<1)=P(z<1)-P(z<-1)[/tex]
And in order to find these probabilities we can using tables for the normal standard distribution, excel or a calculator.
[tex]P(-1<z<1)=P(z<1)-P(z<-1)=0.841-0.159=0.682[/tex]
And that represent 68.2%
Part b
[tex]P(50<X<70)=P(\frac{50-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{70-\mu}{\sigma})=P(\frac{50-60}{5}<Z<\frac{70-60}{5})=P(-2<z<2)[/tex]
And we can find this probability with this difference:
[tex]P(-2<z<2)=P(z<2)-P(z<-2)[/tex]
And in order to find these probabilities we can using tables for the normal standard distribution, excel or a calculator.
[tex]P(-2<z<2)=P(z<2)-P(z<-2)=0.977-0.0228=0.9542[/tex]
And that represent 95.42%
Part c
[tex]P(45<X<75)=P(\frac{45-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{75-\mu}{\sigma})=P(\frac{45-60}{5}<Z<\frac{75-60}{5})=P(-3<z<3)[/tex]
And we can find this probability with this difference:
[tex]P(-3<z<3)=P(z<3)-P(z<-3)[/tex]
And in order to find these probabilities we can using tables for the normal standard distribution, excel or a calculator.
[tex]P(-3<z<3)=P(z<3)-P(z<-3)=0.99865-0.00135=0.9973[/tex]
And that represent 97.73%